Definitions [29]
When we have the values of f near x to the right of a i.e.
\[\lim_{x\to a^{+}}f\left(x\right)\] is the expected value of f at x = a.
If f(x) approaches a real number l, when x approaches a, then l is called the limit of f(x).
Symbolically, \[\lim_{x\to a}f\left(x\right)=l\]
When we have the values of f near x to the left of a, i.e.
\[\lim_{x\to a^{-}}f\left(x\right)\] is the expected value of f at x = a.
A function f(x) is said to be continuous at a point x = a, if the following three conditions are satisfied
- f is defined at every point on an open interval containing a.
- \[\lim_{x\to a}f\left(x\right)\] exists.
- \[\lim_{x\to a}f\left(x\right)=f\left(a\right)\].
A function f(x) is said to be discontinuous at x = a if it is not continuous at x = a, i.e.
- \[\lim_{x\to a}f\left(a\right)\] does not exist.
- The left-hand limit and the right-hand limit are not equal.
- \[\lim_{x\to a}f\left(x\right)\neq f\left(a\right)\].
If \[\lim_{x\to a^{-}}f\left(x\right)=\lim_{x\to a^{+}}f\left(x\right)\neq f\left(a\right),\] then f(x) is said to be removable discontinuous.
If \[\lim_{x\to a^{+}}f\left(x\right)\neq\lim_{x\to a^{-}}f\left(x\right),\] then f(x) is said to be non-removable discontinuous.
The derivative of a real function f at a point c in its domain is defined as:
\[f'(c) = \lim_{h \to 0} \frac{f(c+h) - f(c)}{h}\]
Let \[f\] be a real-valued function which is a composite of two functions \[u\] and \[v\]; i.e., \[f = v \circ u\]. Suppose \[t = u(x)\] and if both \[\frac{dt}{dx}\]and \[\frac{dv}{dt}\]exist, we have
If \(u = g(x)\) and \(y = f(u)\), then \(y = f(g(x))\) is called a composite function. Here, \(g(x)\) is the inner function and \(f(u)\) is the outer function.
If a function reverses the action of another function, it is called its inverse function. For example, if \[y = \sin^{-1} x\], then \[x = \sin y\], which means the inverse function converts a trigonometric value back into an angle.
If a function reverses the action of another function, it is called its inverse function. For example, if \[y = \sin^{-1} x\], then \[x = \sin y\], which means the inverse function converts a trigonometric value back into an angle.
A function of the form \[y = b^x\], where b > 0 and \[b \neq 1\], is called an exponential function.
If b > 0, \[b \neq 1\], and a > 0, then
This means a logarithm tells the exponent to which the base must be raised to obtain the number.
A function of the form \[y = b^x\], where b > 0 and \[b \neq 1\], is called an exponential function.
If b > 0, \[b \neq 1\], and a > 0, then
This means a logarithm tells the exponent to which the base must be raised to obtain the number.
Let \[f\] be a real-valued function which is a composite of two functions \[u\] and \[v\]; i.e., \[f = v \circ u\]. Suppose \[t = u(x)\] and if both \[\frac{dt}{dx}\]and \[\frac{dv}{dt}\]exist, we have
If \(u = g(x)\) and \(y = f(u)\), then \(y = f(g(x))\) is called a composite function. Here, \(g(x)\) is the inner function and \(f(u)\) is the outer function.
Implicit differentiation means differentiating both sides of an equation with respect to x, while remembering that y depends on x. Therefore, whenever a term containing y is differentiated, the factor \[\frac{dy}{dx}\] appears by the chain rule.
When x = f(t) and y = g(t), the relation between x and y is said to be in parametric form.
Let y = f(x). If the first derivative \[\frac{dy}{dx} = f'(x)\] is itself differentiable, then differentiating once again with respect to \[x\] gives the second order derivative.
Evaluating the derivative at a specific point, \[\left.\frac{dy}{dx}\right|_{x=x_0}\], gives the instantaneous rate of change at exactly \[x = x_0\].
If a quantity y varies with another quantity x based on a rule y = f(x), then the derivative \[\frac{dy}{dx}\] (or f'(x)) represents the rate of change of y with respect to x.
If two variables x and y both vary with respect to a third variable t (like time), you can find the rate of change of y with respect to x using:
(Note: This is only valid if \[\frac{dx}{dt} \neq 0\]).
A function ( f ) is said to be monotonic in an interval if it is either increasing or decreasing in that interval.
A function ( f(x) ) is said to be an increasing function on ((a, b)) if x₁ < x₂ ⇒ f(x₁) ≤ f(x₂)
Strictly Increasing Function:
- If x₁ < x₂ ⇒ f(x₁) < f(x₂)
A function ( f(x) ) is said to be a decreasing function on (a, b) if x₁ < x₂ ⇒ f(x₁) ≥ f(x₂)
Strictly Decreasing Function:
- If x₁ < x₂ ⇒ f(x₁) > f(x₂)
A function may attain a maximum value at a point if its value there is greater than nearby values, and it may attain a minimum value if its value there is smaller than nearby values. These are often called extreme values.
A point in the domain of a function is called a critical point if either the derivative is zero there or the derivative does not exist there. Critical points are checked while locating possible maxima or minima.
Formulae [6]
1. \[\lim_{x\to0}\frac{\sin x}{x}=1=\lim_{x\to0}\frac{x}{\sin x}\]
2. $$\lim_{x\to0}\frac{\tan x}{x}=1=\lim_{x\to0}\frac{x}{\tan x}$$
3. \[\lim_{x\to0}\frac{\sin^{-1}x}{x}=1=\lim_{x\to0}\frac{x}{\sin^{-1}x}\]
4. \[\lim_{x\to0}\frac{\tan^{-1}x}{x}=1=\lim_{x\to0}\frac{x}{\tan^{-1}x}\]
5. \[\lim_{x\to0}\frac{\sin x^{\circ}}{x}=\frac{\pi}{180}\]
6. \[\lim_{x\to0}\cos x=1\]
7. \[\lim_{x\to0}\frac{\sin\mathrm{k}x}{x}=\lim_{x\to0}\frac{\tan\mathrm{k}x}{x}=\mathrm{k}\]
8. \[\lim_{x\to\infty}\frac{\sin x}{x}=\lim_{x\to\infty}\frac{\cos x}{x}=0\]
9. \[\lim_{x\to\infty}\frac{\sin\left(\frac{1}{x}\right)}{\frac{1}{x}}=1=\lim_{x\to\infty}\frac{\tan\left(\frac{1}{x}\right)}{\frac{1}{x}}\]
10. \[\lim_{x\to a}\frac{\sin\left(x-a\right)}{x-a}=1=\lim_{x\to a}\frac{\tan\left(x-a\right)}{x-a}\]
| Function | Derivative |
|---|---|
| [f(x)]ⁿ | n[f(x)]ⁿ⁻¹ · f′(x) |
| \[\sqrt{\mathrm{f}(x)}\] | \[\frac{1}{2\sqrt{\mathrm{f}(x)}}\cdot\mathrm{f}^{\prime}(x)\] |
| \[\frac{1}{\mathrm{f}(x)}\] | \[-\frac{1}{\left[\mathrm{f}(x)\right]^{2}}\cdot\mathrm{f}^{\prime}(x)\] |
| sin(f(x)) | cos(f(x)) · f′(x) |
| cos(f(x)) | −sin(f(x)) · f′(x) |
| tan(f(x)) | sec²(f(x)) · f′(x) |
| cot(f(x)) | −cosec²(f(x)) · f′(x) |
| sec(f(x)) | sec(f(x)) tan(f(x)) · f′(x) |
| cosec(f(x)) | −cosec(f(x)) cot(f(x)) · f′(x) |
| \[\mathbf{a}^{\mathbf{f}(x)}\] | \[a^{f(x)}\log a\cdot f^{\prime}(x)\] |
| \[\mathrm{e}^{\mathrm{f}(x)}\] | \[\mathrm{e}^{\mathrm{f}(x)\cdot\mathrm{f}^{\prime}(x)}\] |
| log(f(x)) | \[\frac{1}{\mathrm{f}(x)}\cdot\mathrm{f}^{\prime}(x)\] |
| logₐ(f(x)) | \[\frac{1}{\mathrm{f}(x)\mathrm{loga}}\cdot\mathrm{f}^{\prime}(x)\] |
| Function | Derivative | Condition |
|---|---|---|
| sin⁻¹x | \[\frac{1}{\sqrt{1-x^{2}}}\] | |x| < 1 |
| sin⁻¹(f(x)) | \[\frac{1}{\sqrt{1-\{f\left(x\right)\}^{2}}}\frac{d}{dx}f\left(x\right)\] | |f(x)| < 1 |
| cos⁻¹x | \[-\frac{1}{\sqrt{1-x^{2}}}\] | x| < 1 |
| cos⁻¹(f(x)) | \[-\frac{1}{\sqrt{1-\left\{f\left(x\right)\right\}^{2}}}\frac{d}{dx}f(x)\] | |f(x)| < 1 |
| tan⁻¹x | \[\left(\frac{1}{1+x^{2}}\right)\] | x ∈ R |
| tan⁻¹(f(x)) | \[\frac{1}{1+\left\{f\left(x\right)\right\}^{2}}\frac{d}{dx}f(x)\] | f(x) ∈ R |
| cot⁻¹x | \[-\left(\frac{1}{1+x^{2}}\right)\] | x ∈ R |
| cot⁻¹(f(x)) | \[-\frac{1}{1+\{f(x)\}^{2}}\frac{d}{dx}f(x)\] | f(x) ∈ R |
| sec⁻¹x | \[\frac{1}{|x|\sqrt{x^{2}-1}}\] | |x| > 1 |
| sec⁻¹(f(x)) | \[\frac{1}{|f(x)|\sqrt{\{f(x)\}^{2}-1}}\frac{d}{dx}f(x)\] | |f(x)| > 1 |
| cosec⁻¹x | \[-\left(\frac{1}{|x|\sqrt{x^{2}-1}}\right)\] |
|x| > 1 |
| cosec⁻¹(f(x)) | \[-\frac{1}{|f(x)|\sqrt{\{f(x)\}^{2}-1}}\frac{d}{dx}f(x)\] | |f(x)| > 1 |
| Function | Derivative | Condition |
|---|---|---|
| sin⁻¹x | \[\frac{1}{\sqrt{1-x^{2}}}\] | |x| < 1 |
| sin⁻¹(f(x)) | \[\frac{1}{\sqrt{1-\{f\left(x\right)\}^{2}}}\frac{d}{dx}f\left(x\right)\] | |f(x)| < 1 |
| cos⁻¹x | \[-\frac{1}{\sqrt{1-x^{2}}}\] | x| < 1 |
| cos⁻¹(f(x)) | \[-\frac{1}{\sqrt{1-\left\{f\left(x\right)\right\}^{2}}}\frac{d}{dx}f(x)\] | |f(x)| < 1 |
| tan⁻¹x | \[\left(\frac{1}{1+x^{2}}\right)\] | x ∈ R |
| tan⁻¹(f(x)) | \[\frac{1}{1+\left\{f\left(x\right)\right\}^{2}}\frac{d}{dx}f(x)\] | f(x) ∈ R |
| cot⁻¹x | \[-\left(\frac{1}{1+x^{2}}\right)\] | x ∈ R |
| cot⁻¹(f(x)) | \[-\frac{1}{1+\{f(x)\}^{2}}\frac{d}{dx}f(x)\] | f(x) ∈ R |
| sec⁻¹x | \[\frac{1}{|x|\sqrt{x^{2}-1}}\] | |x| > 1 |
| sec⁻¹(f(x)) | \[\frac{1}{|f(x)|\sqrt{\{f(x)\}^{2}-1}}\frac{d}{dx}f(x)\] | |f(x)| > 1 |
| cosec⁻¹x | \[-\left(\frac{1}{|x|\sqrt{x^{2}-1}}\right)\] |
|x| > 1 |
| cosec⁻¹(f(x)) | \[-\frac{1}{|f(x)|\sqrt{\{f(x)\}^{2}-1}}\frac{d}{dx}f(x)\] | |f(x)| > 1 |
| Function | Derivative |
|---|---|
| [f(x)]ⁿ | n[f(x)]ⁿ⁻¹ · f′(x) |
| \[\sqrt{\mathrm{f}(x)}\] | \[\frac{1}{2\sqrt{\mathrm{f}(x)}}\cdot\mathrm{f}^{\prime}(x)\] |
| \[\frac{1}{\mathrm{f}(x)}\] | \[-\frac{1}{\left[\mathrm{f}(x)\right]^{2}}\cdot\mathrm{f}^{\prime}(x)\] |
| sin(f(x)) | cos(f(x)) · f′(x) |
| cos(f(x)) | −sin(f(x)) · f′(x) |
| tan(f(x)) | sec²(f(x)) · f′(x) |
| cot(f(x)) | −cosec²(f(x)) · f′(x) |
| sec(f(x)) | sec(f(x)) tan(f(x)) · f′(x) |
| cosec(f(x)) | −cosec(f(x)) cot(f(x)) · f′(x) |
| \[\mathbf{a}^{\mathbf{f}(x)}\] | \[a^{f(x)}\log a\cdot f^{\prime}(x)\] |
| \[\mathrm{e}^{\mathrm{f}(x)}\] | \[\mathrm{e}^{\mathrm{f}(x)\cdot\mathrm{f}^{\prime}(x)}\] |
| log(f(x)) | \[\frac{1}{\mathrm{f}(x)}\cdot\mathrm{f}^{\prime}(x)\] |
| logₐ(f(x)) | \[\frac{1}{\mathrm{f}(x)\mathrm{loga}}\cdot\mathrm{f}^{\prime}(x)\] |
\[\mathrm{f(a+h)\approx f(a)+h~f^{\prime}(a)}\]
Theorems and Laws [14]
If f(x) ≤ g(x) ≤ h(x) and \[\lim_{x\to a}\mathrm{f}(x)=l=\lim_{x\to a}\mathrm{h}(x)\]
\[\therefore\lim_{x\to a}g(x)=l\]
If a function \[f\] is differentiable at a point \[c\], then it is also continuous at that point.
Proof: Since \[f\] is differentiable at \[c\], we have
But for \[x \neq c\], we have
Therefore \[\lim_{x \to c} [f(x) - f(c)] = \lim_{x \to c} \left[ \frac{f(x) - f(c)}{x - c} \cdot (x - c) \right]\]
or \[\lim_{x \to c} [f(x)] - \lim_{x \to c} [f(c)] = \lim_{x \to c} \left[ \frac{f(x) - f(c)}{x - c} \right] \cdot \lim_{x \to c} [(x - c)]\]
\[= f'(c) \cdot 0 = 0\]
or \[\lim_{x \to c} f(x) = f(c)\]
Hence \[f\] is continuous at \[x = c\].
If y = `[(f(x), g(x), h(x)),(l, m,n),(a,b,c)]`, prove that `dy/dx = |(f'(x), g'(x), h'(x)),(l,m, n),(a,b,c)|`.
y = `|(f(x), g(x), h(x)),(l, m, n),(a, b, c)|`
`dy/dx= |(d/dx (f(x)), d/dx (g(x)), d/dx (h(x))), (l, m, n), (a, b, c)| + |(f(x), g(x), h(x)),(0, 0, 0),(a, b, c)| + |(f(x), g(x), h(x)),(l, m, n),(0, 0, 0)|`
`= |(f'(x), g'(x), h'(x)),(l, m, n),(a, b, c)|`
Prove that the greatest integer function defined by f(x) = [x], 0 < x < 3 is not differentiable at x = 1 and x = 2.
Any function will not be differentiable if the left-hand limit and the right-hand limit are not equal.
f(x) = [x], 0 < x < 3
(i) At x = 1
Left-side limit:
`lim_(h -> 0) ([1 - h] - [1])/-h`
= `lim_(h -> 0) (0 - 1)/-h`
= `lim_(h -> 0) 1/h`
= Infinite (∞)
Right-hand limit:
`lim_(h -> 0) ([1 + h] - [1])/h`
= `lim_(h -> 0) (1 - 1)/h`
= 0
Left-side limit and right-side limit are not equal.
Hence, f(x) is not differentiable at x = 1.
(ii) At x = 2
Left-side limit:
`lim_(h -> 0) (f(2 + h) - f(2))/h`
= `lim_(h -> 0) ([2 + h]-2)/h`
= `lim_(h -> 0) (2 -2)/h`
= 0
Right-hand limit:
`lim_(h -> 0) (f(2 - h) - f (2))/h`
= `lim_(h -> 0) ([2 - h] - [2])/-h`
= `lim_(h -> 0) (1 - 2)/-h`
= Infinite (∞)
Left-side limit and right-side limit are not equal.
Hence, f(x) is not differentiable at x = 2.
Prove that the function f given by f(x) = |x − 1|, x ∈ R is not differentiable at x = 1.
Any function will not be differentiable if the left-hand limit and the right-hand limit are not equal.
f(x) = |x − 1|, x ∈ R
f(x) = (x − 1), if x − 1 > 0
= −(x − 1), if x − 1 < 0
At x = 1
f(1) = 1 − 1 = 0
left-side limit:
`lim_(h -> 0^-) (f(1 - h) - f(1))/ -h`
= `lim_(h -> 0^-) (1 - (1 - h) - 0)/ (- h)`
= `lim_(h -> 0^-) (+ h)/(- h)`
= −1
Right-side limit:
= `lim_(h -> 0^+) (f(1 + h) - f(1))/h`
= `lim_(h -> 0^+) ((1 + h) - 1 - 0)/ h`
= `lim_(h -> 0^+) h/h`
= 1
Left-side limit and the right-side limit are not equal.
Hence, f(x) is not differentiable at x = 1.
If (x – a)2 + (y – b)2 = c2, for some c > 0, prove that `[1+ (dy/dx)^2]^(3/2)/((d^2y)/dx^2)` is a constant independent of a and b.
Given, (x – a)2 + (y – b)2 = c2 ...(1)
On differentiating with respect to x,
`=> 2 (x - a) + 2(y - b) dy/dx = 0`
`=> (x - a) + (y - b) dy/dx = 0` ...(2)
Differentiating again with respect to x,
`1 + dy/dx * dy/dx + (y - b) (d^2 y)/dx^2` = 0
`1 + (dy/dx)^2 + (y - b) (d^2y)/dx^2` = 0
`=> (y - b) = - {(1 + (dy/dx)^2)/((d^2y)/dx^2)}` ...(3)
Putting the value of (y – b) in (2),
`(x - a) = {(1 + (dy/dx)^2)/((d^2y)/dx^2)}(dy/dx)` ...(4)
Putting the values of (x − a) and (y − b) from (3) and (4) in (1),
`{1 + (dy/dx)^2}^2/((d^2y)/dx^2)^2 * (dy/dx)^2 + {(1 + (dy/dx)^2)/((d^2y)/dx^2)} = c^2`
On multiplying by `((d^2y)/dx^2)^2`,
`[1 + (dy/dx)^2]^2 (dy/dx)^2 + [1 + (dy/dx)^2]^2 = c^2 ((d^2y)/dx)^2`
`=> [1 + (dy/dx)^2]^2 [(dy/dx)^2 + 1] = c^2 ((d^2y)/dx^2)^2`
`=> {1 + (dy/dx)^2}^3 = c^2 ((d^2y)/dx^2)^2`
On taking the square root,
`therefore {1 + (dy/dx)^2}^(3//2)/((d^2y)/dx^2)` = c ...(a constant independent of a and b.)
If x = `e^(x/y)`, then prove that `dy/dx = (x - y)/(xlogx)`.
Given: x = `e^(x/y)`
Taking log on both the sides,
log x = `log e^(x/y)`
⇒ log x = `x/y log e`
⇒ log x = `x/y` ...[∵ log e = 1] ...(i)
Differentiating both sides w.r.t. x:
`d/dx log x = d/dx (x/y)`
⇒ `1/x = (y xx 1 - x xx dy/dx)/y^2`
⇒ `y^2 = xy - x^2 xx dy/dx`
⇒ `x^2 xx dy/dx = xy - y^2`
⇒ `dy/dx = (y(x - y))/x^2`
⇒ `dy/d = y/x xx ((x - y)/x)`
⇒ `dy/dx = 1/logx xx ((x - y)/x) ...[∵ log x = x/y "from equation (i)"]`
`dy/dx = (x - y)/(xlogx)`
Hence proved.
If x = `e^(x/y)`, then prove that `dy/dx = (x - y)/(xlogx)`.
Given: x = `e^(x/y)`
Taking log on both the sides,
log x = `log e^(x/y)`
⇒ log x = `x/y log e`
⇒ log x = `x/y` ...[∵ log e = 1] ...(i)
Differentiating both sides w.r.t. x:
`d/dx log x = d/dx (x/y)`
⇒ `1/x = (y xx 1 - x xx dy/dx)/y^2`
⇒ `y^2 = xy - x^2 xx dy/dx`
⇒ `x^2 xx dy/dx = xy - y^2`
⇒ `dy/dx = (y(x - y))/x^2`
⇒ `dy/d = y/x xx ((x - y)/x)`
⇒ `dy/dx = 1/logx xx ((x - y)/x) ...[∵ log x = x/y "from equation (i)"]`
`dy/dx = (x - y)/(xlogx)`
Hence proved.
If y = 5 cos x – 3 sin x, prove that `(d^2y)/(dx^2) + y = 0`.
Given, y = 5 cos x – 3 sin x
Differentiating both sides with respect to x,
`dy/dx = 5 d/dx cos x - 3 d/dx sin x`
= 5 (−sin x) − 3 cos x
= −5 sin x − 3 cos x
Differentiating both sides again with respect to x,
`(d^2 y)/dx = - 5 d/dx sin x - 3 d/dx cos x`
= −5 cos x − 3 (−sin x)
= 3 sin x − 5 cos x
Hence, `(d^2 y)/dx^2 + y` = 0
(3 sin x − 5 cos x) + (5 cos x − 3 sin x) = 0 ...(On substituting the value of y)
If y `sqrt(x^2 + 1) = log sqrt(x^2 + 1) - x`, show that `(x^2 + 1)(dy)/(dx) + xy + 1 = 0.`
Given:
y `sqrt(x^2 + 1) = log (sqrt(x^2 + 1) - x)`
Differentiate the Left-Hand Side:
Using the product rule (uv)′ = u′v + uv′:
Let u = y and v = `sqrt(x^2 + 1)`
`d/dx (y sqrt(x^2 + 1)) = (dy)/(dx) . sqrt(x^2 + 1) + y . d/dx (sqrt(x^2 + 1))`
= `sqrt(x^2 + 1) (dy)/(dx) + y . (1/(2sqrt(x^2 + 1)) . 2x)`
= `sqrt(x^2 + 1) (dy)/(dx) + (xy)/sqrt(x^2 + 1)` ...(i)
Differentiate the Right-Hand Side:
Using the chain rule for log(u):
`d/dx [log (sqrt(x^2 + 1) - x)] = 1/(sqrt(x^2 + 1) - x) . d/dx (sqrt(x^2 + 1) - x)`
= `1/(sqrt(x^2 + 1) - x) . (x/sqrt(x^2 + 1) - 1)`
Take the LCM in the bracket:
= `1/(sqrt(x^2 + 1) - x) . ((x - sqrt(x^2 + 1))/sqrt(x^2 + 1))`
= `1/(sqrt(x^2 + 1) - x) . ((-sqrt(x^2 + 1) - x)/sqrt(x^2 + 1))`
= `-1/(sqrt(x^2 + 1)` ...(ii)
Equate LHS and RHS
`sqrt(x^2 + 1) (dy)/(dx) + (xy)/sqrt(x^2 + 1) = -1/(sqrt(x^2 + 1)`
Multiply the entire equation by `sqrt(x^2 + 1)` to clear the denominators:
`(sqrt(x^2 + 1) . sqrt(x^2 + 1)) (dy)/(dx) + xy = -1`
`(x^2 + 1) (dy)/(dx) + xy = -1`
`(x^2 + 1) (dy)/(dx) + xy + 1 = 0`
Hence proved
Assume f'(c) = 0 and the second derivative exists at c:
-
Local Maxima: f''(c) < 0
-
Local Minima: f''(c) > 0
-
Test Fails: f''(c) = 0. If this happens, you must go back and use the First Derivative Test to check if it is a maxima, minima, or point of inflection.
Let c be a critical point of a continuous function f:
-
Local Maxima: f'(x) changes sign from positive to negative as x increases through c.
-
Local Minima: f'(x) changes sign from negative to positive as x increases through c.
-
Point of Inflection: f'(x) does not change sign as x passes through c (it is neither a maxima nor a minima).
Statement:
If a function f(x):
-
Is continuous on the closed interval [a,b]
-
Is differentiable on the open interval (a,b)
-
Satisfies f(a) = f(b)
Then there exists at least one c∈(a,b)c \in (a,b) such that:
\[f^{\prime}(c)=0\]
Statement:
If a function f(x):
-
Is continuous on the closed interval [a,b]
-
Is differentiable on the open interval (a,b)
Then there exists at least one number c ∈ (a,b) such that:
\[f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}\]
Key Points
| No. | Rule | Limit Law |
|---|---|---|
| i | Sum | \[\lim_{x\to a}\left(f+g\right)x=\lim_{x\to a}f\left(x\right)+\lim_{x\to a}g\left(x\right)\] |
| ii | Difference | \[\lim_{x\to a}\left(f-g\right)x=\lim_{x\to a}f\left(x\right)-\lim_{x\to a}g\left(x\right)\] |
| iii | Product |
\[\lim_{x\to a}\left[f(x)\cdot g(x)\right]=\lim_{x\to a}f(x)\cdot\lim_{x\to a}g(x)\] |
| iv | Constant multiple | \[\lim_{x\to a}[c\cdot f(x)]=c\cdot\lim_{x\to a}f(x)\] |
| v | Quotient |
\[\lim_{x\to a}\frac{f\left(x\right)}{g\left(x\right)}=\frac{\lim_{x\to a}f\left(x\right)}{\lim_{x\to a}g\left(x\right)}\] where \[\lim_{x\to a}g\left(x\right)\neq0\] |
| vi | Function of function | \[\lim_{x\to a}\mathrm{f}\left[\mathrm{g}(x)\right]=\mathrm{f}\left[\lim_{x\to a}\mathrm{g}\left(x\right)\right]=\mathrm{f}(\mathrm{m})\] |
| vii | Sum with constant | (\lim [f(x)+k] = \lim f(x) + k = l + k) |
| viii | Logarithmic | \[\lim_{x\to a}\log\left[\mathrm{f}(x)\right]=\log\left[\lim_{x\to a}\mathrm{f}(x)\right]=\log l\] |
| ix | Power | \[\lim_{x\to a}[\mathrm{f}(x)]^{\mathrm{g}(x)}=\left[\lim_{x\to a}\mathrm{f}(x)\right]^{\lim_{x\to a}\mathrm{g}(x)}=l^{\mathrm{m}}\] |
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Derivative exists only when the defining limit exists.
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Differentiability at a point means the function has a valid derivative there.
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Every differentiable function is continuous at that point.
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Every continuous function is not necessarily differentiable.
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A composite function has one function inside another function.
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The chain rule formula is \[\frac{df}{dx} = \frac{dv}{dt} \cdot \frac{dt}{dx}\]
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First differentiate the outer function, then multiply by the derivative of the inner function.
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The derivative of an inverse function is usually found using implicit differentiation.
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For \[\sin^{-1} x\] and \[\cos^{-1} x\], the denominator is \[\sqrt{1 - x^2}\].
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For \[\tan^{-1} x\] and \[\cot^{-1} x\], the denominator is \[1 + x^2\].
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For \[\sec^{-1} x\] and \[\csc^{-1} x\], the denominator involves \[|x|\sqrt{x^2 - 1}\].
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Negative signs are especially important in \[\cos^{-1} x\], \[\cot^{-1} x\], and \[\csc^{-1} x\].
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Domain restrictions must be checked before applying formulas.
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The derivative of an inverse function is usually found using implicit differentiation.
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For \[\sin^{-1} x\] and \[\cos^{-1} x\], the denominator is \[\sqrt{1 - x^2}\].
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For \[\tan^{-1} x\] and \[\cot^{-1} x\], the denominator is \[1 + x^2\].
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For \[\sec^{-1} x\] and \[\csc^{-1} x\], the denominator involves \[|x|\sqrt{x^2 - 1}\].
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Negative signs are especially important in \[\cos^{-1} x\], \[\cot^{-1} x\], and \[\csc^{-1} x\].
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Domain restrictions must be checked before applying formulas.
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Exponential function: \[y = b^x\], domain = all real numbers, range = positive real numbers.
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Logarithmic function: \[y = \log_b x\], domain = positive real numbers, range = all real numbers.
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Exponential and logarithmic functions are inverses of each other.
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\[e^x\] and log x are especially important in calculus.
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Main log laws: product, quotient, power, and change of base.
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Standard derivatives: \[\frac{d}{dx}(e^x) = e^x\],
\[\frac{d}{dx}(\log x) = \frac{1}{x}\].
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Exponential function: \[y = b^x\], domain = all real numbers, range = positive real numbers.
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Logarithmic function: \[y = \log_b x\], domain = positive real numbers, range = all real numbers.
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Exponential and logarithmic functions are inverses of each other.
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\[e^x\] and log x are especially important in calculus.
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Main log laws: product, quotient, power, and change of base.
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Standard derivatives: \[\frac{d}{dx}(e^x) = e^x\],
\[\frac{d}{dx}(\log x) = \frac{1}{x}\].
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A composite function has one function inside another function.
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The chain rule formula is \[\frac{df}{dx} = \frac{dv}{dt} \cdot \frac{dt}{dx}\]
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First differentiate the outer function, then multiply by the derivative of the inner function.
- If an equation contains both x and y and cannot be solved directly for y, it is called an implicit function.
- Implicit functions are generally written in the form:
f(x, y) = 0 - To differentiate an implicit function, differentiate both sides with respect to x, treating y as a function of x.
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Parametric form means both x and y are written in terms of a third variable.
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The third variable is called the parameter.
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The main formula is:
\[\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}\] -
This formula is based on the chain rule.
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Always check that \[\frac{dx}{dt} \neq 0\].
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The final answer may remain in terms of the parameter unless the question asks for conversion.
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Second derivative means differentiating the function twice with respect to the same variable.
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It is defined only when the first derivative is differentiable.
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Common notations are \[\frac{d^2y}{dx^2}\], f''(x), y'', \[D^2y\], and \[y_2\].
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Higher order derivatives can be defined similarly.
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Derivative gives instantaneous rate of change.
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Positive derivative means the quantity is increasing.
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Negative derivative means the quantity is decreasing.
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In related rates, first connect the variables by an equation, then differentiate.
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Always substitute the given value only after differentiation.
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Do not forget units in the final answer.
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Marginal cost and marginal revenue are applications of derivatives in economics.
- Increasing means output does not decrease as input increases.
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Strictly increasing means output always increases.
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Decreasing means output does not increase as input increases.
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Monotonic means either increasing or decreasing on an interval.
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f′(x) > 0 implies increasing, f′(x) < 0 implies decreasing, and f′(x) = 0 on an interval implies constant behaviour.
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Maxima and minima are extreme values of a function.
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Critical points occur where \(f'(x)=0\) or \(f'(x)\) is not defined.
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If \(f'(x)\) changes from positive to negative, the function has a local maximum.
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If \(f'(x)\) changes from negative to positive, the function has a local minimum.
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If \(f''(c) < 0\), there is a local maximum at \(x=c\).
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If \(f''(c) > 0\), there is a local minimum at \(x=c\).
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For absolute extrema on \([a,b]\), compare values at critical points and endpoints.
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Not every critical point gives a maximum or minimum.
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The second derivative test is quick, but the first derivative test is often more reliable in detailed reasoning.
Concepts [36]
- Concept of Limits
- Limits by Factorisation, Substitution and Rationalisation
- Algebra of Limits
- Limits of Polynomials and Rational Functions
- Limits of Logarithmic Functions
- Limits of Exponential Functions
- Limits of Trigonometric Functions
- Inverse Functions
- Graphs of Simple Functions
- Continuous and Discontinuous Functions
- Concept of Differentiability
- Differentiation of the Sum, Difference, Product, and Quotient of Two Functions
- Derivatives of Composite Functions
- Derivative of Inverse Function
- Derivative of Inverse Function
- Exponential and Logarithmic Functions
- Exponential and Logarithmic Functions
- Derivatives of Composite Functions
- Derivative of Implicit Functions
- Derivatives of Functions in Parametric Forms
- Second Order Derivative
- Mean Value Theorem
- Simple Problems on Applications of Derivatives
- Rate of Change of Quantities
- Increasing and Decreasing Functions
- Maxima and Minima
- Tangents and Normals
- Limits Using L-hospital's Rule
- Evaluation of Limits
- Infinite Series
- Successive Differentiation
- nth Derivative of Standard Functions
- Algebra of Derivative (Leibnitz or Product Rule)
- Rolle's Theorem
- Lagrange's Mean Value Theorem (LMVT)
- Approximations
