Definitions [15]
The derivative of a real function f at a point c in its domain is defined as:
\[f'(c) = \lim_{h \to 0} \frac{f(c+h) - f(c)}{h}\]
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.
- \[\lim_{x\to a}f\left(x\right)=f\left(a\right)\].
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.
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.
- \[\lim_{x\to a}f\left(x\right)\neq f\left(a\right)\].
The derivative of a real function f at a point c in its domain is defined as:
\[f'(c) = \lim_{h \to 0} \frac{f(c+h) - f(c)}{h}\]
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.
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
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.
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 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 differentiation of a function is performed after taking logarithm on both sides, the process is called logarithmic differentiation.
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.
Formulae [2]
| 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 |
Theorems and Laws [14]
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 \[x = c\].
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.
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.
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)|`
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 g is continuous at c, and f is continuous at g(c), then their composite function \[(f \circ g)\], defined as \[(f \circ g)(x) = f(g(x))\], is also continuous at 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)|`
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.)
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 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 \[x = c\].
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.
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 cos y = x cos (a + y), with cos a ≠ ± 1, prove that `dy/dx = cos^2(a+y)/(sin a)`.
cos y = x cos (a + y)
∴ x = `(cos y)/(cos (a + y))`
On differentiating with respect to y,
`cos (a + y) d/dy cos y - cos y d/dy`
`therefore dx/dy = (cos (a + y))/(cos^2 (a + y))`
`= (- sin y cos (a + y) + cos y sin (a + y))/(cos^2 (a + y))`
`= (sin (a + y) cos y - cos (a + y) sin y)/(cos^2 (a + y))`
`= (sin (a + y - y))/(cos^2 (a + y))` ...[∵ sin (A − B) = sin A cos B − cos A sin B]
`= (sin a)/(cos^2 (a + y))`
`therefore dy/dx = (cos^2 (a + y))/(sin a)`
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)
Key Points
-
Derivative exists only when the defining limit exists.
-
Differentiability at a point means the function has a valid derivative there.
-
Every differentiable function is continuous at that point.
-
Every continuous function is not necessarily differentiable.
| Operation on continuous functions | Result |
|---|---|
| f + g | Continuous |
| f − g | Continuous |
| fg | Continuous |
| \[\dfrac{f}{g}\] | Continuous |
| \[f \circ g\] | Continuous |
-
Derivative exists only when the defining limit exists.
-
Differentiability at a point means the function has a valid derivative there.
-
Every differentiable function is continuous at that point.
-
Every continuous function is not necessarily differentiable.
-
A composite function has one function inside another function.
-
The chain rule formula is \[\frac{df}{dx} = \frac{dv}{dt} \cdot \frac{dt}{dx}\]
-
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.
-
The derivative of an inverse function is usually found using implicit differentiation.
-
For \[\sin^{-1} x\] and \[\cos^{-1} x\], the denominator is \[\sqrt{1 - x^2}\].
-
For \[\tan^{-1} x\] and \[\cot^{-1} x\], the denominator is \[1 + x^2\].
-
For \[\sec^{-1} x\] and \[\csc^{-1} x\], the denominator involves \[|x|\sqrt{x^2 - 1}\].
-
Negative signs are especially important in \[\cos^{-1} x\], \[\cot^{-1} x\], and \[\csc^{-1} x\].
-
Domain restrictions must be checked before applying formulas.
-
Exponential function: \[y = b^x\], domain = all real numbers, range = positive real numbers.
-
Logarithmic function: \[y = \log_b x\], domain = positive real numbers, range = all real numbers.
-
Exponential and logarithmic functions are inverses of each other.
-
\[e^x\] and log x are especially important in calculus.
-
Main log laws: product, quotient, power, and change of base.
-
Standard derivatives: \[\frac{d}{dx}(e^x) = e^x\],
\[\frac{d}{dx}(\log x) = \frac{1}{x}\].
-
Use logarithmic differentiation when the function is a complex product, quotient, or variable exponent form.
-
Write y = function first, then take \[\ln\] on both sides.
-
Apply logarithmic rules before differentiating.
-
Differentiate \[\ln y\] carefully: \[\frac{d}{dx}(\ln y) = \frac{1}{y} \frac{dy}{dx}\].
-
Substitute the original value of y at the end.
-
Ensure the expression inside logarithm remains positive.
-
Parametric form means both x and y are written in terms of a third variable.
-
The third variable is called the parameter.
-
The main formula is:
\[\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}\] -
This formula is based on the chain rule.
-
Always check that \[\frac{dx}{dt} \neq 0\].
-
The final answer may remain in terms of the parameter unless the question asks for conversion.
-
Second derivative means differentiating the function twice with respect to the same variable.
-
It is defined only when the first derivative is differentiable.
-
Common notations are \[\frac{d^2y}{dx^2}\], f''(x), y'', \[D^2y\], and \[y_2\].
-
Higher order derivatives can be defined similarly.
Concepts [12]
- Concept of Differentiability
- Continuous and Discontinuous Functions
- Algebra of Continuous Functions
- Concept of Differentiability
- Derivatives of Composite Functions
- Derivative of Implicit Functions
- Derivative of Inverse Function
- Exponential and Logarithmic Functions
- Logarithmic Differentiation
- Derivatives of Functions in Parametric Forms
- Second Order Derivative
- Mean Value Theorem
