Definitions [5]
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.
- \[\lim_{x\to a}f\left(x\right)=f\left(a\right)\].
The principle which states that for a non-viscous liquid in streamline flow passing through a tube of varying cross-section, the product of the area of cross-section and the velocity of flow remains constant at every point is called the Equation of Continuity.
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.
Formulae [3]
For a non-viscous liquid in streamline flow passing through a tube of varying cross-section:
av = constant
or equivalently:
a ∝ \[\frac {1}{v}\]
where:
- a = area of cross-section of the tube
- v = velocity of flow of the liquid
| 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 [11]
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 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.
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 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
If f & g are two continuous functions at x = a, then
αf is continuous at x = a ∀ α ∈ R
f + g is continuous at x = a
f − g is continuous at x = a
f·g is continuous at x = a
f/g is continuous at x = a, provided g(a) ≠ 0
If y is a differentiable function of u and u is a differentiable function of x, then
\[\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d}y}{\mathrm{d}u}\cdot\frac{\mathrm{d}u}{\mathrm{d}x}\]
- 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.
If y = f(x) is a differentiable function of x such that the inverse function x = f⁻¹(y) exists, then x is a differentiable function of y and
\[\frac{\mathrm{d}x}{\mathrm{d}y}=\frac{1}{\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)}\], where \[\frac{\mathrm{d}y}{\mathrm{d}x}\neq0\].
If differentiation of an expression is done after taking the logarithm on both sides, then it is called logarithmic differentiation. Generally, we apply this method when the given expression is in one of the following forms:
- product of a number of functions,
- a quotient of functions,
- a function which is the power of another function, i.e., \[[f(x)]^{g(x)}\]
If x = f(t) and y = g(t) are differentiable functions of parameter t, then
\[\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\left(\frac{\mathrm{d}y}{\mathrm{d}t}\right)}{\left(\frac{\mathrm{d}x}{\mathrm{d}t}\right)},\frac{\mathrm{d}x}{\mathrm{d}t}\neq0\]
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
