# Derivatives

By Suresh Chand

The derivatives of a function f at a point a is  denoted by f(a) and is defined as

$$f^{‘}(a) = \lim_{h \to 0} \frac{f(x) – f(a)}{x – a}$$

if limit exists

Suppose h = |x – a|. Then (1) can be written as

$$f^{‘}(a) = \lim_{h \to 0} \frac{f(a + h) – f(a)}{h}$$

Example: IF f(x) = x3 – x, then find a formula for f(x).

Solution:

Let f(x) = x3 – x

Then,

$$f^{‘}(x) = \lim_{h \to 0} \frac{f(x + h) – f(x)}{h}$$

$$= \lim_{h \to 0} \left ( \frac{(x + 3)^3 – (x + h) – x^3 + x}{h} \right )$$

$$= \lim_{h \to 0} \left ( \frac{x^3 + h^3 + 3x^2h + 3xh^2 – x – h – x^3 + x}{h} \right )$$

$$= \lim_{h \to 0} \left ( \frac{h(3x^2 + 3xh + h^2 – h)}{h} \right )$$

$$= \lim_{h \to 0} \left ( 3x^2 + 3xh + h^2 – h \right )$$

= 3x2 – 1

Thus, f(x) = 3x2 – 1

Show that f(x) = |x| is not differentiable at x = 0

Solution:

For x > 0, we have |x| = x. Therefore, for  h > 0, |x + h| = x + h

Here, for  > 0

$$f^{‘}(x) = \lim_{h \to 0}\left ( \frac{f(x + h) – f(x)}{h} \right )$$

$$= \lim_{h \to 0}\left ( \frac{|x + h| – |x|}{h} \right )$$

$$= \lim_{h \to 0}\left ( \frac{x + h – x} {h}\right )$$

$$= \lim_{h \to 0}\left ( \frac{h} {h}\right )$$

=1

This means f is differentiable for x > 0

and for x < 0, we have |x| = – x. Then choose h is so small such that |x + h| < 0 and |x + h| = -(x + h).

Here, for x < 0

$$f^{‘}(x) = \lim_{h \to 0}\left ( \frac{f(x + h) – f(x)}{h} \right )$$

$$= \lim_{h \to 0}\left ( \frac{-(x + h) – (-x)} {h}\right )$$

$$= \lim_{h \to 0}\left ( \frac{-h} {h}\right )$$

= -1

This means f is differentiable for x < 0

But,

$$f^{‘}(x)|_{\enspace for \enspace x > 0} \neq f^{‘}(x)|_{\enspace for \enspace x < 0}$$

So, f is not differentiable at x = 0

Geometrically, the curve f(x)  doesn’t have a tangent line at origin i.e. at (0, 0).

### Review of Derivative

#### 1. Multiple Rule

If c be a constant and f is differentiable function then

$$\frac{d}{dx}(cf(x)) = c \frac{d}{dx}f(x)$$

#### 2. Sum and difference rule

If f and g are both differentiable functions then

$$\frac{d}{dx}(f(x) + g(x)) = \frac{d}{dx}f(x) + \frac{d}{dx}g(x)$$

#### 3. Product Rule

If f and g are both differentiable functions then

$$\frac{d}{dx}(f(x) \enspace g(x)) = f(x)\frac{d}{dx}g(x) + g(x)\frac{d}{dx}f(x)$$

Example 1: If f(x) = $$\frac{secx}{1 + tanx}$$ then find f(x)

Solution

Let,

$$f(x) = \frac{secx}{1 + tanx}$$

Then

$$f^{‘}(x) = \frac{(1 + tanx) secx tanx – secx (sec^{2}x)}{(1 + tanx) ^ 2}$$

$$= \frac{secx (tanx + tan^2x – sec^2x)}{(1 + tanx) ^ 2}$$

$$= \frac{secx (tanx – 1)}{(1 + tanx) ^ 2}$$

Example 10: Find an equation of the tangent line to the curve y = 2x sinx at the point $$(\frac{\pi}{2}, \pi)$$.

Solution: Let,

y = 2x sinx

Then

y = 2sinx + 2x cosx

At the point $$(\frac{\pi}{2}, \pi)$$,

$$y^{‘} = 2 sin(\frac{\pi}{2}) + 2\frac{\pi}{2}cos(\frac{\pi}{2})$$

= 2 + 0

= 2

This shows the curve has slope 2 at the point $$(\frac{\pi}{2}, \pi)$$. Since the point $$(\frac{\pi}{2}, \pi)$$, is the common point of the curve and the tangent line to y = 2x sinx at $$(\frac{\pi}{2}, \pi)$$. Therefore the slope of the tangent line is 2

Thus, the equation of tangent line to y = 2x sinx at the point $$(\frac{\pi}{2}, \pi)$$ is

$$y – \pi = 2(x – \frac{\pi}{2})$$

$$y – \pi = 2x – \pi$$

2x – y = 0

### Differentiable Formulas

#### 1. Formula for Trigonometric Functions

1. $$\frac{d}{dx} (sin~ x)= cos\ x$$
2. $$\frac{d}{dx} (cos~ x)= – sin\ x$$
3. $$\frac{d}{dx} (tan ~x)= sec^{2} x$$
4. $$\frac{d}{dx} (cot~ x = -cosec^{2} x$$
5. $$\frac{d}{dx} (sec~ x) = sec\ x\ tan\ x$$
6. $$\frac{d}{dx} (cosec ~x)= -cosec\ x\ cot\ x$$
7. $$\frac{d}{dx} (sinh~ x)= cosh\ x$$
8. $$\frac{d}{dx} (cosh~ x) = sinh\ x$$
9. $$\frac{d}{dx} (tanh ~x)= sech^{2} x$$
10. $$\frac{d}{dx} (coth~ x)=-cosech^{2} x$$
11. $$\frac{d}{dx} (sech~ x)= -sech\ x\ tanh\ x$$
12. $$\frac{d}{dx} (cosech~ x ) = -cosech\ x\ coth\ x$$

#### 2. Formula for Inverse Trigonometric Functions

1. $$\frac{d}{dx}(sin^{-1}~ x)$$ = $$\frac{1}{\sqrt{1 – x^2}}$$
2. $$\frac{d}{dx}(cos^{-1}~ x)$$ = $$-\frac{1}{\sqrt{1 – x^2}}$$
3. $$\frac{d}{dx}(tan^{-1}~ x)$$ = $$\frac{1}{1 + x^2}$$
4. $$\frac{d}{dx}(cot^{-1}~ x)$$ = $$-\frac{1}{1 + x^2}$$
5. $$\frac{d}{dx}(sec^{-1} ~x)$$= $$\frac{1}{|x|\sqrt{x^2 – 1}}$$
6. $$\frac{d}{dx}(cosec^{-1}~x)$$= $$-\frac{1}{|x|\sqrt{x^2 – 1}}$$

#### 3. Other Functions

1. $$\frac{d}{dx}(a^{x}) = a^{x} ln a$$
2. $$\frac{d}{dx}(e^{x}) = e^{x}$$
3. $$\frac{d}{dx}(log_a~ x)$$ = $$\frac{1}{(ln~ a)x}$$
4. $$\frac{d}{dx}(ln~ x) = 1/x$$
5. Chain Rule: $$\frac{dy}{dx}$$ = $$\frac{dy}{du} × \frac{du}{dx}$$ = $$\frac{dy}{dv} × \frac{dv}{du} × \frac{du}{dx}$$

Important Questions