Derivatives of Trigonometric Functions:

Trigonometric and inverse trigonometric functions are differentiable at each point of its domain.

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. $$\frac{d}{dx} cosec x = – cosec x. \cot x$$

Note: remember that the trigonometric functions which are starts with c then their derivate should come with the minus(-) sign preceding the derivative of the functions and also applicable in case of the inverse trigonometric functions Ex: $$\frac{d}{dx} \cos x = – \sin x$$

Derivatives of Inverse Trigonometric Functions:

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

Derivatives of Exponential and Logarithmic Functions:

1. $$\frac{d}{dx}e^x = e^x$$
2. $$\frac{d}{dx}a^x = a^x \log_ea, a>0, a \neq 1$$
3. $$\frac{d}{dx}\log_ex = \frac{1}{x}, x>0$$
4. $$\frac{d}{dx} \log_ax = \frac{1}{xlog_ea} = \frac{log_ae}{x}$$

The logarithmic function is differentiable at each point of its domain. $$a^x$$ is differentiable at each $$x \in , (a >0, a\neq 1)$$

Derivatives of Hyperbolic Functions:

1. $$\frac{d}{dx} \sin h x = \cos h x$$
2. $$\frac{d}{dx} \cos h x = – \sin h x$$
3. $$\frac{d}{dx} \tan h x = \sec h^2 x$$
4. $$\frac{d}{dx} \cot h x = – cosec h^2 x$$
5. $$\frac{d}{dx} \sec h x = \sec h x . \tan h x$$
6. $$\frac{d}{dx} cosec h x = – cosec h x. \cot h x$$

Parametric Differentiation

If x = f(t), y =g(t) are both the derivable functions of ‘t’ then

$$\frac{d}{dx} = \frac{dy/dt}{dx/dt} = \frac{g'(t)}{f'(t)}, (f'(t) \neq 0)$$

Derivative of Infinite series:

1. If $$y = \sqrt{f(x) + \sqrt{f(x) + \sqrt{f(x)+ \cdots + \infty}}}$$ then $$y = \sqrt{f(x)+y}$$

Squaring:

$$y^2 = f(x) +y \\y^2-y = f(x)$$

Differentiating both sides w.r.t. x, we get

$$(2y-1) \frac{dy}{dx} = f'(x)$$

For example let take one trigonometric function:

$$y = \sqrt{\sin x+ \sqrt{ \sin x + \sqrt{\sin x+ \cdots + \infty}}}$$ then $$(2y -1) \frac{dy}{dx} = \cos x$$

Note: do like this for each and every functions to get derivative of the functions.

SHARE