Trigonometric and inverse trigonometric functions are differentiable at each point of its domain.
i. $ \frac{d}{dx} \sin x = \cos x $
ii. $ \frac{d}{dx} \cos x = - \sin x $
iii. $\frac{d}{dx} \tan x = \sec^2 x $
iv. $ \frac{d}{dx} \cot x = - cosec^2 x $
v. $ \frac{d}{dx} \sec x = \sec x . \tan x$
vi. $ \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:
i. $\frac{d}{dx} \sin^{-1}x = \frac{1}{\sqrt{1- x^2}}, (|x|<1)$
ii. $\frac{d}{dx} \cos^{-1}x = - \frac{1}{\sqrt{1- x^2}}, (|x|<1)$
iii. $\frac{d}{dx} \tan^{-1}x = \frac{1}{1+x^2}$, $x \in R $
iv. $\frac{d}{dx} \cot^{-1}x = -\frac{1}{1+x^2}$, $x \in R $
v. $ \frac{d}{dx} \sec^{-1}x = \frac{1}{x \sqrt{x^2-1}}, (|x|>1)$
vi. $ \frac{d}{dx} cosec^{-1}x = -\frac{1}{x \sqrt{x^2- 1}}, (|x|>1)$
Derivatives of Exponential and Logarithmic Functions:
i. $\frac{d}{dx}e^x = e^x$
ii. $ \frac{d}{dx}a^x = a^x \log_ea, a>0, a \neq 1$
iii. $\frac{d}{dx}\log_ex = \frac{1}{x}, x>0$
iv. $\frac{d}{dx} \log_ax = \frac{1}{xlog_ea} = \frac{log_ae}{x}$
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:
i. $ \frac{d}{dx} \sin h x = \cos h x $
ii. $ \frac{d}{dx} \cos h x = - \sin h x $
iii. $\frac{d}{dx} \tan h x = \sec h^2 x $
iv. $ \frac{d}{dx} \cot h x = - cosec h^2 x $
v. $ \frac{d}{dx} \sec h x = \sec h x . \tan h x$
vi. $ \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:
i. 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.
