Question

Compute the derivative of
$\sin(x) and \tan(x)$
​

Answer 1

Question :-

Compute the derivative .

Answer :-

Derivatives :-

$(1) \sf \: \: \frac{d}{dx} {x}^{n} = n {x}^{n - 1} \\ \\ (2) \sf \: \frac{d}{dx} \sin x = \cos x \\ \\ (3) \sf \: \frac{d}{dx} \cos x = - \sin x \\ \\ (4) \sf \: \frac{d}{dx} \tan x = { \sec}^{2} x \\ \\ (5) \sf \: \: \frac{d}{dx} \sec x = \sec x \: . \tan x \\ \\ (6) \sf \frac{d}{dx} \cot x = - \: { \cosec}^{2} x \\ \\ (7) \sf \: \frac{d}{dx} \: \cosec x = - \cosec x \: . \cot x \\ \\(8) \sf \: \frac{d}{dx} \log x = \frac{1}{x} \\ \\ (9) \sf \: \frac{d}{dx} \: {a}^{x} = {a}^{x} \: log_{e}(a) \\ \\ (10) \sf \: \frac{d}{dx} \: {e}^{x} = {e}^{x} \\ \\ (11) \sf \: \: \frac{d}{dx} constant \: = 0 \\ \\ (12) \sf \: \frac{d}{dx} { \sin}^{ - 1} x = \frac{1}{ \sqrt{1 - {x}^{2} } }$

$(13) \sf \: \frac{d}{dx} { \cos}^{ - 1} x = \frac{ - 1}{ \sqrt{1 - {x}^{2} } } \\ \\ (14) \sf \: \frac{d}{dx} { \tan}^{ - 1} x = \frac{1}{1 + {x}^{2} } \\ \\ (15) \sf \: \: \frac{d}{dx} { \sec}^{ - 1} x = \frac{ 1}{x \sqrt{ {x}^{2} - 1 } } \\ \\ (16) \sf \: \frac{d}{dx} { \cosec}^{ - 1} x = \frac{-1}{ x\sqrt{ {x}^{2} - 1} }$