Question

Differentiate with respect to x ​

Answer 1

QUESTION :–

Differentiate the following w.r.t. 'x' $\:\: \bf {tan}^{ - 1} \{ \cot(4x) \} \:\:$ –

(a) -4

(b) ¼

(c) 4

(d) - ¼

ANSWER :–

GIVEN :–

$\\ \:\: \implies \bf y = {tan}^{ - 1} \{ \cot(4x) \} \:\:$

TO FIND :–

$\\ \implies \bf \dfrac{dy}{dx} = \: ?$

SOLUTION :–

$\\ \implies \bf y = {tan}^{-1} \{ \cot(4x) \} \:\:$

• We know that –

$\\ \dashrightarrow \bf { \tan}^{-1} (x) + { \cot }^{ - 1}(x) = \dfrac{\pi}{2} \:\:$

$\\ \dashrightarrow \bf { \tan}^{-1} (x)= \dfrac{\pi}{2} - { \cot }^{ - 1}(x)$

• So that –

$\\ \implies \bf y = \dfrac{\pi}{2} - {cot}^{-1} \{ \cot(4x) \} \:\:$

$\\ \implies \bf y = \dfrac{\pi}{2} - 4x \:\:$

• Now Differentiate with respect to 'x' –

$\\ \implies \bf \dfrac{dy}{dx} = \dfrac{d}{dx} \left(\dfrac{\pi}{2} \right) - \dfrac{d}{dx} (4x) \:\:$

$\\ \implies \bf \dfrac{dy}{dx} = 0 - 4\dfrac{d}{dx} (x) \:\:$

$\\ \implies \large{ \boxed{\bf \dfrac{dy}{dx} = - 4 }}$

▪︎ Hence , Option (a) is correct.

Answer 2

Given ,

The function is

$\sf y = { \tan}^{ - 1} \{ \cot(4x) \}$

Differentiating with respect to x , we get

$\sf \mapsto \frac{dy}{dx} = \frac{d { \tan}^{ - 1} \{ \cot(4x) \}}{dx} \\ \\ \sf \mapsto \frac{dy}{dx} = \frac{ - 1}{1 + {( \cot4x)}^{2} } \times \frac{d \cot(4x)}{dx} \\ \\ \sf \mapsto \frac{dy}{dx} = \frac{ - 1}{1 + {( \cot4x)}^{2} } \times - 4 {(cosec)}^{2}(4x)$

Remmember :

$\star \: \: \sf \frac{d { \tan}^{ - 1} (x) }{dx} = \frac{1}{1 + {(x)}^{2} } \\ \\ \star \: \: \sf \frac{d \cot(x) }{dx} = {(cosec)}^{2}(x)$