Answer to Question #276727 in Calculus for Somu

Question

Differentiate of the following functions with respect to x:

i) 𝑡𝑎𝑛𝑥∙ 𝑙𝑛(𝑠𝑖𝑛𝑥)

ii) √𝑐𝑜𝑡√𝑥

iii) 𝑒 ln (𝑡𝑎𝑛5𝑥)

iv) 𝑆𝑖𝑛2

{ln(𝑠𝑒𝑐𝑥)}

v) ln (𝑐𝑜𝑠𝑒𝑐𝑥)/𝑥

Answer 1

i)

$$(\tan x\cdot\ln(\sin x) )'=\dfrac{\ln(\sin x)}{\cos^2 x}$$

$$+\tan x(\dfrac{1}{\sin x})(\cos x)=\dfrac{\ln(\sin x)}{\cos^2 x}+1$$

ii)

$$\bigg(\sqrt{\cot(\sqrt{x})}\bigg)'=\dfrac{1}{2\sqrt{\cot(\sqrt{x})}}(-\dfrac{1}{\sin ^2 (\sqrt{x})})(\dfrac{1}{2\sqrt{x}})$$

$$=-\dfrac{1}{4\sin ^2 (\sqrt{x})\sqrt{x\cot(\sqrt{x})}}$$

iii)

$$(e^{\ln (\tan x)})'=(\tan x)'=\dfrac{1}{\cos ^2 x}$$

iv)

$$\bigg(\sin^2(\ln(\sec x))\bigg)'=\bigg(\sin^2(\ln(\dfrac{1}{\cos x}))\bigg)'$$

$$=\bigg(\sin^2(\ln(\cos x))\bigg)'$$

$$=2\sin(\ln(\cos x))\cos(\ln(\cos x))(\dfrac{1}{\cos x})(-\sin x)$$

$$=-\tan x\cdot\sin\big(2\ln(\cos x)\big)$$

v)

$$(\dfrac{\ln(\cosec x)}{x})'=(\dfrac{\ln(1/\sin x)}{x})'$$

$$=-(\dfrac{\ln(\sin x)}{x})'=-\dfrac{\dfrac{\cos x}{\sin x}(x)-\ln(\sin x)}{x^2}$$

$$=\dfrac{\ln(\sin x)-x\cot x}{x^2}$$