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Question: If \(y = {(\sin x)^{\tan x}}\), then \(\dfrac{{dy}}{{dx}}\) is equal to A. \[\sin {x^{\tan x}}(1 ...

If y=(sin⁡x)tan⁡xy = {(\sin x)^{\tan x}}, then dydx\dfrac{{dy}}{{dx}} is equal to
A. sin⁡xtan⁡x(1+sec⁡2xlog⁡sin⁡x)\sin {x^{\tan x}}(1 + {\sec ^2}x\log \sin x)
B. tan⁡x(sin⁡x)tan⁡x−1.cos⁡x\tan x{(\sin x)^{\tan x - 1}}.\cos x
C. sin⁡xcos⁡x.sec⁡2xlog⁡sin⁡x\sin {x^{\cos x}}.{\sec ^2}x\log \sin x
D. tan⁡x.(sin⁡x)tan⁡x−1\tan x.{(\sin x)^{\tan x - 1}}

Explanation

Solution

we can find the derivative that is dydx\dfrac{{dy}}{{dx}} of the given function y=(sin⁡x)tan⁡xy = {(\sin x)^{\tan x}} by taking natural logarithm on both sides of the given equation and by using chain rule and product rule of differentiation. We also use properties of logarithms wherever needed as per the requirement of the given problem.

Complete step by step answer:
Given function is y=(sin⁡x)tan⁡xy = {(\sin x)^{\tan x}}
Now before differentiating the above function, we can rewrite the given equation by taking natural logarithm on both sides. then the above equation becomes
⇒log⁡y=log⁡(sin⁡x)tan⁡x\Rightarrow \log y = \log {(\sin x)^{\tan x}}
Now applying exponential property of logarithm for the RHS of the above equation we get
⇒log⁡y=tan⁡xlog⁡(sin⁡x)\Rightarrow \log y = \tan x\log (\sin x)
Now differentiating the above equation with respect to x by applying chain rule of differentiation for RHS and product rule of differentiation for LHS we get
1ydydx=sec⁡2xlog⁡(sin⁡x)+tan⁡x1sin⁡xcos⁡x\dfrac{1}{y}\dfrac{{dy}}{{dx}} = {\sec ^2}x\log (\sin x) + \tan x\dfrac{1}{{\sin x}}\cos x
We know that cot⁡x=cos⁡xsin⁡x\cot x = \dfrac{{\cos x}}{{\sin x}}by using this relation the above equation can be written as
1ydydx=sec⁡2xlog⁡(sin⁡x)+tan⁡xtan⁡x\dfrac{1}{y}\dfrac{{dy}}{{dx}} = {\sec ^2}x\log (\sin x) + \dfrac{{\tan x}}{{\tan x}}
Now we can use the reciprocal relation of tan and cot and is given by cot⁡x=1tan⁡x\cot x = \dfrac{1}{{\tan x}}with this the above equation becomes
1ydydx=sec⁡2xlog⁡(sin⁡x)+tan⁡xtan⁡x\dfrac{1}{y}\dfrac{{dy}}{{dx}} = {\sec ^2}x\log (\sin x) + \dfrac{{\tan x}}{{\tan x}}
⇒1ydydx=sec⁡2xlog⁡(sin⁡x)+1\Rightarrow \dfrac{1}{y}\dfrac{{dy}}{{dx}} = {\sec ^2}x\log (\sin x) + 1
Now cross multiply y we get
⇒dydx=y(sec⁡2xlog⁡(sin⁡x)+1)\Rightarrow \dfrac{{dy}}{{dx}} = y({\sec ^2}x\log (\sin x) + 1)
Now on substituting the value of y from the given problem we get
⇒dydx=(sin⁡x)tan⁡x(sec⁡2xlog⁡sin⁡x+1)\Rightarrow \dfrac{{dy}}{{dx}} = {(\sin x)^{\tan x}}({\sec ^2}x\log \sin x + 1)

So, the correct answer is “Option A”.

Note: ddx\dfrac{d}{{dx}} is an operation that means "take the derivative with respect to x" whereas dydx\dfrac{{dy}}{{dx}} indicates that "the derivative of y was taken with respect to x". further we use the chain rule when differentiating a 'function of a function', like f(g(x))f(g(x)) in general. We use the product rule when differentiating two functions multiplied together, like f(x)g(x)f(x)g(x).