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Question: We are given a trigonometric function as ,\(y={{\left( \cot x \right)}^{\sin x}}+{{\left( \tan x \ri...

We are given a trigonometric function as ,y=(cot⁡x)sin⁡x+(tan⁡x)cos⁡xy={{\left( \cot x \right)}^{\sin x}}+{{\left( \tan x \right)}^{\cos x}}. Then find dydx\dfrac{dy}{dx}.
A. sin⁡x(cot⁡x)sin⁡x−1(−cos⁡ec2x)+(cot⁡x)sin⁡x(log⁡cot⁡x)cos⁡x +cos⁡x(tan⁡x)cos⁡x−1sec2x+(tan⁡x)cos⁡x(log⁡tan⁡x)(−sin⁡x) \begin{aligned} & \sin x{{\left( \cot x \right)}^{\sin x-1}}\left( -\cos e{{c}^{2}}x \right)+{{\left( \cot x \right)}^{\sin x}}\left( \log \cot x \right)\cos x \\\ & +\cos x{{\left( \tan x \right)}^{\cos x-1}}se{{c}^{2}}x+{{\left( \tan x \right)}^{\cos x}}\left( \log \tan x \right)\left( -\sin x \right) \\\ \end{aligned}
B. sin⁡x(cot⁡x)sin⁡x−1(−cos⁡ec2x)+cos⁡x(tan⁡x)cos⁡x−1sec2x\sin x{{\left( \cot x \right)}^{\sin x-1}}\left( -\cos e{{c}^{2}}x \right)+\cos x{{\left( \tan x \right)}^{\cos x-1}}se{{c}^{2}}x
C. sin⁡x(cot⁡x)sin⁡x−1(−sec2x)+(cot⁡x)sin⁡x(log⁡cot⁡x)cos⁡x +cos⁡x(tan⁡x)cos⁡x−1cos⁡ec2x+(tan⁡x)cos⁡x(log⁡tan⁡x)(sin⁡x) \begin{aligned} & \sin x{{\left( \cot x \right)}^{\sin x-1}}\left( -se{{c}^{2}}x \right)+{{\left( \cot x \right)}^{\sin x}}\left( \log \cot x \right)\cos x \\\ & +\cos x{{\left( \tan x \right)}^{\cos x-1}}\cos e{{c}^{2}}x+{{\left( \tan x \right)}^{\cos x}}\left( \log \tan x \right)\left( \sin x \right) \\\ \end{aligned}
D. none of these

Explanation

Solution

To find the differentiation of y we first separate the functions in y. We find their differentiations separately and add them at the end. We use logarithmic formulas to apply the chain rule on those functions. At the end we perform some trigonometric relations to match with the given options.

Complete answer:
Here we have been given y as a sum of two different functions of x. We assume those functions and differentiate separately.
So, y=(cot⁡x)sin⁡x+(tan⁡x)cos⁡x=f(x)+g(x)y={{\left( \cot x \right)}^{\sin x}}+{{\left( \tan x \right)}^{\cos x}}=f\left( x \right)+g\left( x \right) where f(x)=(cot⁡x)sin⁡x,g(x)=(tan⁡x)cos⁡xf\left( x \right)={{\left( \cot x \right)}^{\sin x}},g\left( x \right)={{\left( \tan x \right)}^{\cos x}}.
Differentiating both sides, we get dydx=f′(x)+g′(x)\dfrac{dy}{dx}={{f}^{'}}\left( x \right)+{{g}^{'}}\left( x \right).
We have to differentiate f(x)=(cot⁡x)sin⁡xf\left( x \right)={{\left( \cot x \right)}^{\sin x}} and g(x)=(tan⁡x)cos⁡xg\left( x \right)={{\left( \tan x \right)}^{\cos x}} separately.
We apply logarithm to both cases and then differentiate. We have log⁡ab=blog⁡a\log {{a}^{b}}=b\log a.
For f(x)=(cot⁡x)sin⁡xf\left( x \right)={{\left( \cot x \right)}^{\sin x}}, we get log⁡[f(x)]=log⁡[(cot⁡x)sin⁡x]=sin⁡xlog⁡(cot⁡x)\log \left[ f\left( x \right) \right]=\log \left[ {{\left( \cot x \right)}^{\sin x}} \right]=\sin x\log \left( \cot x \right).
We differentiate to find f′(x){{f}^{'}}\left( x \right).
log⁡[f(x)]=sin⁡xlog⁡(cot⁡x) ⇒f′(x)f(x)=sin⁡x.ddx[log⁡(cot⁡x)]+ddx(sin⁡x).log⁡(cot⁡x) ⇒f′(x)f(x)=sin⁡x.(−cosec⁡2x)cot⁡x+cos⁡x.log⁡(cot⁡x) ⇒f′(x)=f(x)[−sec⁡x+cos⁡x.log⁡(cot⁡x)] \begin{aligned} & \log \left[ f\left( x \right) \right]=\sin x\log \left( \cot x \right) \\\ & \Rightarrow \dfrac{{{f}^{'}}\left( x \right)}{f\left( x \right)}=\sin x.\dfrac{d}{dx}\left[ \log \left( \cot x \right) \right]+\dfrac{d}{dx}\left( \sin x \right).\log \left( \cot x \right) \\\ & \Rightarrow \dfrac{{{f}^{'}}\left( x \right)}{f\left( x \right)}=\sin x.\dfrac{\left( -co{{\sec }^{2}}x \right)}{\cot x}+\cos x.\log \left( \cot x \right) \\\ & \Rightarrow {{f}^{'}}\left( x \right)=f\left( x \right)\left[ -\sec x+\cos x.\log \left( \cot x \right) \right] \\\ \end{aligned}
For g(x)=(tan⁡x)cos⁡xg\left( x \right)={{\left( \tan x \right)}^{\cos x}}, we get log⁡[g(x)]=log⁡[(tan⁡x)cos⁡x]=cos⁡xlog⁡(tan⁡x)\log \left[ g\left( x \right) \right]=\log \left[ {{\left( \tan x \right)}^{\cos x}} \right]=\cos x\log \left( \tan x \right).
We differentiate to find g′(x){{g}^{'}}\left( x \right).
log⁡[g(x)]=cos⁡xlog⁡(tan⁡x) ⇒g′(x)g(x)=cos⁡x.ddx[log⁡(tan⁡x)]+ddx(cos⁡x).log⁡(tan⁡x) ⇒g′(x)g(x)=cos⁡x.(sec⁡2x)tan⁡x+(−sin⁡x).log⁡(tan⁡x) ⇒g′(x)=g(x)[cos⁡ecx−sin⁡x.log⁡(tan⁡x)] \begin{aligned} & \log \left[ g\left( x \right) \right]=\cos x\log \left( \tan x \right) \\\ & \Rightarrow \dfrac{{{g}^{'}}\left( x \right)}{g\left( x \right)}=\cos x.\dfrac{d}{dx}\left[ \log \left( \tan x \right) \right]+\dfrac{d}{dx}\left( \cos x \right).\log \left( \tan x \right) \\\ & \Rightarrow \dfrac{{{g}^{'}}\left( x \right)}{g\left( x \right)}=\cos x.\dfrac{\left( {{\sec }^{2}}x \right)}{\tan x}+\left( -\sin x \right).\log \left( \tan x \right) \\\ & \Rightarrow {{g}^{'}}\left( x \right)=g\left( x \right)\left[ \cos ecx- \sin x.\log \left( \tan x \right) \right] \\\ \end{aligned}
Now we put the differentiation values in dydx=f′(x)+g′(x)\dfrac{dy}{dx}={{f}^{'}}\left( x \right)+{{g}^{'}}\left( x \right).
dydx=f′(x)+g′(x) ⇒dydx=f(x)[−sec⁡x+cos⁡x.log⁡(cot⁡x)]+g(x)[cos⁡ecx−sin⁡x.log⁡(tan⁡x)] \begin{aligned} & \dfrac{dy}{dx}={{f}^{'}}\left( x \right)+{{g}^{'}}\left( x \right) \\\ & \Rightarrow \dfrac{dy}{dx}=f\left( x \right)\left[ -\sec x+\cos x.\log \left( \cot x \right) \right]+g\left( x \right)\left[ \cos ecx- \sin x.\log \left( \tan x \right) \right] \\\ \end{aligned}
Now we put the values of the functions in the equation
dydx=(cot⁡x)sin⁡x[−sec⁡x+cos⁡x.log⁡(cot⁡x)]+(tan⁡x)cos⁡x[cos⁡ecx−sin⁡x.log⁡(tan⁡x)]\dfrac{dy}{dx}={{\left( \cot x \right)}^{\sin x}}\left[ -\sec x+\cos x.\log \left( \cot x \right) \right]+{{\left( \tan x \right)}^{\cos x}}\left[ \cos ecx- \sin x.\log \left( \tan x \right) \right]
To match it with the given options we solve the equation and get

& \dfrac{dy}{dx}={{\left( \cot x \right)}^{\sin x}}\left[ -\sec x+\cos x.\log \left( \cot x \right) \right]+{{\left( \tan x \right)}^{\cos x}}\left[ \cos ecx-\sin x.\log \left( \tan x \right) \right] \\\ & \Rightarrow \dfrac{dy}{dx}=-\sec x{{\left( \cot x \right)}^{\sin x}}+{{\left( \cot x \right)}^{\sin x}}\cos x.\log \left( \cot x \right) \\\ & +\cos ecx{{\left( \tan x \right)}^{\cos x}}-{{\left( \tan x \right)}^{\cos x}}\sin x.\log \left( \tan x \right) \\\ \end{aligned}$$ We used the identity $\cot x.\sec x=\dfrac{\cos x.\sec x}{\sin x}=\dfrac{1}{\sin x}=\cos ecx$ and $\cos ecx. \tan x=\dfrac{\cos ecx.\sin x}{\cos x}=\dfrac{1}{\cos x}=secx$. We also have $\cos x.\sec x=\cos ecx. \sin x=1$. $$\begin{aligned} & \dfrac{dy}{dx}=-\sec x{{\left( \cot x \right)}^{\sin x}}+{{\left( \cot x \right)}^{\sin x}}\cos x.\log \left( \cot x \right) \\\ & +\cos ecx{{\left( \tan x \right)}^{\cos x}}-{{\left( \tan x \right)}^{\cos x}}\sin x.\log \left( \tan x \right) \\\ & \Rightarrow \dfrac{dy}{dx}=-\cos ecx{{\left( \cot x \right)}^{\sin x-1}}+{{\left( \cot x \right)}^{\sin x}}\left( \log \cot x \right)\cos x \\\ & +secx{{\left( \tan x \right)}^{\cos x-1}}+{{\left( \tan x \right)}^{\cos x}}\left( \log \tan x \right)\left( -\sin x \right) \\\ & \Rightarrow \dfrac{dy}{dx}=\sin x{{\left( \cot x \right)}^{\sin x-1}}\left( -\cos e{{c}^{2}}x \right)+{{\left( \cot x \right)}^{\sin x}}\left( \log \cot x \right)\cos x \\\ & +\cos x{{\left( \tan x \right)}^{\cos x-1}}se{{c}^{2}}x+{{\left( \tan x \right)}^{\cos x}}\left( \log \tan x \right)\left( -\sin x \right) \\\ \end{aligned}$$ **The correct option is A.** **Note:** We perform the logarithm or we can remember the following differentiation rule of $$y={{\left[ f\left( x \right) \right]}^{g\left( x \right)}}$$ as $$\dfrac{dy}{dx}={{\left[ f\left( x \right) \right]}^{g\left( x \right)}}\left[ {{g}^{'}}\left( x \right)\log f\left( x \right)+\dfrac{g\left( x \right){{f}^{'}}\left( x \right)}{f\left( x \right)} \right]$$. The Chain Rule tells us how to differentiate composite functions and since so many functions can be written as composites, it is a vitally important tool for computing derivatives rather than remembering difficult derivations.