Question

$y=(\cot x{)}^{\sin x}+(\tan x{)}^{\cos x}$.Find dy/dx

• A. $\sin x\left(\cot x{)}^{\sin x-1}\right(-cose{c}^{2}x)+(\cot x{)}^{\sin x}\left(log\cot x\right)\cos x+\cos x{\left(\tan x\right)}^{\cos x-1}{\sec }^{2}x+{\left(\tan x\right)}^{cosx}(\log tanx\left)\right(-sinx)$
• B. $\sin x\left(\cot x{)}^{\sin x-1}\right(-co{\sec }^{2}x)+\cos x{\left(\tan x\right)}^{\cos x-1}{\sec }^{2}x$
• C. $\sin x\left(\cot x{)}^{\sin x-1}\right(-se{c}^{2}x)+(\cot x{)}^{\sin x}\left(log\cot x\right)\cos x+\cos x{\left(\tan x\right)}^{\cos x+1}co{\sec }^{2}x+{\left(\tan x\right)}^{cosx}(\log tanx\left)\right(sinx)$
• D. None of these

Answer 1

The correct option is A $\sin x\left(\cot x{)}^{\sin x-1}\right(-cose{c}^{2}x)+(\cot x{)}^{\sin x}\left(log\cot x\right)\cos x+\cos x{\left(\tan x\right)}^{\cos x-1}{\sec }^{2}x+{\left(\tan x\right)}^{cosx}(\log tanx\left)\right(-sinx)$

Given $y=(\cot x{)}^{\sin x}+(\tan x{)}^{\cos x}$

Let $u=(\cot x{)}^{\sin x}$

Taking log on both sides

$\log u=\sin x\log \left(\cot x\right)$

Differentiating w.r.t. x, we get,

$\frac{1}{u}\frac{du}{dx}=\frac{\sin x}{\cot x}(-cose{c}^{2}x)+\cos x\log \cot x$

$\Rightarrow \frac{du}{dx}=\sin x\left(\cot x{)}^{\sin x-1}\right(-cose{c}^{2}x)+(\cot x{)}^{\sin x}\cos x\log \cot x$
Now, $v=(\tan x{)}^{\cos x}$

Taking log on both sides

$\log v=\cos x\log \left(\tan x\right)$

Differentiating w.r.t. x, we get,

$\frac{1}{v}\frac{dv}{dx}=\frac{\cos x}{\tan x}\left({\sec }^{2}x\right)-\sin x\log \tan x$

$\Rightarrow \frac{dv}{dx}=\cos x\left(\tan x{)}^{\cos x-1}\right({\sec }^{2}x)+(\tan x{)}^{\cos x}(-\sin x)\log \tan x$
Since, $y=u+v$

$\Rightarrow \frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}$

$\Rightarrow \frac{dy}{dx}=\sin x\left(\cot x{)}^{\sin x-1}\right(-cose{c}^{2}x)+(\cot x{)}^{\sin x}\cos x\log \cot x+$

$\cos x\left(\tan x{)}^{\cos x-1}\right({\sec }^{2}x)+(\tan x{)}^{\cos x}(-\sin x)\log \tan x$