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If y = (\cos x)^{ x }+\tan^{-1}\sqrt{ x}, find \frac{dy}{dx} .

 

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y = (\cos x)^{ x }+\tan^{-1}\sqrt{ x},

$Let u =$ (\cos x)^x , v = \tan^{-1}\sqrt{x}

u = (\cos x)^x
Now, take log on both the sides
\log u = x \log(\cos x)
Now, differentiate w.r.t. x
we get,
\\ \frac{1}{u}\frac{du}{dx} = \log (\cos x) - x.\sin x.\frac{1}{\cos x} \\\\ = \log (\cos x)- x.\tan x \ \ \ (\because \frac{\sin x}{\cos x}=\tan x)\\ \frac{du}{dx}= u.(\log (\cos x)-x.\tan x)\\\\ \frac{du}{dx} =(\cos x)^x(\log (\cos x)-x\tan x)
Similarly, take v = \tan^{-1}\sqrt x
Now, differentiate w.r.t. x
We get,
\\ \frac{dv}{dx} = \frac{1}{1+(\sqrt {x})^2}.\frac{1}{2\sqrt x}= \frac{1}{2\sqrt{x}(1+x)}
Now,
\frac{dy}{dx} = \frac{du}{dx}+\frac{dv}{dx}
\frac{dy}{dx} =(\cos x)^x(\log (\cos x)-x\tan x)+\frac{1}{2\sqrt{x}(1+x)}

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