7. Differentiate the functions with respect to x in 

2 \sqrt { \cot ( x^2 )}

Answers (1)
G Gautam harsolia

Give function is
f(x)=2 \sqrt { \cot ( x^2 )}
Let's take t = x^2
f(t) = 2\sqrt\cot t
Now,  take \cot t = k^2
f(k) = 2k
Differentiation w.r.t. x
\frac{d(f(k))}{dx} = \frac{d(f(k))}{dk}.\frac{dk}{dt}.\frac{dt}{dx}                       -(By chain rule)
\frac{d(f(k))}{dk} = \frac{d(2k)}{dk} = 2
\frac{dk}{dt} = \frac{d(\sqrt{\cot t})}{dt} = \frac{1}{2\sqrt{cot t}}.(-cosec^2 t) = \frac{-cosec^2 x^2}{2\sqrt{cot x^2}} \ \ \ (\because t = x^2)
\frac{dt}{dx} = \frac{d(x^2)}{dx} = 2x
So,
\frac{d(f(k))}{dx} = 2.\frac{-cosec^2 x^2}{2\sqrt{cot x^2}}.2x = \frac{-2\sqrt2x}{\sin^2x^2\sqrt{\frac{2\sin x^2\cos x^2}{\sin^2x^2}} }   ( Multiply and divide by \sqrt 2 and multiply and divide \sqrt {\cot x^2} by  \sqrt{\sin x^2)
                                                                                               (\because \cot x = \frac{\cos x}{\sin x} \ and \ cosec x = \frac{1}{\sin x } )
                                                           =\frac{-2\sqrt2x}{\sin x^2\sqrt{\sin2x^2}} \ \ \ \ (\because 2\sin x\cos x=\sin2x)
There, the answer is   \frac{-2\sqrt2x}{\sin x^2\sqrt{\sin2x^2}}

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