Question

Let f(x)=arctan(squareroot(4x−1)). Find f′(x)

Answer 1

We have
\[f(x)=\tan ^{-1} (\sqrt {4x-1})\]
We know
\[\frac{d}{dx} \tan^{-1} x= \frac{1}{1+x^2}\]
and also we know
\[\frac{d}{dx} f(g(x)= f'(g(x)) \times g'(x)\]
so Let's begin
\[f'(x)= \frac{1}{1+ (\sqrt{4x-1})^2} \frac{d}{dx} (\sqrt{4x-1})\]
\[f'(x)= \frac{1}{1+ (\sqrt{4x-1})^2} \times \frac{1}{2} \frac{1}{\sqrt{4x-1}}\times \frac{d}{dx}(4x)\]
\[f'(x)= \frac{1}{1+ (\sqrt{4x-1})^2} \times \frac{1}{2} \frac{1}{\sqrt{4x-1}}\times 4\]
We get now
\[f'(x)= \frac{1}{1+ {4x-1}} \times \frac{1}{2} \frac{1}{\sqrt{4x-1}}\times 4\]
\[f'(x)= \frac{1}{{4x}} \times \frac{1}{2} \frac{1}{\sqrt{4x-1}}\times 4\]
\[f'(x)= \frac{1}{{\cancel 4x}} \times \frac{1}{2} \frac{1}{\sqrt{4x-1}}\times \cancel4\]
We get finally
\[f'(x)= \frac{1}{2x} \frac{1}{\sqrt{4x-1}}\]