Answer
\[f'(1)=0\]
Work Step by Step
It is given that :- \[f(x)=\cos (\ln x^2)\;\;\;...(1)\]
Differentiate (1) with respect to $x$ using chain rule
\[f'(x)=-\sin (\ln x^2)\cdot (\ln x^2)'\]
\[f'(x)=-\sin (\ln x^2)\cdot\left(\frac{1}{x^2}\right)\cdot (x^2)'\]
\[\Rightarrow f'(x)=-\sin (\ln x^2)\cdot\left(\frac{1}{x^2}\right)\cdot (2x)\]
\[\Rightarrow f'(x)=\frac{-2}{x}\sin (\ln x^2)\]
\[\Rightarrow f'(1)=\frac{-2}{1}\sin (\ln 1^2)\]
\[\Rightarrow f'(1)=\frac{-2}{1}\sin (0)\]
\[\Rightarrow f'(1)=0\]
Hence ,\[f'(1)=0\]
