Answer
\[f'(x)=2x\int_{0}^{x}\sin (t^2)\:dt+x^2\sin(x^2)\]
Work Step by Step
It is given that \[f(x)=\int_{0}^{x}x^2\sin (t^2)\:dt\]
\[f(x)=x^2\int_{0}^{x}\sin (t^2)\:dt\;\;\;...(1)\]
Differentiate (1) with respect to $x$
\[\Rightarrow f'(x)=(x^2)'\int_{0}^{x}\sin (t^2)\:dt+x^2\left(\frac{d}{dx}\int_{0}^{x}\sin (t^2)\:dt\right)\;\;\;...(2)\]
We will use the formula \[\frac{d}{dx}\int_{f(x)}^{g(x)}F(t)\;dt=F(g(x))\cdot g'(x)-F(f(x))\cdot f'(x)\;\;\;...(3)\]
Using (3) in (2)
\[f'(x)=2x\int_{0}^{x}\sin (t^2)\:dt+x^2\left(\sin(x^2)\cdot (x)'\right)\]
\[f'(x)=2x\int_{0}^{x}\sin (t^2)\:dt+x^2\sin(x^2)\]
Hence , \[f'(x)=2x\int_{0}^{x}\sin (t^2)\:dt+x^2\sin(x^2)\]
