Answer
\[g'(x)=4x^3\cos (x^8)\]
Work Step by Step
We will use the rule \[\frac{d}{dx}\int_{f(x)}^{g(x)}F(t)dt=F(g(x))\cdot g'(x)-F(f(x))\cdot f'(x)\;\;\; ...(1)\]
We have given that \[g(x)=\int_{0}^{x^4}\cos (t^2)\;dt\;\;\;...(2)\]
Differentiating (2) with respect to $x$ using (1)
\[g'(x)=\cos ((x^4)^2)\cdot (x^4)'-\cos (0^2)\cdot
(0)'\]
\[\Rightarrow g'(x)=\cos (x^8)\cdot (4x^3)-\cos (0)\cdot (0)\]
\[\Rightarrow g'(x)=4x^3\cos (x^8)\]
Hence \[g'(x)=4x^3\cos (x^8).\]
