Answer
\[y'=3\sin \{(3x+1)^4\}-2\sin (16x^4)\]
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 \[y=\int_{2x}^{3x+1}\sin (t^4)\;dt\;\;\;...(2)\]
Differentiate (2) with respect to $x$ using (1)
\[y'=\sin \{(3x+1)^4\}\cdot (3x+1)'-\sin \{(2x)^4\}\cdot (2x)'\]
\[\Rightarrow y'=3\sin \{(3x+1)^4\}-2\sin (16x^4)\]
Hence ,\[y'=3\sin \{(3x+1)^4\}-2\sin (16x^4)\]
