Answer
\[g'(x)=\frac{\cos^3 x}{1+\sin^4 x}\]
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 \[g(x)=\int_{1}^{\sin x}\frac{1-t^2}{1+t^4}\;dt\;\;\;...(2)\]
Differentiate (2) with respect to $x$ using (1)
\[g'(x)=\left(\frac{1-\sin^2 x}{1+\sin^4 x}\right)\cdot (\sin x)'-\left(\frac{1-1^2}{1+1^4}\right)\cdot (1)'\]
\[\Rightarrow g'(x)=\left(\frac{1-\sin^2 x}{1+\sin^4 x}\right)\cdot (\cos x)-\left(\frac{0}{2}\right)\cdot (0)\]
\[\Rightarrow g'(x)=\left(\frac{\cos^2 x}{1+\sin^4 x}\right)\cdot (\cos x)\]
\[\Rightarrow g'(x)=\frac{\cos^3 x}{1+\sin^4 x}\]
Hence, \[g'(x)=\frac{\cos^3 x}{1+\sin^4 x}\]
