How to find $g'(x)$ when $g(x)=\int_{5x+1}^{x^2} \frac{\sin t}{t}dt$ How do I find $g'(x)$ when $$g(x)=\displaystyle\int\limits_{5x+1}^{x^2}\,\dfrac {\sin\,t}{t}\;dt$$
My attempt$$f(t)=\dfrac{\sin t}{t}$$
applying FTC$$g'(x)=\dfrac{\sin(x^2)^{x^2}}{x^2}-\dfrac{\sin 5x+1}{5x+1}$$
But I got the incorrect answer.

Here is a pretty failsafe method:
Let $F$ be a primitive of your integrand $t\mapsto\sin(t)/t$. Then
$$
g(x)=\int_{5x+1}^{x^2}\frac{\sin t}{t}\,dt=F(x^2)-F(5x+1).
$$
Thus, by the chain rule,
$$
\begin{aligned}
g'(x)&=\frac{d}{dx}F(x^2)-\frac{d}{dx}F(5x+1)\\
&=F'(x^2)\cdot2x-F'(5x+1)\cdot 5\\
&=\frac{\sin(x^2)}{x^2}\cdot 2x-\frac{\sin(5x+1)}{5x+1}\cdot 5.
\end{aligned}
$$
