How to find $g'(x)$ when $g(x)=\int_{5x+1}^{x^2} \frac{\sin t}{t}dt$

Question

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.

Answer 1

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, $$ \begin{aligned} g'(x)&=\frac{d}{dx}F(x^2)-\frac{d}{dx}F(5x+1)\\ &=F'(x^2)\cdot2x-F'(5x+1)\cdot 5\\ &=\frac{\sin(x^2)}{x^2}\cdot 2x-\frac{\sin(5x+1)}{5x+1}\cdot 5. \end{aligned} $$

Answer 2

Use the chain rule with Fundamental Theorem of calculus

$$\begin{equation}\begin{split}g'(x)&=f(h(x))\,h'(x)-f(\phi(x))\,\phi'(x)\\&=f(x^2)×\dfrac {\mathrm d}{\mathrm dx}{x^2}-f(5x+1)×\dfrac {\mathrm d}{\mathrm dx}(5x+1)\\&=\dfrac {\sin\,x^2}{x^2}×(2x)-\dfrac {\sin(5x+1)}{(5x+1)}×5\\&=\dfrac {2x\,\sin\,x^2}{x^2}-\dfrac {-5\,\sin(5x+1)}{(5x+1)}\end{split}\end{equation}$$