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March 19th, 2017, 11:38 PM   #1
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Fundamental Theorem of calculus

What is the $g'(x)$ when $g(x)=\displaystyle \int \limits_{5x+1}^{x^2}\, \dfrac {\sin\,t}{t} \; dt $

I used FTC but I got incorrect answers.

$f(t)=\dfrac{\sin t}{t}$

$g'(x)=\dfrac{\sin(x^2)^{x^2}}{x^2}-\dfrac{\sin 5x+1}{5x+1}$
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March 19th, 2017, 11:55 PM   #2
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Use the chain rule with Fundamental Theorem of calculus

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