March 19th, 2017, 11:38 PM  #1 
Newbie Joined: Mar 2017 From: US Posts: 1 Thanks: 0  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. $f(t)=\dfrac{\sin t}{t}$ $g'(x)=\dfrac{\sin(x^2)^{x^2}}{x^2}\dfrac{\sin 5x+1}{5x+1}$ 
March 19th, 2017, 11:55 PM  #2 
Member Joined: Sep 2016 From: India Posts: 80 Thanks: 24 
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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calculus, fundamental, theorem 
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