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 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: 88 Thanks: 30 Use the chain rule with Fundamental Theorem of calculus $$$\begin{split}g'(x)&=f(h(x))\,h'(x )-f(\phi(x))\,\phi'(x)\\&=f(x^2)×\dfrac {\mathrm d}{\mathrm dx}{x^2}-f(5x+1)×\dfrac {\mathrm d}{\mathrm dx}(5x+1)\\&=\dfrac {\sin\,x^2}{x^2}×(2x)-\dfrac {\sin(5x+1)}{(5x+1)}×5\\&=\dfrac {2x\,\sin\,x^2}{x^2}-\dfrac {-5\,\sin(5x+1)}{(5x+1)}\end{split}$$ \tag*{}$

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