h'(x)=
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$\displaystyle H(x) = \int_c^{f(x)} {G(t)dt} \, \Rightarrow \,H'(x) = G\left( {f(x)} \right)f'(x)$
i cant get it could use some more help
Originally Posted by dat1611 i cant get it could use some more help $\displaystyle \frac{d}{dx}\left[\int_a^u f(t) \, dt \right] = f(u) \cdot \frac{du}{dx}$ substitute $\displaystyle u = \sin{x}$ in for $\displaystyle t$ in the integrand, then multiply that result by the derivative of $\displaystyle \sin{x}$.
Hi! Let $\displaystyle u = sin(x) $ $\displaystyle h(u)=\int_{-5}^{u} (cos(t^{3})+t) \; dt $ $\displaystyle \frac{dh}{dx}=\frac{dh}{du} \cdot \frac{du}{dx} $ $\displaystyle h'(x)=(cos(sin^{3}(x))+sin(x))\cdot cos(x) $
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