f(x)= int[1,x^3] t^3dt f'(x)=
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$\displaystyle \int_{4}^{x^{3}} t^3 dt $ differentiate x using FTC $\displaystyle f '(x) = 3x^{11} $
Originally Posted by thedoge f(x)= int[1,x^3] t^3dt f'(x)= If $\displaystyle u=x^3$, then by the FTC: $\displaystyle \frac{df}{du}=u^3$, and: $\displaystyle \frac{df}{dx}=\frac{df}{du}\frac{du}{dx}=u^3 (3 x^2)=3 x^{11}$ RonL
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