f' (x)= f' (1)= can anyone solve this?
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Originally Posted by dat1611 f' (x)= f' (1)= can anyone solve this? $\displaystyle f\left( x \right) = \int\limits_x^{{x^2}} {{t^2}dt} = \left. {\frac{{{t^3}}}{3}} \right|_x^{{x^2}} = \frac{{{x^6}}}{3} - \frac{{{x^3}}}{3}.$ $\displaystyle f'\left( x \right) = 2{x^5} - {x^2}.$ $\displaystyle f'\left( 1 \right) = 2 \cdot 1 - 1 = 1.$
$\displaystyle f(x)=\int_{x}^{x^{2}}{t^{2}\,dt}\implies f'(x)=x^{4}\cdot 2x-x^{2}=2x^{5}-x^{2}.$
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