Question

Given that, $F\left(x\right)=\frac{1}{x^2}{\int }_4^x(4t^2-2F^{\prime }(t\left)\right)dt$, find F'(4)

Answer 1

$F\left(x\right)=\frac{1}{x^2}{\int }_4^x(4t^2-2F^{\prime }(t\left)\right)dt{F}^{\prime }\left(x\right)=\frac{-1}{x^3}{\int }_4^x(4t^2-2F^{\prime }(t\left)\right)dt-\frac{1}{x^2}(4x^2-2F^{\prime }(x).1-0)$

Now, $F^{\prime }\left(4\right)=\frac{-1}{4^3}{\int }_4^4(4t^2-2F^{\prime }(t\left)\right)dt-\frac{1}{4^2}({4.4}^2-2F^{\prime }(4).1-0)=0-4+2F^{\prime }\left(4\right)\Rightarrow F^{\prime }\left(4\right)=-4+2F^{\prime }\left(4\right)\Rightarrow F^{\prime }\left(4\right)=4$