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Let f(x)=1/x^2 int4^x (4t^2-2f'(t))dt th...

Let `f(x)=1/x^2 int_4^x (4t^2-2f'(t))dt` then find the value of `9f'(4)`

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Verified by Experts

The correct Answer is:
32
`d/(dx)int_(4)^(x)[4t^(2)-2F'(t)]dt=[4x^(2)-2F'(x)].1-0`
or `F'(x)=1/(x^(2))[4x^(2)-2F'(x)]+(-2)/(x^(3))int_(4)^(x)[4t^(2)-2F'(t)]dt`
or `F'(4)=1/16[64-2F'(4)]-1/32int_(4)^(4)g(x)dx`
or `(1+1/8)F'(4)=4`
or `F'(4)=32/9`
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