Discussion Forum

If   $\int f(x)\cos x dx= \frac{1}{2} (f(x))^{2}+C$  and f(0)=0 then f'(0)=

Answer:

Explanation:

We have,

$\int f(x) \cos x dx= \frac{1}{2} (f(x))^{2}+C$

On differentiating  w.r.t 'x' , we get

   $f(x) \cos x=\frac{1}{2} \times 2 f(x).f'(x)$

$\Rightarrow$     $f(x) \cos x= f(x).f'(x)$

$\Rightarrow$   $f'(x)=\cos x$

$\Rightarrow$   $f'(0)= \cos 0$

$\Rightarrow$  $f'(0)=1$