let-give-f-x-x-2x-dt-ln-1-t-2-calculate-f-x-

Question Number 27379 by abdo imad last updated on 05/Jan/18

l e t g i v e f (x) = \int_{x}^{2 x} \frac{d t}{l n (1 + t^{2})} c a l c u l a t e f^{^{'}} (x) .

Commented by abdo imad last updated on 05/Jan/18

i f f (x) = \int_{α (x)}^{β (x)} u (t) d t w i t h α a n d β a r e a r e f u n c t i o n o f x

f^{^{'}} (x) = β^{^{'} (x)} u (β (x)) - α^{^{'}} (x) u (α (x)) i n t h e r x e r c i s e

f^{^{'}} (x) = \frac{{(2 x)}^{^{'}}}{l n (1 + 4 x^{2})} - \frac{{(x)}^{^{'}}}{l n (1 + x^{2})}

= \frac{2}{l n (1 + 4 x^{2})} - \frac{1}{l n (1 + x^{2})} .

Answered by prakash jain last updated on 05/Jan/18

f^{'} (x) = \frac{1}{l n (1 + 4 x^{2})} \frac{d}{d x} 2 x - \frac{1}{\ln (1 + x^{2})} \frac{d}{d x} x + \int_{x}^{2 x} 0 d t

= \frac{2}{l n (1 + 4 x^{2})} - \frac{1}{\ln (1 + x^{2})}