Let f(x)=(6sinx+5cosx)tan^−1x; f ' (x)=

Answer 1

f(x)= [6sin(x)+5cos(x)].tan^−1(x)

U = sin(x)+5cos(x) => U’ = cos(x) - 5sin(x)

V=tan^-1(x) => v’ = 1/(1+x^2)

Therefore f(x)= U.V  using product rule

f’(x)=U’.V+U.V’ = [cos(x) - 5sin(x)].[tan^-1(x)] + [sin(x)+5cos(x)]. [1/(1+x^2)]

= sin(x).{1/(1+x^2) - 5tan^-1(x)} + cos(x).{5/(1+x^2) + tan^-1(x)}