How do you differentiate f(x)=1/x^3-x^4+sinx using the sum rule?

1 Answer
Apr 7, 2016

f'(x)=-3/(x^4)-4x^3+cosx

Explanation:

f(x)=1/(x^3)-x^4+sinx

=>f(x)=x^-3-x^4+sinx

Differentiating both sides w.r.t 'x'

f'(x)=d/(dx)(x^-3-x^4+sinx)

f'(x)=d/(dx)(x^-3)-d/(dx)(x^4)+d/(dx)(sinx)

f'(x)=-3x^-4-4x^3+cosx

f'(x)=-3/(x^4)-4x^3+cosx