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

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Answer 1 · 2016-04-07T05:54:41

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

#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#