How do you differentiate #x^1.7cosx#?

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Answer 1 · 2018-07-30T20:32:23

#d/dx(x^1.7cos(x)) = 1.7x^0.7cos(x)-x^1.7sin(x)#

This problem is an example of the Product Rule where

#f(x) = x^1.7# and #g(x)=cos(x)#

Their respective derivatives are

#f'(x) = 1.7x^0.7# and #g'(x)=-sin(x)#

The Product Rule states that

#d/dx(f(x)*g(x))=f(x)*g'(x)+f'(x)*g(x)#

Plugging the derivatives and original functions in gives

#d/dx(f(x)g(x)) = x^1.7(-sin(x))+1.7x^0.7cos(x)#

Rewriting for simplification and ease of reading gives

#d/dx(f(x)g(x)) = 1.7x^0.7cos(x)-x^1.7sin(x)#