Science

Math

Humanities

... and beyond

Socratic Meta
Featured Answers

Topics

How do you differentiate #g(x) =x^2cosx# using the product rule?

1 Answer

Shwetank Mauria

Mar 16, 2016

#(dg(x))/(dx)=-x^2sinx+2xcosx#

Explanation:

#g(x)=x^2cosx# is product of two functions, say #f_1(x)=x^2# and #f_2(x)=cosx#.

Product rule states that if #g(x)=f_1(x)f_2(x)# then #g'(x)=f_1'(x)f_2(x)+f_1(x)f_2'(x)#.

Here #p'(x)# denotes differential of #p(x)# i.e.

#f_1'(x)# is differential of #f_1(x)# and #f_2'(x)# is differential of #f_2(x)#.

Hence, #(dg(x))/(dx)=x^2*(d(cosx))/dx+(d(x^2))/(dx)*cosx#

= #x^2*(-sinx)+2x*cosx=-x^2sinx+2xcosx#

Related questions

See all questions in Product Rule

Impact of this question

1868 views around the world

You can reuse this answer
Creative Commons License