How do you find the derivative of #q(r)=r^3cosr#?

1 Answer
Jim G.
Jan 15, 2017

#q'(r)=3r^2cosr-r^3sinr#

Explanation:

differentiate using the #color(blue)"product rule"#

#"Given " q(r)=f(r).g(r)" then"#

#color(red)(bar(ul(|color(white)(2/2)color(black)(q'(r)=f(r)g'(r)+g(r)f'(r))color(white)(2/2)|)))larr" product rule"#

#"here " f(r)=r^3rArrf'(r)=3r^2#

#"and " g(r)=cosrrArrg'(r)=-sinr#

#rArrq'(r)=r^3(-sinr)+cosr.3r^2#

#=3r^2cosr-r^3sinr#

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