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How do you find the derivative of #4-(x^2)sinx#?

1 Answer

May 13, 2018

#-2xsin(x)-x^2cos(x)#

Explanation:

By using the product rule
#(f(x)*g(x))'=f(x)'*g(x)+f(x)*g(x)'#

#k(x)=4-x^2sin(x)#
#k'(x)=0-((x^2)' * sin(x)+x^2*(sin(x))')#
#k'(x)=-2xsin(x)-x^2cos(x)#

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