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

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Answer 1 · 2018-05-13T15:59:50

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

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

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

How do you find the derivative of 4-(x^2)sinx4-(x^2)sinx?