How do you differentiate $g (x) = x^{2} cos x$ $g(x) =x^2cosx$ using the product rule?

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Answer 1 · 2016-03-16T12:49:26

$\frac{d g (x)}{d x} = - x^{2} sin x + 2 x cos x$ $(dg(x))/(dx)=-x^2sinx+2xcosx$

$g (x) = x^{2} cos x$ $g(x)=x^2cosx$ is product of two functions, say $f_{1} (x) = x^{2}$ $f_1(x)=x^2$ and $f_{2} (x) = cos x$ $f_2(x)=cosx$ .

Product rule states that if $g (x) = f_{1} (x) f_{2} (x)$ $g(x)=f_1(x)f_2(x)$ 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, $(dg(x))/(dx)=x^2*(d(cosx))/dx+(d(x^2))/(dx)*cosx$

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

How do you differentiate g(x) =x^2cosxg(x)=x2cosxg(x) =x^2cosx using the product rule?