How do you differentiate #(x^2) (sin x)#?

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Answer 1 · 2016-07-04T00:57:26

By using the product rule.

Let #f(x) = (x^2)(sinx)#, then #f(x) = g(x) xx h(x)#.

The derivative of this function is given by #f'(x) = (g'(x) xx h(x)) + (h'(x) xx g(x))#

The derivative of #g(x)# or #x^2# is #g'(x) = 2 xx x^(2 - 1) = 2x#

The derivative of #h(x)# or #sinx# is #h'(x) = cosx#.

Applying the product rule:

#f'(x) = (g'(x) xx h(x)) + (h'(x) xx g(x))#

#f'(x) = (2x(sinx)) + (x^2(cosx))#

#f'(x) = 2xsinx + x^2cosx#

Hence, the derivative of #y = (x^2)(sinx)# is #y' = 2xsinx + x^2cosx#.

Hopefully this helps!