How do find the derivative of #y = x^2 sinx#?

Question

10739 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-08-05T21:27:01

#dy/dx = 2xsin(x) + x^2cos(x)#

You can use the product rule to find the derivative. The product rule says the following:

If #h(x) = f(x)g(x)#, then #h'(x) = f'(x)g(x) + f(x)g'(x)#

In our case,

#h(x) = x^2sinx#

#f(x) = x^2#

#g(x) = sinx#

#f'(x) = 2x#

#g'(x) = cosx#

Plug in those values into our definition for the product rule to get

#dy/dx = h'(x) = 2xsin(x) + x^2cos(x)#