How do you differentiate #f(x)=4-x^2sinx#?

1 Answer
Jul 18, 2017

#d/(dx)[4-x^2sinx] = color(blue)(-x^2cosx - 2xsinx#

Explanation:

We're asked to find the derivative

#d/(dx) [4-x^2sinx]#

Let's differentiate the equation term by term:

#= d/(dx)[4] - d/(dx)[x^2sinx]#

The derivative of #4# (a constant) is #0#:

= #-d/(dx)[x^2sinx]#

We can use the product rule, which states

#d/(dx)[uv] = v(du)/(dx) + u(dv)/(dx)#

where

  • #u = x^2#

  • #v = sinx#:

#= -(x^2d/(dx)[sinx] + sinxd/(dx)[x^2])#

The derivative of #sinx# is #cosx#:

#= -(x^2cosx + sinxd/(dx)[x^2])#

Use the power rule, which states

#d/(dx)[x^n] = nx^(n-1)#

where

#n = 2#:

#= -(x^2(cosx) + sinx(2x))#

or

#= color(blue)(-x^2cosx - 2xsinx#