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

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Answer 1 · 2017-07-18T03:16:54

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

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

#= -(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#