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

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

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