# How do you differentiate y=x^-2cosx-4x^-3?

Oct 24, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = - 2 {x}^{-} 3 \cos x - {x}^{-} 2 \sin x + 12 {x}^{-} 4$

$= \frac{- 2 \cos x}{x} ^ 3 - \sin \frac{x}{x} ^ 2 + \frac{12}{x} ^ 4$

$= \frac{12 - 2 x \cos x - {x}^{2} \sin x}{x} ^ 4$

#### Explanation:

You should know a couple of rules for differentiation:

if $f \left(x\right) = a {x}^{n}$
then $f ' \left(x\right) = n \cdot a {x}^{n - 1}$

if $f \left(x\right) = g \left(x\right) h \left(x\right)$
then $f ' \left(x\right) = g ' \left(x\right) h \left(x\right) + g \left(x\right) h ' \left(x\right)$

And finally, summing derivatives:

$\frac{d}{\mathrm{dx}} \left(a + b\right) = \frac{\mathrm{da}}{\mathrm{dx}} + \frac{\mathrm{db}}{\mathrm{dx}}$

Now we have the knowledge base to tackle the problem. We'll split it up.

$\frac{d}{\mathrm{dx}} {x}^{-} 2 = - 2 {x}^{-} 3$

$\frac{d}{\mathrm{dx}} \cos x = - \sin x$

so

$\frac{d}{\mathrm{dx}} \left({x}^{-} 2 \cos x\right) = - 2 {x}^{-} 3 \cos x - {x}^{-} 2 \sin x$

and also

$\frac{d}{\mathrm{dx}} \left(- 4 {x}^{-} 3\right) = + 12 {x}^{-} 4$

so, in all, we have

$\frac{d}{\mathrm{dx}} \left({x}^{-} 2 \cos x - 4 {x}^{-} 3\right) =$

$- 2 {x}^{-} 3 \cos x - {x}^{-} 2 \sin x + 12 {x}^{-} 4$