Question

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

Answer 1

`dy/dx = -2x^-3cosx - x^-2sinx + 12x^-4`

`= (-2cosx)/x^3 - sinx/x^2 + 12/x^4`

`= (12-2xcosx - x^2sinx)/x^4`

Explanation:

You should know a couple of rules for differentiation:

Power rule:

if `f(x) = ax^n`
then `f'(x) = n*ax^(n-1)`

Product rule:

if `f(x) = g(x)h(x)`
then `f'(x) = g'(x)h(x) + g(x)h'(x)`

And finally, summing derivatives:

`d/dx(a + b) = (da)/dx + (db)/dx`

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

`d/dx x^-2 = -2x^-3`

`d/dx cosx = -sinx`

so

`d/dx (x^-2cosx) = -2x^-3cosx-x^-2sinx`

and also

`d/dx (-4x^-3) = + 12x^-4`

so, in all, we have

`d/dx(x^-2cosx - 4x^-3) =`

`-2x^-3cosx-x^-2sinx+12x^-4`