Question

What is the derivative of `f(x) = x^5 * (x^2-3)^6`?

Answer 1

`5x^4(x^2-3)^6+12x^6(x^2-3)^5`

Explanation:

Here:

`d/dx x^5 * (x^2-3)^6`

we can use product rule:

`d/dx color(red)a * color(blue)b = (color(red)(a))'(color(blue)b) + (color(red)a)(color(blue)b)'`

So:

`d/dx color(red)(x^5) * color(blue)((x^2-3)^6)`

becomes:

`(color(red)x^5)'color(blue)((x^2-3)^6)+(color(red)x^5)(color(blue)((x^2-3)^6))'`

Simplifying:

`(5x^4)color(blue)((x^2-3)^6)+(color(red)x^5)(color(blue)((x^2-3)^6))'`

`d/dx color(blue)((x^2-3)^6)`

We can use chain rule here:

`d/dxf(x) = d/(du)f(u) * d/dx (x)`

`->d/dx(x^2-3)^6`

becomes:

`d/dx (u)^6 * d/dx (x^2-3)`

`=6u^5*2x`

Since `u=(x^2-3)`:

`=6(x^2-3)^5*2x`

`d/dx color(blue)((x^2-3)^6)=12x(x^2-3)^5`

Simplifying our former equation:

`(5x^4)color(blue)((x^2-3)^6)+(color(red)x^5)(color(blue)((x^2-3)^6))'`

becomes:

`(5x^4)(x^2-3)^6+(x^5)(12x)(x^2-3)^5`

Multiplying it out:

`=5x^4(x^2-3)^6+12x^6(x^2-3)^5`

And there we have our answer