(x^3 + 8x^2 +6x) ^(-4)

Then.. multiply by the outside exponent and decrease it by one…

-4(x^3 + 8x^2 + 6x) ^(-5)

Then multiply by the derivative of the inside…

-4(x^3 + 8x^2 + 6x) ^(-5) * (3x^2 + 16x + 6)

Rewrite the problem…

___-4 (3x^2 + 16x +6)_____

……(x^3 + 8x^2 + 6x)^5

So what you can do is to bring the denominator up, just change the power 4 to a -4.

then you have (x^3 + 8x^2 +6x) ^ – 4

Then is you find the differentiation of that, all you have to do is bring the -4 down, then the new overall power becomes -4 -1 = -5 and then differentiation within the bracket

By differentiating within the bracket, you have x^3 to become 3x^2

Then the next 8x^2 becomes 16x^1

and lastly 6x becomes 6

All of this shcould be simple differentiations of each term within the brackets cos its addition so you can diff it term by term

Finally you have your ans

-4(x^3 + 8x^2 +6x)^-5(3x^2 + 16x +6)

_______1_________=

(x^3 + 8x^2 +6x) ^4

(x^3 + 8x^2 +6x) ^-4

g'(x) =

-4(x^3 + 8x^2 +6x)^-5 (3x^2 + 16x +6)