# Thread: Help with a question involving chain rule + product rule (derivative)

1. ## Help with a question involving chain rule + product rule (derivative)

I'm having trouble finding the derivative of this function: x(3x - 9)^3

The answer is 27(x-3)^2 (4x - 3) but I don't know how to get there. It would be very much appreciated if anyone can provide a step-by-step solution to how they got to the answer. It doesn't have to be too detailed.

2. Originally Posted by Archduke01
I'm having trouble finding the derivative of this function: x(3x - 9)^3

The answer is 27(x-3)^2 (4x - 3) but I don't know how to get there. It would be very much appreciated if anyone can provide a step-by-step solution to how they got to the answer. It doesn't have to be too detailed.

Set f(x) = x and g(x) = (3x - 9)^3. First calculate f'(x) and g'(x) individually using the chain rule. Then use the product rule to get d/dx f(x)g(x) = f'(x)g(x) + f(x)g'(x), substituting in the values you found for all of these functions.

3. Originally Posted by Archduke01
I'm having trouble finding the derivative of this function: x(3x - 9)^3

The answer is 27(x-3)^2 (4x - 3) but I don't know how to get there. It would be very much appreciated if anyone can provide a step-by-step solution to how they got to the answer. It doesn't have to be too detailed.

$(x(3x-9)^3)' =1*(3x-9)^3+x*9(3x-9)^2=(3x-9)^2(3x-9+9x)=$
$9(x-3)^2(12x-9)=27(x-3)^2(4x-3)$

Tonio

4. Thanks so much for clarifying, guys. I appreciate it.

5. Originally Posted by tonio
$(x(3x-9)^3)' =1*(3x-9)^3+x*9(3x-9)^2=(3x-9)^2(3x-9+9x)=$
$9(x-3)^2(12x-9)=27(x-3)^2(4x-3)$

Tonio
I don't understand that second line. Why is it $1*(3x-9)^3+x*9(3x-9)^2$?

6. Originally Posted by Archduke01
I don't understand that second line. Why is it $1*(3x-9)^3+x*9(3x-9)^2$?

Derivative of the first times the second plus the first times the derivative of the 2nd...the formula for derivative of a product! Of course, the second factor is composed, so you've to apply the chain rule ot it.

Tonio