Calculus 1 Differentiation Problem: Chain Rule with Binomial Theorem Application

[Husaaved]

I'm not entirely sure if this belongs in homework or elsewhere -- I'm self-teaching working through a basic calculus text, so it's not homework per se. In any case it's a simple differentiation problem wherein I am supposed to differentiate:

f(x) = x(3x-9)^3
f'(x) = 3x(3)(3x-9)^2 Applying chain rule
f'(x) = 9x(3x-9)^2

I know this isn't the correct answer.

I was half tempted to multiply out using the binomial theorem but I suspect there's a more efficient way to solve this. How am I to treat the x coefficient? Evidently not as a constant.

 

[R136a1]

Did you see the product rule?

(fg)^\prime = f^\prime g + fg^\prime

 

[HallsofIvy]

You have to use the product rule first, then the chain rule.

Of, since x= (x^{1/3})^3[/tex], f(x)= (x^{1/3}(3x- 9))^3= (3x^{4/3}- 9x^{1/3})^3<br /> and <b>now</b> use the chain rule.

 

[NascentOxygen]

You have to use the product rule first, then the chain rule.

Or, you could skip the need for the product rule by looking at it this way ...

Since x= (x^{1/3})^3,
then f(x)= (x^{1/3}(3x- 9))^3= (3x^{4/3}- 9x^{1/3})^3
and now use the chain rule.

There, itex fixed.

 

[HallsofIvy]

Thanks!