Question

Differentiate by rule.
y = (3x^4 - 5)^5

Answer 1

Power rule into chain rule:\[\Large\rm y=(stuff)^n\]\[\Large\rm y'=n(stuff)^{n-1}(stuff)'\]k? :3

Answer 2

\[\Large\rm y=(3x^4-5)^5\]Hmmm... so we have this...

Answer 3

\(\Large\rm 3x^4-5\) is our stuff, yah?
Our inner function.

Answer 4

Understand the first step, or no?

Answer 5

@zepdrix should call it blob lol

Answer 6

Yes.

Answer 7

Xd

Answer 8

\[\Large\rm y=(3x^4-5)^5\]So you power rule and setup your chain:\[\Large\rm y=5(3x^4-5)^4\cdot(3x^4-5)'\]We've only `setup` our chain rule at this point.
When you get more comfortable with these, you can skip this step.
So the prime is telling us that we still need to take the derivative of that part.

Answer 9

So how bout it, what do you think senor oz?
What's the derivative of that second part?

Answer 10

Hmmm .... I am not sure.

Answer 11

Just take the derivative of each term separately.
\[\Large\rm (3x^4-5)'=(3x^4)'-(5)'\]5 is a constant. Derivative of a constant?

Answer 12

Okay

Answer 13

It is 0

Answer 14

\[\Large\rm (3x^4-5)'=(3x^4)'-(5)'\]\[\Large\rm (3x^4-5)'=(3x^4)'-0\]Mmmm k good.
Just apply your power rule to the first term.
What do you get?

Answer 15

12x^3

Answer 16

Good good good.
So we have:\[\Large\rm y=(3x^4-5)^5\]\[\Large\rm y'=5(3x^4-5)^4\cdot(3x^4-5)'\]\[\Large\rm y'=5(3x^4-5)^4\cdot(12x^3-0)\]\[\Large\rm y'=5(3x^4-5)^4\cdot(12x^3)\]

Answer 17

Realize that multiplication is `commutative` (you can multiply things in any order),
so bring the 12x^3 to the front and multiply it with the 5.
It will look a little nicer that way.