Question

Calculus 1 help! Derivatives!
y=(1/6)(8x+11)^3 + (1-(1/x^3))^-2
Answer choices:
A) y'=(1/2)(8x)²-(3/x^4)^-2
B) y'= (1/2)(8x+11)^² - (1-(1/x^3))^-2
C) y'= (4/3)(8x+11)² + (3/x^4)(1-(1/x^3))^-2
D) y'= (4(8x+11)² - (3/x^4)(1-(1/x^3))^-2

We just earned these today and now my teacher is throwing this at us for homework! Can someone show me the steps? I cannot figure out where to start and although I have tried, I am not getting the answers! Please help! I will give a medal if someone shows the steps as it will help me with the rest of my homework!

Answer 1

I would love to help. The most important thing to remember when finding derivatives is the power rule:

\[f(x) = x ^{n}\]

Then\[f'(x) = nx ^{n-1}\]

Try using this simple rule, and let me know if you have any other questions.

Answer 2

I forgot to mention: you can count each term in the statement as a different function, and thus the derivative of the whole statement is the sum of derivatives of each term.

Answer 3

How about d/dx notation on the right side of each answer choice ?

Answer 4

\[\Large\rm y=\frac{1}{6}(8x+11)^3 + \left(1-\frac{1}{x^3}\right)^{-2}\]Applying our power rules gives us:\[\large\rm y'=\frac{1}{6}\cdot3(8x+11)^2\color{royalblue}{\left(8x+11\right)'} -2\left(1-\frac{1}{x^3}\right)^{-3}\color{royalblue}{\left(1-\frac{1}{x^3}\right)'}\]Understand the power rule portion of it?
The blue parts are popping up because we have to apply the chain rule.
We have to differentiate the blue parts still.

Answer 5

ok I see a problem! I put y=(1/6)(8x+11)^3 + (1-(1/x^3))^-2 as the problem. and that "-2" exponent is supposed to be a -1. so the problem is; y=(1/6)(8x+11)^3 + (1-(1/x^3))^-1

But still that is not what is throwing me off. I keep applying different rules and I never just seem to do the right ones or I am not doing it correctly

Answer 6

The chain rule can be tricky, you just need to remember that when you have a something other than just the independent variable as the input as a different functions... i.e.

\[f(x) = 8x + 11\]
And\[g(x) = 1/6(f(x))^{3}\]

you must take the derivative of the outside function multiplied by the derivative of the inner function. And you continue this process until you've reached the inner most function. In this case, there are two layers of functions as I have displayed.

So, find the derivative of g(x) and multiply by the derivative of f(x). Do this for every term in the equation.

Answer 7

Ugh sorry I can't seem to stay connected to the site today :( So weird....\[\large\rm y'=\frac{1}{6}\cdot3(8x+11)^2\color{royalblue}{\left(8x+11\right)'} -1\left(1-\frac{1}{x^3}\right)^{-2}\color{royalblue}{\left(1-\frac{1}{x^3}\right)'}\]So applying the power rule, do you understand why the -1 turns into a -2?
When you subtract 1 from a negative, it actually gets "larger" in the negative direction.

Answer 8

Yes I understand! :)

Answer 9

So from here, I still do not know what to do. D: He did not show us many examples on these and we have a whole bunch of these questions due tomorrow D:

Answer 10

So then applying your chain rule gives you something like:\[\large\rm y'=\frac{1}{6}\cdot3(8x+11)^2\color{orangered}{\left(8\right)} -1\left(1-\frac{1}{x^3}\right)^{-2}\color{orangered}{\left(1+\frac{3}{x^4}\right)}\]ya?

If you're having trouble with that second blue part,
Rewrite your x term using negative exponent so you can apply the power rule easier:\[\Large\rm \frac{1}{x^3}=x^{-3}\]

Answer 11

Blah that 1 should be 0, woops...\[\large\rm y'=\frac{1}{6}\cdot3(8x+11)^2\color{orangered}{\left(8\right)} -1\left(1-\frac{1}{x^3}\right)^{-2}\color{orangered}{\left(0+\frac{3}{x^4}\right)}\]

Answer 12

\[\frac{ 1 }{ 2 }(8x+11)^{2}(8) - 1(1-\frac{ 1 }{ x ^{3} })^{-2}(0+\frac{ 3 }{ x ^{4} })\]

Answer 13

It turns into this ?? So far??

Answer 14

Yes, and that would take care of all of the differentiation.
It's just simplifying from there!

Answer 15

Ok, so for the (8) and (0+3 / x^4) I cannot see how to work with those. I know i seem stupid right now, it's been a year since I took precal or any math course at that D: lol I was doing well in my work up to here.

Answer 16

Just like zepdrix said, there is no calculus after this point. Just algebra. Multiply everything out and you should arrive at an answer that matches one of your options.

And as a note, you do not seem stupid. Calculus can be difficult at times, no doubt!

Answer 17

Remember that multiplication is `commutative`.
Meaning ~ you can multiply in any order.
That means we can drag the 8 to the front, and combine it with the 1/2.\[\large\rm y'=8\cdot\frac{1}{2}(8x+11)^2-\left(1-\frac{1}{x^3}\right)^{-2}\left(\frac{3}{x^4}\right)\]

Answer 18

Do the same with the other term:\[\large\rm y'=8\cdot\frac{1}{2}(8x+11)^2-\frac{3}{x^4}\left(1-\frac{1}{x^3}\right)^{-2}\]

Answer 19

ok ok! That helps alot lol My common sense has left my brain this evening uhh ok so... I have

Answer 20

\[4(8x+11)^{2} - \frac{ 3 }{ x ^{4} } (1-\frac{ 1 }{ x ^{3} })^{-2}\]

Answer 21

Which matches choice D???? :O

Answer 22

Ya looks like it's D
:)

D is a missing a bracket somewhere.. must just be a typo though.

Answer 23

Yea lol Thats my fault D: lol Thank you so much! I truly appreciate your time and patience!