Question

A couple of Calculus questions. Can anyone help?

Answer 1

First find the first derivative. Use the chain rule (basically use the definition of the chain rule)

Answer 2

This is the spot where you say things and attempt things.

Answer 3

She's not a lazy one, maybe she's just thinking :3

Answer 4

Thank you @math&ing001, I like to take my time on things and make sure I understand. Math doesn't come super easy to me... Sorry I can't meet your expectations @agent0smith

Answer 5

My expectations are that people asking questions make an effort, show work etc.

Answer 6

I understand that... You may move onto the next person. @agent0smith

Answer 7

Prove me wrong. Show the definition of the chain rule (easy to find on google). That's really all you need to get the first derivative.

Answer 8

No thank you. I appreciate your help but now I actually feel pressured. I'll figure it out.

Answer 9

okay... Pressured to google "the chain rule" and post a pic that shows it

Answer 10

@zepdrix Hey Zep, do you think you could help?

Answer 11

@math&ing001 Do you think you could give me pointers?

Answer 12

Chain rule says, if \(\large\rm y=f(g(x))\) then \(\large\rm y'=f'(g(x))\cdot g'(x)\).

Really important rule, hard to understand at first.
Try to practice it a lot.

Anyway, if we have this \(\large\rm y=2f(g(x))\)
What do we get for \(\large\rm y'\) ?

Hint: Ignore the 2.

Answer 13

f'(g(x))? or am I looking for a number?

Answer 14

Hmm, you didn't follow the rule I posted.
You should have ended up with something almost identical to that rule. :p

Answer 15

y′=f′(g(x))⋅g′(x) This?

Answer 16

That's the rule for differentiating f(g(x)).
If we start with a 2 in front, then we should still end up with a 2, ya?
\(\large\rm y=2f(g(x))\)
\(\large\rm y'=2f'(g(x))\cdot g'(x)\)

Answer 17

Oh I thought you meant ignore the 2 as in take it out oops

Answer 18

Nah, he doesn't contribute to the process at all,
but he still comes along for the ride.

Answer 19

ahh I'm on board now

Answer 20

The second derivative is going to be a lot more complicated.
We'll start by `setting up` our product rule.

Answer 21

I'm going to use a little color, and Leibniz notation just to try and keep things clear.

Answer 22

\[\large\rm y'=2f'(g(x))\cdot g'(x)\]Then,\[\large\rm y''=\color{royalblue}{\frac{d}{dx}2f'(g(x))}g'(x)+2f'(g(x))\color{royalblue}{\frac{d}{dx}g'(x)}\]The blue parts are the derivatives that we need to take.
All I've done so far is "set up" the product rule.

Answer 23

Ok, got it

Answer 24

First one is a little tricky.
Maybe start with the other one.
derivative of g'(x) ?

Answer 25

g"(x)? lol

Answer 26

\[\large\rm y''=\color{royalblue}{\frac{d}{dx}2f'(g(x))}g'(x)+2f'(g(x))\color{orangered}{g''(x)}\]k good.

Answer 27

How bout the other blue guy, any ideas? :O

Answer 28

Hmm 2f''(g(x))

Answer 29

Close, again, this is an application of our chain rule.
Chain rule tells us that we not only take the derivative of the outer function (as you did), but that we also must multiply by a copy of the inner function, and take its derivative.
2f''(g(x))*g'(x)

Answer 30

\[\large\rm y''=\color{orangered}{2f''(g(x))g'(x)}g'(x)+2f'(g(x))\color{orangered}{g''(x)}\]So that takes care of the other blue guy, good.

Answer 31

Now simplify, and it should match one of your options.

Answer 32

So d? :)

Answer 33

Yay good job

Answer 34

You're the best :)

Answer 35

Seriously though,
chain rule is the most important concept of calculus (in my opinion),
do a thousand problems involving chain rule,

otherwise the rest of calculus wont make any sense to you.

Answer 36

You're absolutely right haha. I'm glad calc is coming close to an end