Question

Help with understanding the Chain Rule, anyone?

Answer 1

Find f'(x) of \[\cot^3\sqrt{x}\]

Answer 2

Trid to break it up into pieces, but it seems to only confuse me more.

Answer 3

The chain rule is essentially to find the derivative of an expression to a power.

Answer 4

So you can rewrite yours as:

\[(\cot \sqrt{x})^3\]

Answer 5

let \[\sqrt{x}\] = y
then we have \[\cot ^3 y\]

Answer 6

Yeah, I've done that much, @Zelda. Then tried breaking that up into functions. f(x) = x^3 ; g(x) = cot(x); h(x)= x^(1/2)

Answer 7

Rather, try this:

\[Say \ you \ have: \]\[f(x) = t^n(\theta)\]\[\rightarrow f'(x) = n(t(\theta))^{n-1} * t' (\theta)*\theta'\]

Answer 8

That's how the chain rule works, with t being a trig function.

Answer 9

So, it's like the derivative of the outside (bigger function) times teh derivative of the inside (smaller) function?

Answer 10

Essentially that's what happens :D so try it here with steps please.

Answer 11

Okay. So, for the function:\[f(x) = \cot^3\sqrt{x}\]
= \[(\cot \sqrt{x})^3\]

Answer 12

Good, keep going.

Answer 13

= \[\frac{ d }{ dx }[(\cot(\sqrt{x}))^3] * \frac{ d }{ dx }[\cot(\sqrt{x)}] * \frac{ d }{ dx }[\sqrt{x})]\]

Answer 14

Yup

Answer 15

Still trying to figure this out, lol.

= \[3(-\csc^2x)^2 * (-\csc^2x) * \frac{ 1 }{ 2\sqrt{x} }\]

Answer 16

It's finding out how to find the derivative of a function with a function inside it.

Answer 17

Hmm, that seems right so far.

Answer 18

Blah, it's the square root of x; not x in the fist two parts. :/ So easy to make mistakes.

Answer 19

Oh lol, yea. I didn't see that either :p

Answer 20

Oh another error btw, in the first part, you need the original trig function, not the derivative inside.

Answer 21

So rather it should be: \[f'(x) = 3\cot^2 (\sqrt(x)) * \csc^2(\sqrt(x)) * \frac{ 1 }{ 2 \sqrt x }\]

Answer 22

So, when I'm taking these insane derivatives, I just "ignore" the inside function? If that makes sense?

Answer 23

So you reduce the power by 1, but you use the orignal function still.

Answer 24

Oh, and that should be negative up there what I posted >.>

Answer 25

@MoonlitFate, essentially, you go out to inside. Taking the derivative of essentially one layer at a time.

Answer 26

Kind of like an onion? o_o

Answer 27

no. like the chain rule.

Answer 28

@Abbot-- I meant how to remember the whole "layer" thing, lol. :p

Answer 29

@MoonlitFate, yes you can sort of use that analogy.

Answer 30

Anyway, can I post any problem in here try to work and see if I actually get this?

Answer 31

Sure, we'll help out. Hope I helped so far.

Answer 32

Found a scary looking one, lol.\[h(x) = 2\cot^2(\pi t +2)\]

Answer 33

Have to rewrite that: \[h(x) = 2\cot(\pi t + 2)^2\]

Answer 34

You have to enclose it all in parentheses actually.

\[h(x) = 2(\cot(\pi*t+2))^2\]

Answer 35

Oh, lol.

Answer 36

Hopefully, I have this right now:\[\frac{ d }{ dx }[(2\cot(\pi t +2))^2] *\frac{ d }{ dx }[\cot(\pi t +2)] * \frac{ d }{ dx }[\pi t +2)]\]

Answer 37

seems right.

Answer 38

something like that?