Question

Your friend is having difficulty applying the chain rule for differentiation. Using the example g(t)=cos^2(x^3+x^2) , explain how to apply the chain rule to differentiate g(t)

Answer 1

first of all you need to reduce the order then differantiate the inside the cos so
2.cos(x^3+x^2).(3x^2+2x).(-)sin(x^3+x^2)

Answer 2

Is the function supposed to be g(x) ?

Answer 3

probably

Answer 4

oh i didnt realize that if it is g(t) solution will change

Answer 5

if it doesnt change the solution, im sure whether its x or t doesnt matter. if it does change, keep it at g(t)

Answer 6

Chain rule can be kind of tricky.
I would try to get the "friend" to understand that a function consists of `layers`.

`functions within functions`.

Our outermost function is \(\Large g(x)=x^2\)
If we stuff the function \(\Large h(x)=\cos x\) into g(x) we get,\[\Large g(h(x))=\left[\cos(x)\right]^2=\cos^2x\]If we then stuff the function \(\Large j(x)=x^3+x^2\) inside of g(h(x)) we get,\[\Large g(h(j(x)))=\cos^2(x^3+x^2)\]

Then we just apply the chain rule as the rule tells us:\[\Large \frac{d}{dx}g(h(j(x)))\quad=\quad g'(h(j(x)))\cdot h'(j(x))\cdot j'(x)\]

Hmm I think I just made it more confusing :(
But if you want to be technical, and explain it using the definition for chain rule, then this might help...

Answer 7

oh and its not actually my friend its an open ended promt

Answer 8

that's why I said "friend" hehe

Answer 9

ok and im not exactly sure what im supposed to do. its open ended so i assume i have to explain it using the definition and mathematically. its kind of vague.

Answer 10

Let,\[\Large g(x)=x^2\]\[\Large h(x)=\cos x\]\[\Large j(x)=x^3+x^2\]Then,\[\Large f(x)\quad=\quad g(h(j(x)))\]and by the definition of chain rule,\[\Large f'(x)\quad=\quad g'(h(j(x)))\cdot h'(j(x))\cdot j'(x)\]And then list out these derivatives individually,\[\Large g'(h(j(x)))\quad=\quad 2\left[\cos(x^3+x^2)\right]\]\[\Large h'(j(x))\quad=\quad -\sin(x^3+x^2)\]\[\Large j'(x)\quad=\quad 3x^2+2x\]So our final answer f'(x) is the product of these three thingies.
So we just need to multiply them together.

Mmmmmmm I dunno D: That's how I would approach it maybe.
Blah I kinda did all the work for you.. sorry bout that...
takes the fun away.

Answer 11

i multiplied them together and got: -2(3x^2+2x)sin(x^3+x^2)cos(x^3+x^2). would that be g'(t)?

Answer 12

Ahhh crap, i was calling the original function f(x).

I didn't notice that they labeled it as g(x). Sorry about that..
That makes things a little confusing lol.

But yes, that would be your .... f'(x) or g'(x) as they labeled it.
But not g'(t). I think that was a typo, hopefully..

Answer 13

ok thanks!

Answer 14

If it IS supposed to be g'(t), then we would have one more `chain`. We would throw a \(\Large \dfrac{dx}{dt}\) on the end.

Answer 15

ok