Question

(Calc Q > Derivatives > Chain Rule > Double chain rule scenarios?!)
So I'm working on some problems for my homework and come across a problem that requires me to apply the chain rule twice. I have no clue where to begin, even though I do understand how to apply the chain rule. The problem in question is:
g(x) = 2tan^3(4πx).
The answer is 24π(tan^2(4πx))(sec^2(4πx).

However having the answer helps me none since I do not know how to apply the chain rule twice... Help?! (MUCH thanks in advance!!!)

Answer 1

it might be good to rewrite the trig
\[ 2~[\tan(4πx)]^3\]
it helps to see the chaining parts
\[ 2*3~[\tan(4πx)]^2*\tan'(4πx)\]
\[ 2*3~[\tan(4πx)]^2*\sec^2(4πx)*4πx'\]

Answer 2

the chain rule is just a matter of popping out the derivative of an inner function, like a set of russian dolls

worry about the outer most function, before focusing on the inner functions

Answer 3

\[ 2~tan^3(4πx)~:~let~a=tan(4\pi x)\]
\[ 2~a^3\to~2*3a^2~a'\]

since a = tan(4pi x) , let b = 4pi x
a = tan(b)
a' = sec^2(b) * b'

\[ 2~a^3\to~2*3a^2~a'\]
\[2*3a^2~a'\to~2*3a^2~sec^2(b) * b'\]
so, what is b'? etc ....

Answer 4

i still did'n get it...
im going to ask simpler question
what is the derivative of [2xcos(x ^{2})\]

Answer 5

show me how you would try to work it so i can see if you are introducing an error in the process

Answer 6

you mentioned that derivate of 2a^3 is 2*3a*a'
how do you get a' from derivative?

Answer 7

thats what the chain rule does, it pops out a derivative of an inner function right? You did say: " I do understand how to apply the chain rule"

Answer 8

the derivative of f(a) = f'(a) * a'

Answer 9

one issue with learning this is that you are not told that the chain rule has always been present from the start of you derivative rules ....

Answer 10

we know that the derivative wrtx of y is y'; or dy/dx

the derivative wrtx of x is simply x'; or dx/dx which equals 1

the derivative of a wrtx is da/dx, or simply a'

Answer 11

2tan^3(4πx)

2 (a)^3

the derivative is 6 a^2 a'

Answer 12

the chain rule pops out a derivative of the inside function

Answer 13

in general\[[f(g(h(n(m(...))))))]'=f'(g)*g'(h)*h'(n)*n'(m)*m'(...)~*~...\]

Answer 14

plz check my answer...

2x cos (x^2)

assume : a= 2x a'=2
b= cos(c) b'=-sin(c)
c=x^2 c'=2x
derivative:
a'=2 b'
= 2(-sin(c))
= -2 sin(x^2) c'
= -2 sin(x^2)*2x
then the answer is -4x sin(x^2)..?

Answer 15

dont forget that you started out with a product ...

2x cos (x^2) the product rule gives us

2x' cos (x^2) + 2x cos'(x^2)

the leftside is simple enough since x' = 1

2cos (x^2) + 2x cos'(x^2)
^^^
inner function pops out a derivative
afterwards

2cos (x^2) - 2x sin(x^2) * (x^2)'

2cos (x^2) - 2x sin(x^2) * (2x)

2cos (x^2) - 4x^2 sin(x^2)

Answer 16

how you organize the results will depend on what best suits you :)

Answer 17

ok2...
i get it now...
that's mean my second question is not double chain rule...

Answer 18

right, it can be expressed as a single popout :)