# Find derivative of: sin^3(4x) (using chain rule)

• Nov 21st 2010, 05:11 AM
kessel81
Find derivative of: sin^3(4x) (using chain rule)
Hi, I am trying to use chain rule to find the derivative of this but I can't seem to get the right answer. I used the same method with other questions and I get the right answer but not with this one here. I took a screenshot of what I did....I'm not sure what I'm doing wrong though. Thanks!

http://i52.tinypic.com/18oqk1.jpg
• Nov 21st 2010, 05:50 AM
e^(i*pi)
It looks like you've forgot about the power on the sin(4x)

$\displaystyle f(x) = y = \sin^3(4x)$

$\displaystyle u = 4x \: \rightarrow \dfrac{du}{dx} = 4$

$\displaystyle v = \sin(u) \: \rightarrow \: \dfrac{dv}{du} = -\cos(u)$

$\displaystyle y = v^3 \: \rightarrow \: \dfrac{dy}{dv} = 3v^2$

Using the chain rule:

$\displaystyle \dfrac{dy}{dx} = \dfrac{du}{dx} \times \dfrac{dv}{du} \times \dfrac{dy}{dv}$

Spoiler:
$\displaystyle \dfrac{d}{dx}f(x) = 4 \times -\cos(4x) \times 3\sin^2(4x)$

• Nov 21st 2010, 06:10 AM
kessel81
Quote:

Originally Posted by e^(i*pi)
It looks like you've forgot about the power on the sin(4x)

$\displaystyle f(x) = y = \sin^3(4x)$

$\displaystyle u = 4x \: \rightarrow \dfrac{du}{dx} = 4$

$\displaystyle v = \sin(u) \: \rightarrow \: \dfrac{dv}{du} = -\cos(u)$

$\displaystyle y = v^3 \: \rightarrow \: \dfrac{dy}{dv} = 3v^2$

Using the chain rule:

$\displaystyle \dfrac{dy}{dx} = \dfrac{du}{dx} \times \dfrac{dv}{du} \times \dfrac{dy}{dv}$

Spoiler:
$\displaystyle \dfrac{d}{dx}f(x) = 4 \times -\cos(4x) \times 3\sin^2(4x)$

Hey thanks. I don't quite understand this whole "dy/dx = du/dx X dv/du X dy/dv" stuff I have this form in my textbook too but I'm not sure what it means or how their solving it...I know about Dx and F'(x). DxF(x) means that Dx is operating on F(x) to get another function F'(x)...

• Nov 21st 2010, 06:19 AM
e^(i*pi)
You can think of it like a fraction:

Consider: $\displaystyle \dfrac{1}{2} \times \dfrac{2}{3} \times \dfrac{3}{4}$ - can you see that 2 and 3 will cancel because they are on the top and the bottom which leaves $\displaystyle \dfrac{1}{4}$

It is much the same thing but although technically you don't cancel them but it works in the same way (confusing yes but you can treat as a fraction).

In what I did before
$\displaystyle \dfrac{dy}{dx} = \dfrac{du}{dx} \times \dfrac{dv}{du} \times \dfrac{dy}{dv}$ note that $\displaystyle dv$ and $\displaystyle du$ will cancel.

Of course since that sum gives an answer of terms of u and v you need to go back to the original substitutions which I made at the top of my post to get in terms of x.

As for DxF(x) I have no idea how that notation works
• Nov 21st 2010, 06:44 AM
kessel81
I'm still not sure how you solved it....sorry I'm really bad at calculus :| What I don't understand is that the chain rule is F'(x) = F'(g(x)) X g'(x) which only has two functions F(x) and g(x)...but you have y (or F(x) i think), u and v....could you possibly solve it in a different way? I really don't like this dy/dx notation and even before starting the chain rule I always (and so did my teacher) used either F'(x) or Dx..

thanks again for helping..
• Nov 21st 2010, 07:01 AM
e^(i*pi)
Since we need to use the chain rule three times I shall use f(x), g(x) and h(x)

$\displaystyle f(x) = \sin^3(4x)$

If $\displaystyle g(x) = 4x$ and $\displaystyle h(x) = sin^3[g(x)]$

Hence: $\displaystyle f(x) = h[g(x)]$ and $\displaystyle f'(x) =h'[g(x)] \cdot g'(x) = 3\sin^2(4x) \cos(4x) \times 4 = 12\sin^2(4x)\cos(4x)$

I cannot use the Dx notation unless you explain it to me because I have no idea what it means

edit: wolfram uses much the same method: WolframAlpha
• Nov 21st 2010, 07:02 AM
Plato
Here is the derivative: $\displaystyle \displaystyle 3\left[ {\sin ^2 (4x)} \right]\left\{ {\cos (4x)} \right\}\left( 4 \right)$
Forget u-substitution. Look at the chain of derivatives.
First a cube, then a sine, then 4x.
• Nov 21st 2010, 08:13 AM
kessel81
Quote:

Originally Posted by e^(i*pi)
Since we need to use the chain rule three times I shall use f(x), g(x) and h(x)

$\displaystyle f(x) = \sin^3(4x)$

If $\displaystyle g(x) = 4x$ and $\displaystyle h(x) = sin^3[g(x)]$

Hence: $\displaystyle f(x) = h[g(x)]$ and $\displaystyle f'(x) =h'[g(x)] \cdot g'(x) = 3\sin^2(4x) \cos(4x) \times 4 = 12\sin^2(4x)\cos(4x)$

I cannot use the Dx notation unless you explain it to me because I have no idea what it means

edit: wolfram uses much the same method: WolframAlpha

I GOT IT!! thanks a lot! lol now that makes sense :)
• Nov 21st 2010, 08:34 AM
kessel81
Quote:

Originally Posted by e^(i*pi)
Since we need to use the chain rule three times I shall use f(x), g(x) and h(x)

$\displaystyle f(x) = \sin^3(4x)$

If $\displaystyle g(x) = 4x$ and $\displaystyle h(x) = sin^3[g(x)]$

Hence: $\displaystyle f(x) = h[g(x)]$ and $\displaystyle f'(x) =h'[g(x)] \cdot g'(x) = 3\sin^2(4x) \cos(4x) \times 4 = 12\sin^2(4x)\cos(4x)$

I cannot use the Dx notation unless you explain it to me because I have no idea what it means

edit: wolfram uses much the same method: WolframAlpha

OMG I tried doing it myself again and now this one thing doesn't make sense. When you evaluated h'[g(x)], which is just 3sin^2[g(x)] = 3sin^2(4x) right....how did you get that cos(4x) right after it?
• Nov 21st 2010, 08:51 AM
e^(i*pi)
Quote:

Originally Posted by kessel81
OMG I tried doing it myself again and now this one thing doesn't make sense. When you evaluated h'[g(x)], which is just 3sin^2[g(x)] = 3sin^2(4x) right....how did you get that cos(4x) right after it?

Chain Rule: the derivative of$\displaystyle \sin(ax)$ is $\displaystyle a\cos(ax)$

$\displaystyle f(x) = i(h[g(x)])$

$\displaystyle f'(x) = i'(h[g(x)]) \times h'[g(x)] \times g'(x)$

where $\displaystyle i(h[g(x)]) = (h[g(x)])^3$

$\displaystyle f'(x) = 3\sin^2(4x) \times \cos(4x) \times 4 = 12\sin^2(4x)\cos(4x)$

There could be a mistake with the derivation (but not the answer) because it really is easier to use Plato's method
• Nov 21st 2010, 08:55 AM
Plato
If you were to evaluate $\displaystyle \sin^3(4x)$ at $\displaystyle x=2.7$, how would you do it?
One would first multiply $\displaystyle 4\times 2.7$; then what.
Find the sine of that number, RIGHT?
Then take that result and raise it to the third power.
So to find the derivative of $\displaystyle \sin^3(4x)$ we take a chain of derivatives in exactly the reverse order:
the derivative of the third power; the derivative of sine; the derivative of 4x.
$\displaystyle 3\sin^2(4x)\cos(4x)[4]$. The derivative of sine is cosine.