# Math Help - Chain rule not working here?

1. ## Chain rule not working here?

I'm applying the chain rule to a problem I'm working on, but it doesn't seem to get the same answer as the one specified in the back of the book.

The problem is:
$p(x) = sin^3(pi*x)$

Chain rule states:

$f'(g(x))g'(x)$

From this problem, I noted:

$f(x) = sin^3(x)$
$f'(x) = 3sin^2(x)$
$g(x) = pi*x$
$g'(x) = pi$

So applying these to the chain rule, we get:

$3sin^2(pi*x) * (pi)$

But this doesn't seem to be the correct answer. I've applied the same methodology for other problems and the chain rule seems to work there, but not here. Does anyone know what I'm doing wrong? Any help would be appreciated. Thanks.

2. Originally Posted by bobbooey
I'm applying the chain rule to a problem I'm working on, but it doesn't seem to get the same answer as the one specified in the back of the book.

The problem is:
$p(x) = sin^3(pi*x)$

Chain rule states:

$f'(g(x))g'(x)$

From this problem, I noted:

$f(x) = sin^3(x)$
$f'(x) = 3sin^2(x)$
$g(x) = pi*x$
$g'(x) = pi$

So applying these to the chain rule, we get:

$3sin^2(pi*x) * (pi)$

But this doesn't seem to be the correct answer. I've applied the same methodology for other problems and the chain rule seems to work there, but not here. Does anyone know what I'm doing wrong? Any help would be appreciated. Thanks.
Dear bobbooey,

$p(x)=sin^{3}\pi{x}$

$\frac{d}{dx}p(x)=\frac{d}{d(sin\pi{x})}(sin^{3}\pi {x})~\frac{d}{d(\pi{x})}(sin\pi{x})~\frac{d}{dx}\p i{x}$ ; using the Chain rule

$\frac{d}{dx}p(x)=3sin^{2}\pi{x}\times{cos\pi{x}}\t imes{\pi}$

Hope this will help you to understand.

3. Originally Posted by bobbooey
I'm applying the chain rule to a problem I'm working on, but it doesn't seem to get the same answer as the one specified in the back of the book.

The problem is:
$p(x) = sin^3(pi*x)$

Chain rule states:

$f'(g(x))g'(x)$

From this problem, I noted:

$f(x) = sin^3(x)$ you left out "pi"
$f'(x) = 3sin^2(x)$ no! even if "pi" was included
$g(x) = pi*x$
$g'(x) = pi$ yes

So applying these to the chain rule, we get:

$3sin^2(pi*x) * (pi)$

But this doesn't seem to be the correct answer. I've applied the same methodology for other problems and the chain rule seems to work there, but not here. Does anyone know what I'm doing wrong? Any help would be appreciated. Thanks.
$f(x)=sin^3({\pi}x)$

$u={\pi}x$

$v=sin({\pi}x)=sin(u)$

$w=sin^3u=v^3$

$\frac{d}{dx}f(x)=\frac{dw}{dx}=\frac{dw}{dv}\frac{ dv}{du}\frac{du}{dx}$

$=3v^2cosu({\pi})=3sin^2({\pi}x)cos({\pi}x)({\pi})$