# Derivative

• Dec 4th 2010, 02:48 PM
Hellbent
Derivative
Hi,

I'm not arriving at the correct answer, kindly point out my mistake(s).

Derivative of:
$\displaystyle \displaystyle y = \sin (\cos\,x)$

$\displaystyle \displaystyle u = \cos\,x$

$\displaystyle \displaystyle \frac{du}{dx} = -\sin\,x$

$\displaystyle \displaystyle v = \sin\,u$

$\displaystyle \displaystyle \frac{dv}{du} = \cos\,u$

$\displaystyle \displaystyle \frac{dy}{dx} = v\frac{du}{dx}\,+\,u\frac{dv}{du}$

$\displaystyle \displaystyle = \sin\,u(-\sin\,x) + \cos\,x(\cos\,u)$

$\displaystyle \displaystyle = -\sin^{2}x[\cos(x)] + \cos x [\cos(\cos x)]$
• Dec 4th 2010, 03:01 PM
TheCoffeeMachine
Quote:

Originally Posted by Hellbent
$\displaystyle \displaystyle \frac{dy}{dx} = v\frac{du}{dx}\,+\,u\frac{dv}{du}$

That's the product rule, and what you have isn't a product.
What you've is a composite function, so use the chain rule:

$\displaystyle (f \circ g)'(x) = f'(g(x))g'(x)$ or $\displaystyle \displaystyle \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}.$
• Dec 4th 2010, 03:09 PM
e^(i*pi)
This problem requires the chain rule:

$\displaystyle f(x) = \cos(x)$

$\displaystyle g(x) = \sin[f(x)]$

Hence $\displaystyle y^{\prime} = g'(x) \cdot f'(x)$

If you like you can let $\displaystyle u = \cos(x)$ giving $\displaystyle y = \sin(u)$ and then using the chain rule: $\displaystyle \dfrac{dy}{dx} = \dfrac{dy}{du} \cdot \dfrac{du}{dx}$
• Dec 4th 2010, 03:26 PM
Creebe
Okay, just use the chain rule. If you're not familiar with the chain rule, read this:
Chain rule - Wikipedia, the free encyclopedia

Anyway, here is the math:

$\displaystyle y = sin(cos(x))$

1.$\displaystyle cos(x) = u(x)$

Now $\displaystyle y$ becomes $\displaystyle y(u(x))$

2. Differentiate:

$\displaystyle \frac {dy}{dx} = \frac {dy}{du} * \frac {du}{dx}$

$\displaystyle y = cos(cos(x))*-sin(x)$

What you did is that you used the product rule. Only use the product rule when you have something like this: $\displaystyle y(x) = p(x)q(x)$ Here, $\displaystyle p$ is a function of $\displaystyle x$, and $\displaystyle q$ is a function of $\displaystyle x$ as well.

Your question is however: $\displaystyle y(x) = p(q(x))$ It means that $\displaystyle p$ is a function of $\displaystyle q(x)$, and $\displaystyle q$ is a function of $\displaystyle x$
• Dec 4th 2010, 04:29 PM
Hellbent
$\displaystyle u = \cos x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y = \sin u$

$\displaystyle u^{'} = -\sin\,x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y^{'} = \cos u$

$\displaystyle (u^{'})(y^{'}) = -\sin x(\cos(\cos x))$
• Dec 4th 2010, 05:27 PM
harish21
Quote:

Originally Posted by Hellbent
$\displaystyle u = \cos x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y = \sin u$

$\displaystyle u^{'} = -\sin\,x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y^{'} = \cos u$

$\displaystyle (u^{'})(y^{'}) = -\sin x(\cos(\cos x))$

correct! (Yes)