Derivative

**Hellbent** · December 4th 2010, 02:48 PM

Hi,

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

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

$\displaystyle u = \cos\,x$

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

$\displaystyle v = \sin\,u$

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

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

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

$\displaystyle = -\sin^{2}x[\cos(x)] + \cos x [\cos(\cos x)]$

**TheCoffeeMachine** · December 4th 2010, 03:01 PM

Originally Posted by Hellbent

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

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:

$(f \circ g)'(x) = f'(g(x))g'(x)$ or $\displaystyle \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}.$

**e^(i*pi)** · December 4th 2010, 03:09 PM

This problem requires the chain rule:

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

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

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

If you like you can let $u = \cos(x)$ giving $y = \sin(u)$ and then using the chain rule: $\dfrac{dy}{dx} = \dfrac{dy}{du} \cdot \dfrac{du}{dx}$

**Creebe** · December 4th 2010, 03:26 PM

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:

$y = sin(cos(x))$

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

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

2. Differentiate:

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

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

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

Your question is however: $y(x) = p(q(x))$ It means that $p$ is a function of $q(x)$ , and $q$ is a function of $x$

**Hellbent** · December 4th 2010, 04:29 PM

$u = \cos x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y = \sin u$

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

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

**harish21** · December 4th 2010, 05:27 PM

Originally Posted by Hellbent

$u = \cos x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y = \sin u$

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

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

correct!

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