Differentiation of sin and cos

[maccaman]

i can't work out how to do this question, i have the answer, i just don't know how to get it
Q. differentiate
y = sin (cos x)

A. - cos (cos) sin x

If anyone could show me the solution, it would be greatly appreciated.

 

[jamesrc]

Use the chain rule defined something like this:

$$\frac{df(g(x))}{dx} = \frac{df(g)}{dg} \frac{dg(x)}{dx}$$

Here are the steps:

$$y = \sin \left( \cos x\right )$$
$$\frac{dy}{dx} = \cos\left(\cos x\right) \frac{d}{dx}\left(\cos x\right)$$
$$\frac{dy}{dx} = \cos\left(\cos x\right) \left(-\sin x\right)$$

Let me know if anything is still unclear.

 

[maccaman]

we do the chain rule like this
sin (cos x) = uv

where u = sin, v = cos x

dy/dx = v (du/dx) + u (dv/dx)

= cos x . cos + sin . -sin x

Please tell me where i went wrong in this. i know its in my working, that's what i need help with, i get what your saying.

 

[maccaman]

how stupid of me, i didnt even need to use the chain rule, sorry man, i totally screwed it up, i get how to do it now, thanks

 

[Zorodius]

Yes, you do need to use the chain rule.

we do the chain rule like this
sin (cos x) = uv

where u = sin, v = cos x

dy/dx = v (du/dx) + u (dv/dx)

= cos x . cos + sin . -sin x

Please tell me where i went wrong in this. i know its in my working, that's what i need help with, i get what your saying.

This is the product rule, not the chain rule. sin (cos x) isn't the product of sine and cosine, it's a composite function.

 

[maccaman]

yeh sorry bout the wrong rule: i just did it like we normally differentiate

sin (cos x)
dy/dx = cos (cos x) x (derivative of brackets) which is -sin x
sorry for such a debate