Find the derivative of:

$\displaystyle y = \sqrt{x} Csc(x + 1)^3$

Tbh the Csc is throwing me off here. I'm unsure what to apply here. I know that the chain rule and product rule will be used but I'm stuck.

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- Oct 11th 2009, 02:25 PMVitaXFind the Derivative
Find the derivative of:

$\displaystyle y = \sqrt{x} Csc(x + 1)^3$

Tbh the Csc is throwing me off here. I'm unsure what to apply here. I know that the chain rule and product rule will be used but I'm stuck. - Oct 11th 2009, 02:36 PMDBA
d/dx(csc u) = - csc u * cot u * du/dx

Your y is that

y = sqrt(x)* [csc(x+1)]^3 --> csc cubed

or

y = sqrt(x)* [csc(x+1)^3] --> x+1 cubed ? - Oct 11th 2009, 03:04 PMVitaX
Little confusing the way you wrote it out here. When applying the product rule $\displaystyle y' = uv' + vu'$ what would $\displaystyle u$ and $\displaystyle v$ be in the original function. $\displaystyle u = \sqrt{x}$ and $\displaystyle v = Csc(x + 1)^3$ Correct so far? If so then my problem is finding the derivative of $\displaystyle v = Csc(x + 1)^3$ I know how to apply the chain rule but the $\displaystyle Csc$ is just confusing me.

- Oct 11th 2009, 03:20 PMDBA
Yes, that would be the product rule.

But then for v' you need to use the Chain Rule.

I am still not sure if the CSC function is cubed or the x+1 since it would be a difference.

v = [csc(x+1)]^3

v' = 3*[csc(x+1)]^2 * -csc(x+1)*cot(x+1) * 1

v = csc((x+1)^3)

v' = -csc(x+1)*cot(x+1) * 3*(x+1)^2 * 1

^2 means squared

^3 means cubed

Sorry, I do not know yet, how to write the functions better. - Oct 11th 2009, 03:57 PMVitaX
I believe the Csc function is not cubed.

Here's what I did:

$\displaystyle

y = \sqrt{x} Csc(x + 1)^3

$

$\displaystyle y' = Csc(x+1)^3*\frac{1}{2}x^{-\frac{1}{2}} + x^{\frac{1}{2}}(-CscxCotx)3(x+1)^2$

$\displaystyle u=x^{\frac{1}{2}}$

$\displaystyle u'=\frac{1}{2}x^{-\frac{1}{2}}$

$\displaystyle v=Csc(x+1)^3$

$\displaystyle v'=(-CscxCotx)3(x+1)^2$

Note $\displaystyle \frac{d}{dx}Cscx=-CscxCotx$

Hows this look?

I took the derivative of $\displaystyle Cscx$ seperately from $\displaystyle (x+1)^3$ Was I supposed to do this?

Edit: $\displaystyle y' = Csc(x+1)^3\frac{1}{2}x^{-\frac{1}{2}} + x^{\frac{1}{2}}(-Csc(x+1)^3)(Cot(x+1)^3)3(x+1)^2$

$\displaystyle y' = \frac{1}{2}x^{-\frac{1}{2}} Csc(x+1)^3 + 3x^{\frac{1}{2}}(x+1)^2(-Csc(x+1)^3)(Cot(x+1)^3)$ - Oct 11th 2009, 06:31 PMDBA
http://www.mathhelpforum.com/math-he...c0bf6ae7-1.gif

http://www.mathhelpforum.com/math-he...b9010e97-1.gif

Note http://www.mathhelpforum.com/math-he...aaf44e5c-1.gif

Yes, but you have CSC[(x+1)^3] and not CSC(x), so for your v' you need to use the entire term in [ ]

The x in your case is (x+1)^3

Example:

f= csc(x+1)

f'= -csc(x+1)cot(x+1)

f= csc[(x+1)^3]

f'= -csc[(x+1)^3]cot[(x+1)^3]

"Hows this look?

I took the derivative of http://www.mathhelpforum.com/math-he...1e8a9361-1.gif seperately from http://www.mathhelpforum.com/math-he...bb9227b3-1.gif Was I supposed to do this?"

So yes but fom csc[(x+1)^3]

You always have inner and outer function

Example

f= (2-x)^3

outer function ()^3

inner function (2-x)

So,

f'= 3*(2-x)^2 * (0-1)

Hope that helps - Oct 11th 2009, 06:39 PMVitaX
- Oct 11th 2009, 08:16 PMDBA
Yes, I think that is right.