# Find the Derivative

• Oct 11th 2009, 02:25 PM
VitaX
Find the Derivative
Find the derivative of:

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

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 PM
DBA
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 PM
VitaX
Quote:

Originally Posted by DBA
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 ?

Little confusing the way you wrote it out here. When applying the product rule $y' = uv' + vu'$ what would $u$ and $v$ be in the original function. $u = \sqrt{x}$ and $v = Csc(x + 1)^3$ Correct so far? If so then my problem is finding the derivative of $v = Csc(x + 1)^3$ I know how to apply the chain rule but the $Csc$ is just confusing me.
• Oct 11th 2009, 03:20 PM
DBA
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 PM
VitaX
Quote:

Originally Posted by DBA
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.

I believe the Csc function is not cubed.

Here's what I did:
$
y = \sqrt{x} Csc(x + 1)^3
$

$y' = Csc(x+1)^3*\frac{1}{2}x^{-\frac{1}{2}} + x^{\frac{1}{2}}(-CscxCotx)3(x+1)^2$
$u=x^{\frac{1}{2}}$
$u'=\frac{1}{2}x^{-\frac{1}{2}}$
$v=Csc(x+1)^3$
$v'=(-CscxCotx)3(x+1)^2$
Note $\frac{d}{dx}Cscx=-CscxCotx$

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

Edit: $y' = Csc(x+1)^3\frac{1}{2}x^{-\frac{1}{2}} + x^{\frac{1}{2}}(-Csc(x+1)^3)(Cot(x+1)^3)3(x+1)^2$
$y' = \frac{1}{2}x^{-\frac{1}{2}} Csc(x+1)^3 + 3x^{\frac{1}{2}}(x+1)^2(-Csc(x+1)^3)(Cot(x+1)^3)$
• Oct 11th 2009, 06:31 PM
DBA
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 PM
VitaX
Quote:

Originally Posted by DBA
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

So would the final answer be as I said above?
$y' = \frac{1}{2}x^{-\frac{1}{2}} Csc(x+1)^3 + 3x^{\frac{1}{2}}(x+1)^2(-Csc(x+1)^3)(Cot(x+1)^3$
• Oct 11th 2009, 08:16 PM
DBA
Yes, I think that is right.