1. ## Find 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.

2. d/dx(csc u) = - csc u * cot u * du/dx

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

3. Originally Posted by DBA
d/dx(csc u) = - csc u * cot u * du/dx

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 $\displaystyle y' = uv' + vu'$ 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.

4. 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.

5. 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:
$\displaystyle y = \sqrt{x} Csc(x + 1)^3$
$\displaystyle y' = Csc(x+1)^3*\frac{1}{2}x^{-\frac{1}{2}} + x^{\frac{1}{2}}(-CscxCotx)3(x+1)^2$
$\displaystyle u=x^{\frac{1}{2}}$
$\displaystyle u'=\frac{1}{2}x^{-\frac{1}{2}}$
$\displaystyle v=Csc(x+1)^3$
$\displaystyle v'=(-CscxCotx)3(x+1)^2$
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: $\displaystyle y' = Csc(x+1)^3\frac{1}{2}x^{-\frac{1}{2}} + x^{\frac{1}{2}}(-Csc(x+1)^3)(Cot(x+1)^3)3(x+1)^2$
$\displaystyle y' = \frac{1}{2}x^{-\frac{1}{2}} Csc(x+1)^3 + 3x^{\frac{1}{2}}(x+1)^2(-Csc(x+1)^3)(Cot(x+1)^3)$

6. Note

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 seperately from 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

7. Originally Posted by DBA

Note

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

8. Yes, I think that is right.