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