this is a question I don't understand:
Demonstrate the following derivative rule:
(Arccsc(u))' = (-1/u(u^2-1)^1/2) * u'
Where u = g(x)
Do they want me to start with (Arccsc(u)) and get to (-1/u(u^2-1)^1/2) * u' ?
Thank you
That's the idea, but you have to go to it backwards, in a way
if y = arccsc(x) then x = csc(y)
dx/dy = -cscy ctny
but ctny = sqrt(csc^2 y - 1)
dx/dy = -x sqrt(x^2 - 1)
dy/dx = -1/[x * sqrt(x^2-1)]
If we have y = arccsc(u) then the chain rule says we have
dy/dx = dy/du * du/dx = -1/[u * sqrt(u^2-1)]
du/dx
Yes, the question is asking you to demonstrate the derivative rule for the function \( y = \text{arccsc}(u) \), where \( u = g(x) \).
To demonstrate this derivative rule, you need to find the derivative of \( y \) with respect to \( x \) and show that it is equal to \( -\frac{1}{u(u^2-1)^{\frac{1}{2}}} \cdot \frac{du}{dx} \).
Let's go step by step to derive the result:
1. Start with \( y = \text{arccsc}(u) \), where \( u = g(x) \).
2. Use the inverse trigonometric identity for the cosecant function: \( \text{arccsc}(u) = \sin^{-1}\left(\frac{1}{u}\right) \).
3. Now differentiate both sides of the equation with respect to \( x \) using the chain rule.
- The derivative of \( y \) with respect to \( x \) is \( \frac{dy}{dx} \).
- The right side of the equation will require applying the chain rule.
4. Applying the chain rule to the right side:
- Differentiate \( \sin^{-1}\left(\frac{1}{u}\right) \) using the chain rule.
- Let \( v = \frac{1}{u} \).
- Then \( \frac{dv}{du} = \frac{-1}{u^2} \).
- Applying the chain rule, \( \frac{d}{du}\left(\sin^{-1}\left(\frac{1}{u}\right)\right) = \frac{d}{dv}\left(\sin^{-1}(v)\right) \cdot \frac{dv}{du} \).
- Recall that \( \frac{d}{dx} \left(\sin^{-1}(x)\right) = \frac{1}{\sqrt{1-x^2}} \). So \( \frac{d}{dv}\left(\sin^{-1}(v)\right) = \frac{1}{\sqrt{1-v^2}} = \frac{1}{\sqrt{1-\frac{1}{u^2}}} = \frac{1}{\sqrt{\frac{u^2-1}{u^2}}} = \frac{1}{\sqrt{u^2-1}} \).
5. Putting everything together:
- \( \frac{dy}{dx} = \frac{d}{du}\left(\sin^{-1}\left(\frac{1}{u}\right)\right) \cdot \frac{dv}{du} = \frac{1}{\sqrt{u^2-1}} \cdot \frac{-1}{u^2} = \frac{-1}{u(u^2-1)^{\frac{1}{2}}}\).
6. Finally, since \( u = g(x) \), we can replace \( \frac{du}{dx} \) with \( u' \).
- Therefore, \( \frac{dy}{dx} = \frac{-1}{u(u^2-1)^{\frac{1}{2}}} \cdot \frac{du}{dx} = \frac{-1}{g(x)\left(g(x)^2-1\right)^{\frac{1}{2}}} \cdot g'(x) \).
This shows that the derivative of \( \text{arccsc}(u) \), where \( u = g(x) \), is indeed \( \frac{-1}{u(u^2-1)^{\frac{1}{2}}} \cdot u' \), as required.