Posted by Jim on Tuesday, November 15, 2011 at 1:33am.
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' ?
- Math (Calculus) - Steve, Tuesday, November 15, 2011 at 5:02pm
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)]
Answer this Question
More Related Questions
- pre cal\trig - Find the following cot(Arccsc 25/24) show your work
- Trigonometry - In this problem, you will describe in detail how we arrive at the...
- trig - cos(arccsc(-20)) Im not allowed to have a decimal answer
- Trig (inverse functions) - Problem: tan[arccsc(-5/3) + arctan(1/4)] My work: let...
- Calculus - y= [(x-3)/(x^2+1)]^2 find the derivative. I know i would start off ...
- Calculus - 1.) Find the derivative of tan (sec x). 2.) Find the derivative if 1/...
- Math (Calculus) - Hello, Could somebody please help me with the following ...
- CALCULUS - what is the derivative of ln(x-1)? If you are studying the calculus ...
- math - can some one please help me with the following questions as i dont ...
- Calculus - find the derivative of f(x)=((x^2+3)^5+x)^2 using the chain rule. I ...