Author Topic: Derivatives of Inverse Functions - HELP! (Read 1906 times) Tweet Share

squance · « **on:** August 18, 2008, 06:31:21 pm »

Hey. Im having issues with this question and would like some help.

Using implicit differentiation, show that..

d/dx(arccosec x) = -1/(x sqrt(x^2-1)), x>1

So far I have...

Let y = arccosec x

x = cosec y

d/dx(x) = d/dx(cosec y)
1 = d/dx(1/sin y)
1 = -cosy (dydx)/sin^2 y (after applying quoitent rule to 1/sin y)

I think i have to use the cos^2 y+ sin^2 y = 1 thing but I can't seem to get it.

Some help please?

Mao · « **Reply #1 on:** August 18, 2008, 06:47:22 pm »

$y=\mbox{csc}^{-1}x$

$\implies x=\frac{1}{\sin y}$

from this: $\sin y=\frac{1}{x}$ , $\cos y = \sqrt{1-\sin^2 y}=\frac{\sqrt{x^2-1}}{x}$

$\implies \frac{d}{dx}(x)=\frac{d}{dx}\left(\frac{1}{\sin y}\right)$

$\implies 1=-\frac{\cos y}{\sin^2 y}\cdot \frac{dy}{dx}$

$\implies \frac{dy}{dx}=-\frac{\sin^2 y}{\cos y}$

substituting the above expressions for sin y and cos y:

$\implies \frac{dy}{dx}=-\frac{x^{-2}}{\frac{\sqrt{x^2-1}}{x}}$

$\implies \frac{dy}{dx}=-\frac{1}{x\sqrt{x^2-1}}$ as required

squance · « **Reply #2 on:** August 18, 2008, 07:08:09 pm »

Quote from: Mao on August 18, 2008, 06:47:22 pm

$\implies 1=\frac{\sin y - \cos y}{\sin^2 y}\cdot \frac{dy}{dx}$

*typing*

Im not sure how you got 1=\frac{\sin y - \cos y}{\sin^2 y}\cdot \frac{dy}{dx} because i got -cos y (dy/dx)/ (sin^2(y)) using quotient rule....

Mao · « **Reply #3 on:** August 18, 2008, 07:19:35 pm »

yeah, mistake

fixed now

i forgot what $\frac{d}{dx}(1)$ was momentarily

squance · « **Reply #4 on:** August 18, 2008, 07:34:15 pm »

Heheh
thanks for the help

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Author Topic: Derivatives of Inverse Functions - HELP! (Read 1906 times) Tweet Share

squance

Derivatives of Inverse Functions - HELP!

Mao

Re: Derivatives of Inverse Functions - HELP!

squance

Re: Derivatives of Inverse Functions - HELP!

Mao

Re: Derivatives of Inverse Functions - HELP!

squance

Re: Derivatives of Inverse Functions - HELP!

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