derivative of y with respect to x given x and y

[rubing]

Again I am at odds with an answer in my book. The last time members on the forum felt that I was correct and that the info in the answer key must be a typo, but the next problem I am again obtaining a different answer. Question as follows:

find dy/dx

x = √(t²-4) y = √(t²+4)

the rule dy/dx = (dy/dt ) / (dx/dt), I get the following:

√(t²-4) / √(t²+4)

,however my answer key says

√(t²-4) / 3t√(t+2)

i am suspicious maybe I am not using the correct rule to solve these? Thanks for any advice!

 

[srmichael]

Again I am at odds with an answer in my book. The last time members on the forum felt that I was correct and that the info in the answer key must be a typo, but the next problem I am again obtaining a different answer. Question as follows:

find dy/dx

x = √(t²-4) y = √(t²+4)

the rule dy/dx = (dy/dt ) / (dx/dt), I get the following:

√(t²-4) / √(t²+4)

,however my answer key says

√(t²-4) / 3t√(t+2)

i am suspicious maybe I am not using the correct rule to solve these? Thanks for any advice!

Hmmm, that book has issues:

\(\displaystyle \displaystyle \frac{dy}{dt}=\frac{t}{\sqrt{t^2+4}}\)

\(\displaystyle \displaystyle \frac{dx}{dt}=\frac{t}{\sqrt{t^2-4}}\)

\(\displaystyle \displaystyle \frac{dy}{dx}=\frac{\frac{t}{\sqrt{t^2+4}}}{\frac{t}{\sqrt{t^2-4}}}\)

\(\displaystyle \displaystyle \frac{dy}{dx}=\frac{\sqrt{t^2-4}}{\sqrt{t^2+4}}\)

At which point you can simplify by rationalizing the denominator to:

\(\displaystyle \displaystyle \frac{dy}{dx}=\frac{(\sqrt{t^2-4})(\sqrt{t^2+4})}{t^2+4}\)

\(\displaystyle \displaystyle \frac{dy}{dx}=\frac{\sqrt{t^4-16}}{t^2+4}\)

But I don't see how you can get to the answer the book has....