Question

PLEASE CALCULATE DERIVATIVE WITH RESPECT TO X: TAN^-1(X/SQRT(1-X^2). (From Unit 1 Problem Set 2 Section 5A.3G)

If we set (x/sqrt(1-x^2) to equal y, then I understand how to derive dy/dx. However, according to the solutions method, we divide dy/dx by 1+y^2 to get d/dx tan^-1. I don't understand why we do this.

According to another method we could use the chain rule where the derivative of the function is d/du tan^-1 * du/dx, where u = y, from above, and d/du tan^-1 = 1/(1=u^2). How do we work out this last derivative?

Answer 1

First we should take advantage of the definition for the inverse tangent function:

\(tan^{-1}\left(\cfrac{x}{\sqrt{1-x^2}}\right) = y\)

\(\cfrac{x}{\sqrt{1-x^2}} = tan(y)\)

Now its just a matter of applying the rules for derivatives:

\([ x * (1-x^2)^{-1/2} = tan(y) ]'\)

[fg]' = fg'+f'g

f = x ; f' = 1
g = \((1-x^2)^{-1/2}\); g' = \(\cfrac{-2x}{-2}(1-x^2)^{-3/2}\)

[tan(y)]' = \(sec^2(y) * y'\)
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\(\cfrac{x^2}{(1-x^2)^{3/2}} + \cfrac{1}{(1-x^2)^{1/2}} = sec^2(y)*\frac{dy}{dx}\)

\(\cfrac{x^2}{(1-x^2)^{3/2}} + \cfrac{(1-x^2)^{2/2}}{(1-x^2)^{1/2}(1-x^2)^{2/2}} = sec^2(y)*\frac{dy}{dx}\)

\(\cfrac{x^2}{(1-x^2)^{3/2}} + \cfrac{(1-x^2)}{(1-x^2)^{3/2}} = sec^2(y)*\frac{dy}{dx}\)

\(\cfrac{x^2 + (1-x^2)}{(1-x^2)^{3/2}} = sec^2(y)*\frac{dy}{dx}\)

\(\cfrac{1}{(1-x^2)^{3/2}} = sec^2(y)*\frac{dy}{dx}\)

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Now that the algebra is done, we can start to mess with that sec^2

\(\cfrac{1}{sec^2(y)(1-x^2)^{3/2}} = \cfrac{dy}{dx}\)

Ive tried to draw it out in the picture attached how:

sec^2(y) = 1/(1-x^2)

\(1\div \cfrac{(1-x^2)^{3/2}}{(1-x^2)^{2/2}} = \cfrac{dy}{dx}\)

\(\cfrac{1}{(1-x^2)^{1/2}}=\cfrac{dy}{dx}\)

http://www.wolframalpha.com/input/?i=d%2Fdx+tan^-1%28x%2Fsqr%281-x^2%29%29

Answer 2

it lost the last bit of me reply ...
\[1\div \cfrac{(1-x^2)^{3/2}}{(1-x^2)^{2/2}} = \cfrac{dy}{dx}\]
\[1\div {(1-x^2)^{1/2}} = \cfrac{dy}{dx}\]
\[\frac{1}{(1-x^2)^{1/2}} = \cfrac{dy}{dx}\]

Answer 3

I believe that the other method is a way to "clean up" the work so that you can keep track of it better

Answer 4

Very clear explanation, thank you very much again. I prefer your method over the other one

Answer 5

y'=\frac{1}{1+\left(\frac{x}{\sqrt{1-x^{2}}}\right)^{2}}*\frac{\sqrt{1-x^{2}}-\frac{-2x^{2}}{2\sqrt{1-x^{2}}}}{\left(\sqrt{1-x^{2}}\right)^{2}}

y'=\frac{1}{1+\frac{x^{2}}{1-x^{2}}}*\frac{\sqrt{1-x^{2}}+\frac{x^{2}}{\sqrt{1-x^{2}}}}{1-x^{2}}

y'=\frac{1}{\frac{1-x^{2}+x^{2}}{1-x^{2}}}*\frac{\frac{\left(\sqrt{1-x^{2}}\right)^{2}+x^{2}}{\sqrt{1-x^{2}}}}{1-x^{2}}

y'=\frac{1}{\frac{1}{1-x^{2}}}*\frac{\frac{1-x^{2}+x^{2}}{\sqrt{1-x^{2}}}}{1-x^{2}}

y'=\frac{1}{\frac{1-x^{2}}{1-x^{2}}}*\frac{1}{\sqrt{1-x^{2}}}

y'=\frac{1}{\sqrt{1-x^{2}}}