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OCW Scholar - Single Variable Calculus 70 Online
OpenStudy (anonymous):

Hi, I've been struggling with problem 5A-3G in the problem set. The problem asks to derive the following: tan^−1(x/sqroot(1 − x^2)) I pretty much understand nothing from here. I can't even start on this problem, even knowing what the derivative of inverse tan is.

OpenStudy (anonymous):

Still lost. I understand the derivative of inverse tan, but I don't understand how the chain rule is applied, nor how the derivative of inverse tan leads to the value the MIT solutions reach.

OpenStudy (phi):

can you do this problem \[ \frac{d}{dx} \left( \frac{x}{\sqrt{1-x^2}}\right) \]

OpenStudy (phi):

you can use the quotient rule, or write it as \[ x\ (1-x^2)^{-\frac{1}{2}} \] and use the product rule d (u v) = u dv + v du

OpenStudy (anonymous):

Thanks, I'll check those out.

OpenStudy (anonymous):

I tried using the product rule early with: \[x (1 - x^{2})^{-1/2}\] earlier and that's where some difficulty started. Is there no use of the chain rule there? Don't I have to differentiate the inside of the parenthesis as well?

OpenStudy (phi):

yes, you have to use the chain rule if we let u=x and v = (1-x^2)^(- 1 / 2) the d ( u v) = u dv + v du du with respect to x i.e. du/dx = d/dx x = 1 so the v du part is (1-x^2)^(- 1 / 2) the messy part is u dv u is x dv with respect to x, i.e. dv/dx is d/dx (1-x^2)^(- 1 / 2)

OpenStudy (phi):

\[ \frac{d}{dx} (1-x^2)^{-\frac{1}{2}} \] is the derivative of e.g. u^n (where n is the stuff inside the parens) \[ \frac{d}{dx} u^n = n \ u^{n-1} \frac{d}{dx} u \] usually we do not introduce the variable "u", we just write \[ d u^n = n u^{n-1} du \] which is short-hand for : to take the derivative of u^n with respect of "x" (or whatever variable) we write \[ n u^{n-1} \] and then multiply that by the derivative of u (with respect to "x") \[ n u^{n-1} du\]

OpenStudy (anonymous):

Whew. Alright, thanks, that clarified quite a bit. So at the end of it all, I've got \[X ^{2} * (1-X ^ {2} ) ^ {-3/2} + (1 - X^{-2})^{-1/2}\] Now where do I go from here? I'm not sure how to combine these terms.

OpenStudy (phi):

in other words \[ \frac{d}{dx} (1-x^2)^{-\frac{1}{2}} =-\frac{1}{2} (1-x^2)^{-\frac{1}{2}-1} \frac{d}{dx}(1-x^2)\] now we find the derivative of 1-x^2 \[ \frac{d}{dx}(1-x^2)= \frac{d}{dx} 1 - \frac{d}{dx} x^2\\ = 0 - 2 x^{2-1} \frac{d}{dx} x\\ = 0 - 2x \\= -2x \] and all together \[ \frac{d}{dx} (1-x^2)^{-\frac{1}{2}} =-\frac{1}{2} (1-x^2)^{-\frac{1}{2}-1} ( -2x) \] which simplifies to \[ x (1-x^2)^{-\frac{3}{2}}\] and that goes into \[ \frac{d}{dx} x(1-x^2)^{-\frac{1}{2}} =(x )(x ) (1-x^2)^{-\frac{3}{2}} + (1-x^2)^{-\frac{1}{2}} \\ = \frac{x^2}{ (1-x^2)^{\frac{3}{2}}} + \frac{1}{ (1-x^2)^{\frac{1}{2}} } \]

OpenStudy (phi):

**typo up above: is the derivative of e.g. u^n (where u is the stuff inside the parens)

OpenStudy (anonymous):

Alright, so it seems like I got about where you got. Now I'll try to fit that all into the whole problem and come back if I have any issues.

OpenStudy (phi):

now we take a step backward (or up) we are doing \[ \frac{d}{dx} \tan^{-1}u = \frac{1}{1+u^2} \frac{d}{dx} u \] where \[ u = x(1-x^2)^{-\frac{1}{2}} \] we found du = \[ \left(\frac{x^2}{ (1-x^2)^{\frac{3}{2}}} + \frac{1}{ (1-x^2)^{\frac{1}{2}} } \right)\] we multiply that mess by 1/(1+u^2) but first, replace u with its definition (as a function of "x") and simplify 1/(1+u^2) then multiply the result times du it simplifies to 1/sqr(1-x^2)

OpenStudy (anonymous):

Thanks for the complicated explanation. My head hurts so much now though. I'm hoping this wouldn't be considered an easy problem? I just can't imagine figuring out how to solve it without help.

OpenStudy (anonymous):

Alright, well, trying to do all the algebra in that mess, I ended up lost somewhere anyways and might've done something wrong. I simplified 1/(1+u^2) to 1/(1+x(1-x^2)^-1). From there, I tried multiplying and just got a whole mess, ending with X2 / ((1-x^2)^3/2 + x(1-X^2)^1/2 + 1/(1-X^2)^1/2 + x(1-X^2)^-1/2. I have zero idea what to do from here, and am inclined to think I screwed up somewhere.

OpenStudy (anonymous):

The problem set answers also state that: dy/dx = (1 - x^2)^-3/2, whereas we got something different I believe. I'm tempted to throw in the towel at this point and just move on, I think I've spent 2 hours on this problem.

OpenStudy (phi):

you must be looking at something different.

OpenStudy (phi):

or go to wolfram http://www.wolframalpha.com and type in d/dx tan^−1(x/sqrt(1 − x^2))

OpenStudy (phi):

if you see funny characters, re-load your browser (OpenStudy is having some difficulties)

OpenStudy (phi):

*** I simplified \( 1/(1+u^2) \text{ to } 1/(1+x(1-x^2)^-1) \) **** you should write \[ u= \frac{x}{\sqrt{1-x^2}} \\ u^2 = \frac{x^2}{1-x^2} \] the denominator \[ 1+u^2 = 1+\frac{x^2}{1-x^2} \\ = \frac{1-x^2}{1-x^2}+\frac{x^2}{1-x^2} \\ = \frac{1-x^2+x^2}{1-x^2} \\ = \frac{1}{1-x^2} \] and therefore \[ \frac{1}{1+u^2} = \frac{1-x^2}{1}= (1-x^2) \] now multiply that times du

OpenStudy (anonymous):

Alright, I understand how you simplified it now. i was screwing up somewhere, I'm struggling with segmenting and simplifying the parts in the right order. I also get it now, for some reason I didn't notice how easy it was to simplify du/dx from the previous form into what MIT had as its value. I'm going to try and redo this problem now by myself. Thanks for the help.

OpenStudy (anonymous):

Oookay! I completely understand how this problem is done now. Thanks! Only took me 3 hours :P

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