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OCW Scholar - Single Variable Calculus 43 Online
OpenStudy (anonymous):

In session 15 of this course I ran in to a wall with the problem. The problem given is: Use imlipicit differentiation to find the derivative of the inverse of \[f(x)=x^2\] for \[x>0\]. Now the way I'm doing it is as follows: \[y = f(x) \rightarrow y=x^2 \rightarrow \sqrt{y}=x\] \[\frac{ d }{ dy }(\sqrt{y}) \frac{ dy }{ dx } = \frac{ d }{ dx } x \rightarrow \frac{y'}{2\sqrt{y}}=1\] \[y'=\frac{1}{2\sqrt{y}} =\frac{1}{2x}\] Of course the answer should be: \[\frac{ 1 }{ 2\sqrt{x} }\] So could someone tell me exactly what I'm doing wrong and explain how I should view the process of implicit differentiation instead? PS: How can I open a question without being bound by length, so that I won't have to edit it?

OpenStudy (phi):

as you know \[ f^{-1}(x)= \sqrt{x} = x^\frac{1}{2} \] and \[ \frac{df^{-1}(x)}{dx} = \frac{1}{2\sqrt{x}}\] If we are given \[ f(x)= x^2 \\ y= x^2 \] then by implicit differentiation \[ dy = 2 x \ dx \] the derivative of the *inverse of f(x)* will be dx/dy i.e. the reciprocal of the derivative of f(x) thus, dividing by the infinitesimal dy, we have \[ 1= 2 x \frac{dx}{dy} \\ \frac{dx}{dy}= \frac{1}{2x} \] using \[ x = \sqrt{y}\] we have \[ \frac{dx}{dy}= \frac{1}{2\sqrt{y}} \] or renaming x to y and vice versa (this is the standard convention) \[ \frac{dy}{dx}= \frac{1}{2\sqrt{x}} \]

OpenStudy (phi):

we could also take the derivative with respect to x: \[ \frac{d}{dx} y= \frac{d}{dx}x^2 \\ \frac{dy}{dx}= 2x \] now flip both sides because we want the reciprocal \[ \frac{dx}{dy}= \frac{1}{2x} \] replace x with \( \sqrt{y}\) \[ \frac{dx}{dy}= \frac{1}{2\sqrt{y}} \] finally because we assume x is the input variable and y is the result, rename x to y and vice versa \[ \frac{d f^{-1}(x)}{dx} = \frac{1}{2\sqrt{x}}\]

OpenStudy (anonymous):

That clears up a lot! So to summarize: For Implicit differentiation you differentiate both "Variable" sides (the function y or f(x) and the variable of that function. You then work the differential product (dy/dx) to "switch" dy with dx (That clears up the ^-1 notation a bunch), simplify and simply exchange y and x for the sake of simplicity. To your second post: This looks a lot more like something I would do. So really all you're doing is differentiate both sides and raise them to the -1th power. After which you exchange symbols (y and x). Is this correct?

OpenStudy (phi):

yes, that is it. I assume you saw the derivation for \[ \frac{d \sin^{-1}(x)}{dx}\] ?

OpenStudy (anonymous):

I did, yes. But it didn't really clarify anything for me.

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