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OCW Scholar - Single Variable Calculus 8 Online
OpenStudy (anonymous):

Can anyone help me with Problem Set 1, problem 1F-7? I'm confused. Is the derivative for part (a) just 1/2 ((x - a)^2 - y0^2)^(-1/2)?

OpenStudy (larseighner):

No. If you work this by the chain rule you have to do the inside functions as well. You have the derivative of the outside function. Now you have to multiply by the derivative of the inside function. The inside function is (x-a)^2-y_0^2. You could treat that as a function with an inside-er function (x-a)^2, or you can just expand the square and take the derivative of the polynomial in one fell swoop.

OpenStudy (phi):

\[ \frac{d}{dx}\sqrt{stuff} = \frac{d}{dx}\left(stuff\right)^{\frac{1}{2}} =\frac{1}{2}\left(stuff\right)^{-\frac{1}{2}} \frac{d}{dx}\left(stuff\right) \] so you have done half of the problem. you must multiply your result by \[ \frac{d}{dx} \left( (x-a)^2 + y_0^2\right) = \frac{d}{dx} (x-a)^2 + \frac{d}{dx}y_0^2 \] the y0 is a constant, so its derivative is 0 you have \[ \frac{d}{dx} (x-a)^2 = 2(x-a)^1 \frac{d}{dx} (x-a) \\ 2(x-a) \left(\frac{d}{dx} x - \frac{d}{dx} a\right) \\ 2(x-a) \frac{dx}{dx} \\ 2(x-a)\]

OpenStudy (phi):

the entire answer is \[ \frac{x-a}{\sqrt{(x+a)^2 + y_0^2}} \]

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