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## Main Question or Discussion Point

First off: I think I understand the chain rule and how it derives from

[tex]

\lim_{h \to 0} \frac{ f(x+h)-f(x)}{h}

[/tex]

and how to apply the chain rule when taking the derivative of an implicit function. The textbook I am reading Applied Calculus (by B. Rockett) uses the following example on differentiating implicitly:

we are given the following function

[tex]x^2+y^2 = 25;[/tex] We differentiate both sides with respect to [itex]x[/itex]:

[tex]\frac{d}{dx}x^2 + \frac{d}{dx} y^2 = \frac{d}{dx} 25[/tex] then comes the part that confuses me. Taking the derivative of [itex]x^2[/itex] and [itex]25[/itex] is not the problem, the thing that I cant seem to get into my head is that middle part. OK [itex]y[/itex] is a function of [itex]x[/itex], so we ca apply the power rule (on the right).

[tex]2x+ 2y \frac{dy}{dx} = 0;\hspace{20mm} \frac{d}{dx} y^n = n \cdot y^{n-1} \frac{dy}{dx}[/tex] 1. but how does the [itex]y[/itex] in d

2. how can one express [itex]\frac{dy}{dx}[/itex] using limits as in the first equation in this post?

3. what does 'with respect to _' mean formally?

[tex]

\lim_{h \to 0} \frac{ f(x+h)-f(x)}{h}

[/tex]

and how to apply the chain rule when taking the derivative of an implicit function. The textbook I am reading Applied Calculus (by B. Rockett) uses the following example on differentiating implicitly:

we are given the following function

[tex]x^2+y^2 = 25;[/tex] We differentiate both sides with respect to [itex]x[/itex]:

[tex]\frac{d}{dx}x^2 + \frac{d}{dx} y^2 = \frac{d}{dx} 25[/tex] then comes the part that confuses me. Taking the derivative of [itex]x^2[/itex] and [itex]25[/itex] is not the problem, the thing that I cant seem to get into my head is that middle part. OK [itex]y[/itex] is a function of [itex]x[/itex], so we ca apply the power rule (on the right).

[tex]2x+ 2y \frac{dy}{dx} = 0;\hspace{20mm} \frac{d}{dx} y^n = n \cdot y^{n-1} \frac{dy}{dx}[/tex] 1. but how does the [itex]y[/itex] in d

**y**/dx come about on the right side of the equation? why isn't it simply [itex] 2y \frac{d}{dx} [/itex] (without the y)?2. how can one express [itex]\frac{dy}{dx}[/itex] using limits as in the first equation in this post?

3. what does 'with respect to _' mean formally?

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