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    What is e^x differentiated w.r.t.y? and why?
    Thanks in advance
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    \frac{\mathrm{d}}{\mathrm{d}y}(e  ^x) = \frac{\mathrm{d}x}{\mathrm{d}y}e  ^x by the chain rule.

    Let z = e^x

    We require dz/dy, but that's dz/dx * dx/dy. But dz/dx = e^x.
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    0, it is the derivative of a constant...

    I'm sure you are looking for a more subtle answer that that though...

    edit: Well, that is unless y is a function of x in which case Swayum's post is the right.
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    I think this thread is crying out for the classic differential operator story!

    The cocky exponential function e^x is strolling along the road insulting the functions he sees walking by. He scoffs at a wandering polynomial for the shortness of its Taylor series. He snickers at a passing smooth function of compact support and its glaring lack of a convergent power series about many of its points. He positively laughs as he passes |x| for being nondifferentiable at the origin. He smiles, thinking to himself, "Damn, it's great to be e^x. I'm real analytic everywhere. I'm my own derivative. I blow up faster than anybody and shrink faster too. All the other functions suck."

    Lost in his own egomania, he collides with the constant function 3, who is running in terror in the opposite direction.

    "What's wrong with you? Why don't you look where you're going?" demands e^x. He then sees the fear in 3's eyes and says "You look terrified!"

    "I am!" says the panicky 3. "There's a differential operator just around the corner. If he differentiates me, I'll be reduced to nothing! I've got to get away!" With that, 3 continues to dash off.

    "Stupid constant," thinks e^x. "I've got nothing to fear from a differential operator. He can keep differentiating me as long as he wants, and I'll still be there."

    So he scouts off to find the operator and gloat in his smooth glory. He rounds the corner and defiantly introduces himself to the operator. "Hi. I'm e^x."

    "Hi. I'm \frac{d}{dy} ."
 
 
 
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Updated: November 11, 2008

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