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tea222
ANSWERS: 5
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assume, y=e^xy. take its ln... =>lny=xy lne now differentiate it with respect to x...
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The derivative of e^A is e^A times the derivative of A. If you're deriving wrt t you'd get e^(xy) (x' y + x y') where x' and y' are the derivatives wrt t of x and y and I used the product rule to differentiate xy.
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Like a function of x {f(x)} implicitly introduces a second variable (y), a function of x and y {f(x,y)} implicitly introduces a third variable (z). This takes us into the topic of Multivariable Calculus http://en.wikipedia.org/wiki/List_of_multivariable_calculus_topics In single variable calculus, one can ask for "the derivative" and, because there is only one independent variable, it is implicit that the questioner is asking the derivitave with respect to the independent variable. But, in multivariable calculus, the derivative becomes a partial derivative with resepect to one of the independent variables. http://en.wikipedia.org/wiki/Partial_derivative Therefore, you must ask for a partial derivative with respect a specific independent variable. The partial derivative of e^(xy) with respect to x [δ{f(x,y)}/δx] is computed as if y is a constant value and the partial derivative with respect to y is computed as if the x value is a constant [δ{f(x,y)}/δy]
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or the proof goes like this: 2. derivative of e^x = e^x times x' (that is, e to the x times the derivative of x) = e^x times 1; Thus, the textbooks always show that the derivative of e^x is e^x.
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derivative of e^xy would technically be the gradient of a scalar function of x and y (like an electrical potential field, or a function of density over position). To "differentiate" e^xy with the respect to x means that you "chop" the function along a constant y value (set y=constant to turn a "slice" of the function into z=f(x) rather than z=f(x,y)), and then differentiate with respect to x. You do this with y, and you have the "gradient," which gives the DIRECTION and magnitude of greatest change for a function (so the gradient of a "hill" would always point radially outward).
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