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  • 4 years ago

This question refers to lecture 6 of the Single Variable Calculus course, where Dr. Jerison is showing how to calculate d/dx a^x. (18-01sc-single-variable-calculus-fall-2010 Session 17: The Exponential Function, its Derivative, and its Inverse) Isn't there an error at 18:12 of the lecture? The equation d/dx f(kx) = k f'(kx) mixes Newton's and Leibnitz's notation, but, using Newton's, the equation is f'(kx) = k f'(kx). It seems this could only be correct for k=1 (or f'(kx) = 0). But k is "any number", not just 1; and f'(kx) = 2^(kx) ln(kx), so f'(kx) could only be 0 if kx = 1. Well, that rules out k being "any number - it has to be the reciprocal of x - and it couldn't be 0. Any help explaining this apparent difficulty with the proof would be appreciated. Thanks.

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  1. anonymous
    • 4 years ago
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    d/dx f(kx) is not equivalent to f'(kx), but rather (f(kx))'. In this case f(kx) is a composition of functions where f( ) is the outer function and kx is the inner function. Apply the chain rule. Multiply the derivative of the outer function - f'(kx) - by the derivative of the inner function - k. So d/dx f(kx) = kf'(kx). Or (f(kx))' = kf'(kx).

  2. anonymous
    • 4 years ago
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    I think I understand what you mean: When Dr Jerison writes f'(kx), that means the deriv with respect to kx of f(kx). [Would that be d/d(kx) f(kx) in Leibnitz notation?] And if u = kx, then, by the chain rule, d/dx f(u) = d/du f(u) * du/dx d/dx f(u) = d/du f(u) * d/dx kx d/dx f(u) = d/du f(u) * k which, replacing the rest of the u's with kx's gives: d/dx f(kx) = d/d(kx) f(kx) * k or as Dr Jerison correctly wrote: d/dx f(kx) = k * f'(kx) Please tell me if I got anything wrong here. And thanks so much for your taking the time to answer.

  3. anonymous
    • 4 years ago
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    Looks good to me. Don't know enough about notation to say if it's conventional to write d/d(kx). Conceptually though, you seem to have it as far as I can tell.

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