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How to differentiate functions with x in their exponents

  1. Jul 14, 2011 #1
    I'm not sure how to differentiate y with respect to x for:

    y = (7x^2)^[x^2]

    Any ideas?
     
  2. jcsd
  3. Jul 14, 2011 #2
    [itex] y = {(7x^2)}^{x^2} [/itex]

    Mission: To find [itex] \frac{dy}{dx} [/itex]

    We want to eliminate the problem of the exponent [itex] x^2 [/itex].
    There are several ways, but here's one neat way. Take the natural logarithm of both sides. In each step i will write in red what mathematical identity I have used to get to the expression.

    [itex] \ln(y) = \ln({(7x^2)}^{x^2}) [/itex]


    [itex] \ln(y) = x^2\ln(7x^2) [/itex] [itex] \ln{(a^b)} = b\ln(a) [/itex]

    [itex] \ln(y) = x^2\ln{({(\sqrt{7}x)}^2)} [/itex] [itex] ab^2 = {(\sqrt{a}b)}^2 [/itex]

    [itex] \ln(y) = 2x^2\ln{(\sqrt{7}x)} [/itex] [itex] \ln{(a^b)} = b\ln{a} [/itex]

    Now raise [itex] e [/itex] to the power of each side to get.

    [itex] y = e^{2x^2\ln(\sqrt{7}x)} [/itex]

    What remains is just deriving this expression, and to do so you only need to know the chain rule and product rule, and how to derive [itex] e^x [/itex].

    Ok, so lets do the remaining.

    [itex] \frac{dy}{dx} = \frac{d}{dx}e^{2x^2\ln(\sqrt{7}x)} = e^{2x^2\ln(\sqrt{7}x)}\cdot \frac{d}{dx}2x^2\ln(\sqrt{7}x) [/itex] Chain rule

    Now

    [itex] \frac{d}{dx}2x^2\ln(\sqrt{7}x) = 4x\ln(\sqrt{7}x) + 2x^2 \frac{1}{x} = 2x(\ln(7x^2) +1) [/itex] Product rule

    Thus

    [itex] \frac{dy}{dx} = 2x(\ln(7x^2) +1)e^{2x^2\ln(\sqrt{7}x)} = 2x(\ln(7x^2) +1){(7x^2)}^{x^2} [/itex]
     
    Last edited: Jul 14, 2011
  4. Jul 14, 2011 #3
    Use the chain rule with u=x^2, and v=x^2. Factor out the constant, and your answer should appear quickly.
     
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