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Hyperbolic Differentiation

  • Thread starter DiamondV
  • Start date
  • #1
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Homework Statement


Differentiate
gif.gif


Homework Equations


Chain Rule: dg/dx = du/dx . dv/du . dg/dv

The Attempt at a Solution


My answer(wrong):
e33ab0e3d5.jpg


Correct answer provided to us(not mine):
2be4cb75b9.png


I understand the correct solution that was provided to us, but what I don't understand is why my method isnt correct? Also can you check in particular my dv/du . I suspect theres something wrong there.
 

Answers and Replies

  • #2
pasmith
Homework Helper
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The derivative of [itex]3^v = e^{v \log 3}[/itex] is [itex](\log 3) e^{v \log 3} = (\log 3)3^v[/itex], not [itex]3v^{v-1}[/itex] which appears to be what you have.

If you set [itex]v = u\log x[/itex] then you want to calculate [tex]\frac{dv}{du} = \log x + \frac ux \frac{dx}{du}[/tex] and the [itex]\frac{dx}{du}[/itex] cancels with [itex]\frac{du}{dx}[/itex] when you apply the chain rule.

Differentiating functions in exponents is straightforward: By the chain rule, with [itex]a > 0[/itex], [tex]
\frac{d}{dx} a^{f(x)} = (\log a)a^{f(x)} f'(x).[/tex]
 

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