Hyperbolic Differentiation: How Do We Differentiate Functions with Exponents?

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SUMMARY

This discussion focuses on differentiating functions with exponents using the Chain Rule. The correct application of the Chain Rule is highlighted, specifically the formula for differentiating exponential functions: d/dx a^{f(x)} = (\log a)a^{f(x)} f'(x). A common mistake identified is the incorrect calculation of dv/du when differentiating 3^v, which should be (\log 3)3^v instead of 3v^{v-1}. The discussion emphasizes the importance of correctly applying the Chain Rule to avoid errors in differentiation.

PREREQUISITES
  • Understanding of the Chain Rule in calculus
  • Familiarity with exponential functions and their properties
  • Knowledge of logarithmic differentiation
  • Basic skills in calculus, particularly differentiation techniques
NEXT STEPS
  • Study the Chain Rule in more depth, focusing on its applications in differentiation
  • Learn about logarithmic differentiation and its advantages in complex functions
  • Practice differentiating various exponential functions with different bases
  • Explore common mistakes in calculus to improve accuracy in differentiation
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Students studying calculus, particularly those focusing on differentiation techniques, as well as educators looking for examples of common errors in applying the Chain Rule.

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Homework Statement


Differentiate
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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 isn't correct? Also can you check in particular my dv/du . I suspect there's something wrong there.
 
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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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