Do i just use the chain rule to differentiate 3^2x

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Discussion Overview

The discussion revolves around the differentiation of the expression \(3^{2x}\). Participants explore various methods for applying differentiation rules, particularly the chain rule, and provide alternative approaches to reach the derivative.

Discussion Character

  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant asks if the expression is \(3^{2x}\) and confirms that the chain rule can be used for differentiation.
  • Another participant suggests rewriting \(3^{2x}\) as \(9^x\) to simplify the differentiation process.
  • A different participant introduces the formula for differentiating \(a^u\) and emphasizes that \(a\) is a constant and \(u\) is a function of \(x\).
  • Another approach is presented where the natural logarithm is applied to both sides of the equation \(y=3^{2x}\), leading to a differentiation method that incorporates constants and results in a derivative expressed in terms of \(3^{2x}\) and \(\ln(3)\).

Areas of Agreement / Disagreement

Participants do not reach a consensus on a single method for differentiation, as multiple approaches are suggested and discussed.

Contextual Notes

Some participants reference specific differentiation rules and methods without establishing a common understanding of their applicability or limitations in this context.

Who May Find This Useful

Individuals interested in differentiation techniques, particularly those involving exponential functions and the chain rule, may find this discussion relevant.

Angus
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Do i just use the chain rule to differentiate 3^2x
 
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Is that [tex]3^{2x}[/tex]? If so, yes, you can use the chain rule, or you could write it as [tex](3^2)^x = 9^x[/tex] and differentiate that instead.
 
You might not know the chain rule for forms a^u. I certainly didn't encounter it a lot. Use this:

[tex]\frac{d}{dx}a^u = a^u(\ln a)\frac{du}{dx}[/tex]

a is a constant. u is a function of x.
 
Last edited:
Or you can do the following
[tex]y=3^{2x} => ln(y)=2x*ln(3)[/tex]
Now differentiate both sides (note : ln(3) is just a constant) :
[tex]\frac {d} {dx} ln(y) = log(3)*{\frac{d} {dx}} 2x = 2*ln(3)[/tex]

But [tex]\frac {d} {dx} ln(y) =\frac{1} {y} \frac {dy} {dx} = \frac {1} {3^{2x}} \frac {dy} {dx}[/tex]

So [tex]\frac {dy} {dx} = 3^{2x}*2ln(3)[/tex]
 

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