Differential, logarithm, exponent

[ray5450]

My question is about differentiating the following...thanks...

Given: ln(y)=-axln(a^x).
Given: My question comes before the final answer.

Since ln(a^x)=xln(a), why can it not be immediately substituted above as ln(y)=-aln(a)x², where:

dy . 1
dx y ⁼ -2aln(a)x ?

It is not the same as multiplying out as:

dy . 1
dx y ⁼-axa^xln(a) + -aln(a^x) .

 

[soroban]

Hello, ray5450!

Where did this problem come from?

\(\displaystyle \text{Differentiate: }\:\ln(y)\:=\:-ax\ln(a^x)\)

Yes, this can be written: .\(\displaystyle \ln(y) \:=\:-ax^2\ln(a)\)

Note that \(\displaystyle \ln(a)\) is a constant.
Hence, we have: .\(\displaystyle \ln(y) \:=\:-a\ln(a)\cdot x^2\)

Differentiate implicitly: .\(\displaystyle \dfrac{1}{y}y' \:=\:-a\ln(a)\cdot 2x\)

Then: .\(\displaystyle y' \;=\;-2a\ln(a)\cdot x \cdot y\) .[1]


Back to the beginning . . .
\(\displaystyle \ln(y) \:=\:-ax\ln(a^x) \quad\Rightarrow\quad \ln(y) \:=\:\ln\left(a^{-ax^2}\right) \)
. . Hence: .\(\displaystyle y \:=\:a^{-ax^2}\)

Substitute into [1]: .\(\displaystyle y' \;=\;-2a\ln(a)\cdot x \cdot a^{-ax^2}\)


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Here is my question . . .

If the original function was: .\(\displaystyle y \:=\:a^{-ax^2}\)
. . it can be differentiated directly.

. . \(\displaystyle y' \;=\;a^{-ax^2}\cdot (-2ax)\cdot \ln(a)\)

which produces the same result.

 

[ray5450]

Thank-you, I did exactly what you did. I got this from an old calculus book that I used to practice. The 2nd solution that I posted is what the book had. After posting here, I found another edition of the book and it must have been corrected as the solution was the same thing as we did. (Darn old book!)