# Thread: Differentiating The natural Exponential Function

1. ## Differentiating The natural Exponential Function

Differentiate the function:
$y = (x^5)(e^{-3lnx})$
Howw??

Differentiate the function:
$y = (x^5)(e^{-3lnx})$
Howw??
The definition of logarithms is such that:

$a=\ln(b)$

means that $e^a=b$, or what is the same thing $e^{\ln(b)}=b$, then:

$e^{-3\ln(x)}=e^{\ln(1/x^3)}=1/x^3$

so:

$y=x^2$,

hence:

$\frac{dy}{dx}=2x$

RonL

I'll give it a try . . .

Differentiate the function: . $y \:= \:(x^5)(e^{-3\ln x})$

The answer is $2x$

Let $e^{-3\ln x} \:=\:z$

Take logs: . $\ln\left(e^{-3\ln x}\right)\:=\:\ln z\quad\Rightarrow\quad -3\ln x\cdot\ln(e)\:=\:\ln z\quad\Rightarrow\quad-3\ln x\:=\:\ln z$

. . $\ln z \:=\:-3\ln x\:=\:\ln\left(x^{-3}\right)\quad\Rightarrow\quad z \:=\:x^{-3}$

Hence: . $e^{-3\ln x} \:=\:x^{-3}$

The function becomes: . $y \:=\:x^5\!\cdot\!x^{-3} \:=\:x^2$

Therefore: . $y' \:=\:2x$

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We can do it head-on . . . (not recommended).

We have: . $y \;=\;\left(x^5\right)\left(e^{-3\ln x}\right)$

Product Rule: . $y' \;=\;\left(x^5\right)\left(e^{-3\ln x}\right)\left(-3\cdot\frac{1}{x}\right) + \left(5x^4\right)\left(e^{-3\ln x}\right)$

We have: . $y' \;=\;-3x^4e^{-3\ln x} + 5x^4e^{-3\ln x} \;=\;2x^4e^{-3\ln x}$

Since $e^{-3\ln x}\:=\:x^{-3}$, we have: . $y' \;=\;2x^4\!\cdot\!x^{-3}\;=\;2x$