Differentiating The natural Exponential Function

• Jan 27th 2007, 12:47 AM
Differentiating The natural Exponential Function
Differentiate the function:
$\displaystyle y = (x^5)(e^{-3lnx})$
Howw?? :(
• Jan 27th 2007, 01:18 AM
CaptainBlack
Quote:

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

The definition of logarithms is such that:

$\displaystyle a=\ln(b)$

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

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

so:

$\displaystyle y=x^2$,

hence:

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

RonL
• Jan 27th 2007, 06:53 AM
Soroban

I'll give it a try . . .

Quote:

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

The answer is $\displaystyle 2x$

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

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

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

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

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

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

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

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

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

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

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