Differentiate with e^x

**KURO** · October 9th 2008, 04:32 PM

Hi everyone need some help with my homework, I've been trying to solve these for awhile. Thanks

1) y=xe^(-kx)

2) f(x)= xe^(x)cscx

**jpatrie** · October 9th 2008, 07:20 PM

this is my answer for number 1, but im a bit rusty, so dont take it as the correct answer.

$lny=lnx-kxlne$

$lny=lnx -kx$

$\frac{1}{y}\frac{dy}{dx}=\frac{1}{x}+-k-x\frac{dy}{dx}$

$\frac{dy}{dx}(\frac{1}{y}+x)=\frac{1-kx}{x}$

$\frac{dy}{dx}(\frac{1+xy}{y})=\frac{1-kx}{x}$

$\frac{dy}{dx}=\frac{\frac{1-kx}{x}}{\frac{1+xy}{y}}$

$\frac{dy}{dx}=\frac{y-kxy}{y+xy^2}$

$\frac{dy}{dx}=\frac{(xe^{-kx})-kx(xe^{-kx})}{(xe^{-kx})+x(xe^{-kx})^2}$

$\frac{dy}{dx}=\frac{1-kx}{1+x(xe^{-kx})}$

**Jhevon** · October 9th 2008, 07:21 PM

Originally Posted by jpatrie

this is my answer for number 1, but im a bit rusty, so dont take it as the correct answer.

$lny=lnx-kxlne$

$lny=lnx -kx$

${\color{red}\frac{1}{y}\frac{dy}{dx}=\frac{1}{x}+-k-x\frac{dy}{dx}}$

$\frac{dy}{dx}(\frac{1}{y}+x)=\frac{1-kx}{x}$

$\frac{dy}{dx}(\frac{1+xy}{y})=\frac{1-kx}{x}$

$\frac{dy}{dx}=\frac{\frac{1-kx}{x}}{\frac{1+xy}{y}}$

$\frac{dy}{dx}=\frac{y-kxy}{y+xy^2}$

$\frac{dy}{dx}=\frac{(xe^{-kx})-kx(xe^{-kx})}{(xe^{-kx})+x(xe^{-kx})^2}$

$\frac{dy}{dx}=\frac{1-kx}{1+x(xe^{-kx})}$

this is overkill. logarithmic differentiation is not necessary and is too messy for this problem. simply using the product rule will suffice

you also made a mistake where i put the red. so everything after that is wrong

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