Differentiation of e^(-x) / ln(x)

[Petra de Ruyter]

Hi there
Getting stuck on this equation.
e^(-x) / ln(x)

Solving it by quotient rule, however answer has extra x in numerator.
using the dy/dx = (v(du/dx)-u(dv/dx))/v^2

dy/dx = (ln(x)*-e^(-x) - (-e^(-x)*1/x)/ln(x)^2

Answer = (-e^(-x) (x ln(x)-1))/x(ln(x))^2

Would appreciate help with breaking this down. Cheers

 

[fzero]

Your result has a factor
$$ \ln x - \frac{1}{x} = \frac{ x\ln x -1}{x}$$
so it is equivalent to the form given as the answer.

 

[Petra de Ruyter]

Thank you for the reply. Sorry, not sure that I understand your response. Can you please break it down, to show where the "x" value is coming from.

Cheers.

 

[fzero]

You wrote dy/dx = (ln(x)*-e^(-x) - (-e^(-x)*1/x)/ln(x)^2, so if I format this we have
$$ \frac{ \ln x (-e^{-x}) - (-e^{-x}) \frac{1}{x} }{ (\ln x)^2} = - \frac{e^{-x}}{ (\ln x)^2} \left( \ln x - \frac{1}{x} \right).$$
Now we can write
$$ \ln x - \frac{1}{x} = \frac{x}{x} \ln x - \frac{1}{x} = \frac{ x \ln x - 1}{x}.$$
When we put this into the previous expression, we get
$$ - \frac{e^{-x}}{ (\ln x)^2} \frac{ x \ln x - 1}{x} = - e^{-x}\frac{ x \ln x - 1}{x(\ln x)^2},$$
which is the same as the answer that you wrote as (-e^(-x) (x ln(x)-1))/x(ln(x))^2.

 

[Petra de Ruyter]

Many thanks for the response. Thank you. Now I get it.
Cheers

 

[Petra de Ruyter]

You wrote dy/dx = (ln(x)*-e^(-x) - (-e^(-x)*1/x)/ln(x)^2, so if I format this we have
$$ \frac{ \ln x (-e^{-x}) - (-e^{-x}) \frac{1}{x} }{ (\ln x)^2} = - \frac{e^{-x}}{ (\ln x)^2} \left( \ln x - \frac{1}{x} \right).$$
Now we can write
$$ \ln x - \frac{1}{x} = \frac{x}{x} \ln x - \frac{1}{x} = \frac{ x \ln x - 1}{x}.$$
When we put this into the previous expression, we get
$$ - \frac{e^{-x}}{ (\ln x)^2} \frac{ x \ln x - 1}{x} = - e^{-x}\frac{ x \ln x - 1}{x(\ln x)^2},$$
which is the same as the answer that you wrote as (-e^(-x) (x ln(x)-1))/x(ln(x))^2.

You wrote dy/dx = (ln(x)*-e^(-x) - (-e^(-x)*1/x)/ln(x)^2, so if I format this we have
$$ \frac{ \ln x (-e^{-x}) - (-e^{-x}) \frac{1}{x} }{ (\ln x)^2} = - \frac{e^{-x}}{ (\ln x)^2} \left( \ln x - \frac{1}{x} \right).$$
Now we can write
$$ \ln x - \frac{1}{x} = \frac{x}{x} \ln x - \frac{1}{x} = \frac{ x \ln x - 1}{x}.$$
When we put this into the previous expression, we get
$$ - \frac{e^{-x}}{ (\ln x)^2} \frac{ x \ln x - 1}{x} = - e^{-x}\frac{ x \ln x - 1}{x(\ln x)^2},$$
which is the same as the answer that you wrote as (-e^(-x) (x ln(x)-1))/x(ln(x))^2.

I made a mistake...should be
You wrote dy/dx = (ln(x)*-e^(-x) - (e^(-x)*1/x)/ln(x)^2

Cheers Petra

 

[fzero]

I made a mistake...should be
You wrote dy/dx = (ln(x)*-e^(-x) - (e^(-x)*1/x)/ln(x)^2

Cheers Petra

Sorry I should have caught that. From the start we have
$$ \begin{split} \frac{d}{dx} \frac{e^{-x}}{\ln x} &= \frac{1}{\ln x} \frac{d}{dx} e^{-x}+e^{-x} \frac{d}{dx} \frac{1}{\ln x} \\
& = - \frac{e^{-x}}{\ln x} - e^{-x} \frac{1}{x (\ln x)^2} \\
& = - e^{-x} \frac{1}{x (\ln x)^2} ( x\ln x +1). \end{split}$$
Are you sure there isn't a + sign in the answer that was given?

 

[Petra de Ruyter]

No, I've double checked the answer. It has a "-" which I don't think is correct.

Cheers Petra

 

[Mark44]

Getting stuck on this equation.
e^(-x) / ln(x)

Minor point -- the above is not an equation, with the main clue being that there is no =. An equation indicates that two quantities have the same value; i.e., are equal.

 

[Petra de Ruyter]

My mistake, used the wrong forum category and terminology.