Exponential Differentiation with 1 Over x

[Jason76]

Formula for exponential differentiation: \(\displaystyle e^{u}(du)\)

\(\displaystyle y = e^{\frac{1}{x}}\) That is e to the 1 over x, if too small to read.

Couldn't you change it to \(\displaystyle e^{x^{-2}}\) (e to the x to the -2 power) and proceed from there, or do something different?

 

[DrPhil]

Formula for exponential differentiation: \(\displaystyle e^{u}(du)\)

\(\displaystyle y = e^{\frac{1}{x}}\) That is e to the 1 over x, if too small to read.

Couldn't you change it to \(\displaystyle e^{x^{-2}}\) (e to the x to the -2 power) and proceed from there, or do something different?

\(\displaystyle \displaystyle 1/x = x^{-1}\)

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

There is a -1 power, NOT -2, in the function y(x).
After you apply the chain rule, differentiating 1/x, then you will have a -2 power in the derivative y'(x).

 

[JeffM]

Formula for exponential differentiation: \(\displaystyle e^{u}(du)\)

\(\displaystyle y = e^{\frac{1}{x}}\) That is e to the 1 over x, if too small to read.

Couldn't you change it to \(\displaystyle e^{x^{-2}}\) (e to the x to the -2 power) and proceed from there, or do something different?

Jason

What in the world are you talking about?

\(\displaystyle e^{(x^{(-2)})} = e^{(1/x^2)} \ne e^{(1/x)}\ if\ x \ne 1.\)

\(\displaystyle u = \dfrac{1}{x} \implies \dfrac{du}{dx} = - \dfrac{1}{x^2}.\)

\(\displaystyle y = e^{(1/x)} \implies y = e^u \implies \dfrac{dy}{du} = e^u \implies \dfrac{dy}{dx} = \dfrac{dy}{du} * \dfrac{du}{dx} = e^u \left(- \dfrac{1}{x^2}\right) = - \dfrac{e^{(1/x)}}{x^2}.\)

 

[Jason76]

The best route to go (and maybe the only one) is to use the quotient rule on \(\displaystyle du\) Because, rewriting the fraction doesn't lead to the same answer as in the book. After that, you just do \(\displaystyle e^{u}(du)\) I'll give an example shortly.

 

[DrPhil]

The best route to go (and maybe the only one) is to use the quotient rule on \(\displaystyle du\) Because, rewriting the fraction doesn't lead to the same answer as in the book. After that, you just do \(\displaystyle e^{u}(du)\) I'll give an example shortly.

There is only one answer for du/dx.

Quotient: \(\displaystyle \displaystyle u = \dfrac{1}{x} \Longrightarrow \dfrac{du}{dx} = -\dfrac{1}{x^2}\)

Power Law: \(\displaystyle \displaystyle u = x^{-1} \Longrightarrow \dfrac{du}{dx} = -\ x^{-2}\)

These are identical.

 

[Jason76]

Oh, sorry, mistake, NOT 1 over x. 1 over x squared (that was what was in my book)

\(\displaystyle y = e^{\frac{1}{x^{2}}}\)

Quotient rule where \(\displaystyle 1\) is \(\displaystyle f\) and \(\displaystyle x^{2}\) is \(\displaystyle g\).

Quotient Rule: \(\displaystyle \dfrac{g(f') - f(g')}{g^{2}}\)

\(\displaystyle u = e^{\frac{1}{x^{2}}}\)

\(\displaystyle du = \dfrac{x^{2}(0) - 1(2x)}{x^{4}}\)

\(\displaystyle du = -\dfrac{2x}{x^{4}}\)

Final answer:

\(\displaystyle e^{\frac{1}{x^{2}}}( -\frac{2}{x^{3}})\)

But let's go ahead and do one over x.

\(\displaystyle y = e^{\frac{1}{x}}\)

Quotient rule where \(\displaystyle 1\) is \(\displaystyle f\) and \(\displaystyle x\) is \(\displaystyle g\).

\(\displaystyle u = e^{\frac{1}{x}}\)

\(\displaystyle du = \dfrac{x(0) - 1(1)}{x^{2}}\)

\(\displaystyle du = -\dfrac{1}{x^{2}}\)

Final answer:

\(\displaystyle e^{\frac{1}{x^{2}}}(-\dfrac{1}{x^{2}})\)

 

[daon2]

^ Please be careful about what you write. \(\displaystyle du\) is not the same as \(\displaystyle u'=\dfrac{du}{dx}\) (what you meant). Also, you let \(\displaystyle u=e^{1/x}\), and then used \(\displaystyle u\) as the exponent! It's all very confusing. Write each equation as you would a sentence in English. One letter can change the entire meaning!

 

[Deleted member 4993]

Formula for exponential differentiation: \(\displaystyle e^{u}(du)\)

\(\displaystyle y = e^{\frac{1}{x^2}}\) That is e to the 1 over x, if too small to read.

Couldn't you change it to \(\displaystyle e^{x^{-2}}\) (e to the x to the -2 power) and proceed from there, or do something different?

To salvage the whole thing:

\(\displaystyle \dfrac{d}{dx}\left [e^{\frac{1}{x^2}}\right ]\)

\(\displaystyle =\left [e^{\frac{1}{x^2}}\right ] * \dfrac{d}{dx}\left [{\frac{1}{x^2}}\right ]\)

\(\displaystyle =\left [e^{\frac{1}{x^2}}\right ] * (-2) * \left [{\dfrac{1}{x^3}}\right ]\)

\(\displaystyle = -\dfrac{2 * e^{\frac{1}{x^2}}}{x^3}\)

\(\displaystyle = -2 * x^{-3} * e^{\frac{1}{x^2}}\)