derivatives

**turtle** · November 12th 2006, 03:25 PM

can somebody please help me find the second derivative of y = x squared times e to the 1/x squared

thanks

**Soroban** · November 12th 2006, 05:44 PM

Hello, turtle!

This problem is a killer . . .

Find the second derivative of: . $y \:= \:x^2\!\cdot\!e^{1/x^2}$

We have: . $y \;= \;x^2\cdot e^{x^{-2}}$

Product Rule: . $y'\;=\;x^2\cdot e^{x^{-2}}(-2x^{-3}) + 2x\cdot e^{x^{-2}} \;=\;-2x^{-1}e^{x^{-2}} + 2xe^{x^{-2}}$

. . . . Factor: . $y'\;=\;2e^{x^{-2}}\left(x - x^{-1}\right)$

Product Rule: . $y''\;= \;2e^{x^{-2}}\left(1 + x^{-2}\right) + 2e^{x^{-2}}\left(-2x^{-3}\right)\left(x - x^{-1}\right)$

. . . . . . . . . . $y''\;=\;2e^{x^{-2}}\left(1 + x^{-2}\right) - 4x^{-3}e^{x^{-2}}\left(x - x^{-1}\right)$

Factor: . $y'' \;= \;2e^{x^{-2}}\left[1 + x^{-2} - 2x^{-3}\left(x - x^{-1}\right)\right]$

. . . . . . $y'' \;=\;2e^{x^{-2}}\left[1 + x^{-2} - 2x^{-2} + 2x^{-4}\right]$

. . . . . . $y''\;=\;2e^{x^{-2}}\left[1 - x^{-2} + 2x^{-4}\right]$

. . . . . . $y''\;=\;2e^{x^{-2}}\left[1 - \frac{1}{x^2} + \frac{2}{x^4}\right]$

. . . . . . $y''\;=\;2e^{x^{-2}}\left[\frac{x^4 - x^2 + 2}{x^4}\right]$

**Soltras** · November 12th 2006, 06:18 PM

You can also try a different approach that is sometimes helpful, but maybe not in this case which is pretty complicated.

Start with $y = x^2 e^{1/x^2}$ .

Take the natural log of both sides:

$\ln y = \ln \left( x^2 e^{1/x^2} \right)$

Expand the logs on the right side:

$\ln y = 2 \ln x + x^{-2}$

Take the derivative (noting the chain rule):

$\frac{1}{y}\,y' = 2x^{-1} - 2x^{-3}$

Move the y over:

$y' = y \left( 2x^{-1} - 2x^{-3} \right)$

Now take the second derivative, applying the product rule on the right side:

$y'' = y' \left( 2x^{-1} - 2x^{-3} \right) + y \left( -2x^{-2} + 6x^{-4} \right)$

Substitute in the y' which was found one step earlier:

$y'' = y \left( 2x^{-1} - 2x^{-3} \right)^2 + y \left( -2x^{-2} + 6x^{-4} \right)$

Factor out the y and simplify the garbage that's left over:

$y'' = y \left[ 2x^{-2} - 2x^{-4} + 4x^{-6} \right]$

Substitute in y (which was the original equation):

$y'' = x^2 e^{1/x^2} \left[ 2x^{-2} - 2x^{-4} + 4x^{-6} \right]$

Oh look, that x^2 can get passed in for a little simplification.
Also, a 2 can get pulled out:

$y'' = 2e^{1/x^2} \left( 1 - x^{-2} + 2x^{-4} \right)$

In hindsight, this seems to be no less complicated than the first solution provided. Sometimes this technique can save considerable time though, and other times it can over complicate. I like this one though because you can see how with an exponential, the derivative is closely related to the function itself (I enjoy differential equations).

**Soroban** · November 13th 2006, 08:02 AM

I like it, Soltras!

I considered log differentiation but decided to do it head-on.

I especially like the part where $\ln\left(e^{^{\frac{1}{x^2}}}\right)$ becomes $x^{-2}$ . . . delightful!

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