How would I derive this?

**kmjt** · June 5th 2010, 02:15 PM

$y=x^2(e^{-2x})$

What steps would I have to do to get the derivative?

**mr fantastic** · June 5th 2010, 02:18 PM

Originally Posted by kmjt

$y=x^2(e^{-2x})$

What steps would I have to do to get the derivative?

Use the product rule. Please show all your work and say where you're stuck if you need more help.

**e^(i*pi)** · June 5th 2010, 02:19 PM

Originally Posted by kmjt

$y=x^2(e^{-2x})$

What steps would I have to do to get the derivative?

Use the product rule (and don't forget the chain rule on $e^{-2x}$ )

**kmjt** · June 5th 2010, 02:50 PM

y'= (2x)(e^-2x) + (x^2)(-2e^-2x)

I haven't derived for so long, how would I simplify

**mr fantastic** · June 5th 2010, 02:59 PM

Originally Posted by kmjt

y'= (2x)(e^-2x) + (x^2)(-2e^-2x)

I haven't derived for so long, how would I simplify

Factorise - There is a common factor ....

**AllanCuz** · June 5th 2010, 03:13 PM

Originally Posted by kmjt

$y=x^2(e^{-2x})$

What steps would I have to do to get the derivative?

You can also use logs!

$lny = lnx^2 + lne^{-2x}$

$lny = 2lnx -2x$

Implicitly derive

$\frac{1}{y} y^{ \prime } = \frac{2}{x} - 2$

$y^{ \prime } = ( \frac{2}{x} - 2 )x^2(e^{-2x})$

**kmjt** · June 5th 2010, 05:43 PM

I ended up getting xe^-2x [ 2 - 2x ] but how do I set this equal to 0 to find the intervals of increase and decrease?

**mr fantastic** · June 5th 2010, 06:41 PM

Originally Posted by kmjt

I ended up getting xe^-2x [ 2 - 2x ] but how do I set this equal to 0 to find the intervals of increase and decrease?

From the null factor law: $xe^{-2x} (2 - 2x) = 0 \Rightarrow$ $x = 0$ or $2 - 2x = 0$ . Note that one of the solutions does NOT correspond to a turning point.

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