# How would I derive this?

• Jun 5th 2010, 02:15 PM
kmjt
How would I derive this?
$\displaystyle y=x^2(e^{-2x})$

What steps would I have to do to get the derivative?
• Jun 5th 2010, 02:18 PM
mr fantastic
Quote:

Originally Posted by kmjt
$\displaystyle y=x^2(e^{-2x})$

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.
• Jun 5th 2010, 02:19 PM
e^(i*pi)
Quote:

Originally Posted by kmjt
$\displaystyle y=x^2(e^{-2x})$

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

Use the product rule (and don't forget the chain rule on $\displaystyle e^{-2x}$)
• Jun 5th 2010, 02:50 PM
kmjt
y'= (2x)(e^-2x) + (x^2)(-2e^-2x)

I haven't derived for so long, how would I simplify (Speechless)
• Jun 5th 2010, 02:59 PM
mr fantastic
Quote:

Originally Posted by kmjt
y'= (2x)(e^-2x) + (x^2)(-2e^-2x)

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

Factorise - There is a common factor ....
• Jun 5th 2010, 03:13 PM
AllanCuz
Quote:

Originally Posted by kmjt
$\displaystyle y=x^2(e^{-2x})$

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

You can also use logs!

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

$\displaystyle lny = 2lnx -2x$

Implicitly derive

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

$\displaystyle y^{ \prime } = ( \frac{2}{x} - 2 )x^2(e^{-2x})$
• Jun 5th 2010, 05:43 PM
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?
• Jun 5th 2010, 06:41 PM
mr fantastic
Quote:

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