# Thread: How would I derive this?

1. ## How would I derive this?

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

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

2. 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.

3. 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 $e^{-2x}$)

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

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

5. 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 ....

6. 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

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

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

7. 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?

8. 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.