# Math Help - Find the derivative of this product

1. ## Find the derivative of this product

$\sqrt{z} * e^{-z}$

2. Originally Posted by sk8erboyla2004
$\sqrt{z} * e^{-z}$
$\sqrt{z} * e^{-z}$

*Chain rule & product rule

First rewrite as:

$(z)^{\frac{1}{2}} \cdot e^{-z}$

Take a stab at it and come back if you're still stuck!

3. $\sqrt{z} * e^{-z}$

Find derivative

This is the answer I came up with

$\frac{1}{2\sqrt{z}}e^{-z}+\sqrt{z}e^{-z}$

however the solution the manual was rather subtraction instead of addition ?

Also

$\sqrt{(x^{2}*5^{x})^{3}}$

Urgent response needed
find the derivative

I got this not really simplified and im not sure how to

$\frac{3}{2}\sqrt{(x^{2}*5^{x})}[5^{x}(2x+x^{2}(ln5)]$

4. Almost, note my correction....

Originally Posted by sk8erboyla2004
$\sqrt{z} * e^{-z}$

Find derivative

This is the answer I came up with

$\frac{1}{2\sqrt{z}}e^{-z}-\sqrt{z}e^{-z}$

however the solution the manual was rather subtraction instead of addition ?

Also

$\sqrt{(x^{2}*5^{x})^{3}}$

Urgent response needed
find the derivative

I got this not really simplified and im not sure how to

$\frac{3}{2}\sqrt{(x^{2}*5^{x})}[5^{x}(2x+x^{2}(ln5)]$
Because the derivative of $e^{-z}$ is $e^{-z}$ times the derivative of it's exponent
which is $-z$.

As for the second one I would simplify first. Note that

$(x^2\cdot 5^x)^{3/2}= x^3\cdot 5^{3x/2}$, which is easier to work with.

5. Originally Posted by sk8erboyla2004
$\sqrt{z} * e^{-z}$

find derivative

this is the answer i came up with

$\frac{1}{2\sqrt{z}}e^{-z}+\sqrt{z}e^{-z}$

however the solution the manual was rather subtraction instead of addition ? the reason the book answer was subtraction, is because they moved the $e^{-z}$ to the denominator and turned it into $\frac{\sqrt{z}}{e^z}$ & used the quotient rule.
so you would have:
$\frac{\sqrt{z}}{e^z}$

$= \frac{z^{\frac{1}{2}}e^z - \frac{1}{2}z^{-\frac{1}{2}}e^z}{e^{2z}}$

also

$\sqrt{(x^{2}*5^{x})^{3}}$

urgent response needed
find the derivative

i got this not really simplified and im not sure how to

$\frac{3}{2}\sqrt{(x^{2}*5^{x})}[5^{x}(2x+x^{2}(ln5)]$

see above in red! For the second one, change the square root into $^{\frac{1}{2}}$ and then do your chain rule. Then when you are finished taking the derivative, you can change any of those left back into roots.

6. I would not use the chain rule on the second and there's no reason to do a quotient rule on the first.
He/she left out the derivative of -z.

7. Originally Posted by sk8erboyla2004
$\sqrt{z} * e^{-z}$

Find derivative

This is the answer I came up with

$\frac{1}{2\sqrt{z}}e^{-z}+\sqrt{z}e^{-z}$

however the solution the manual was rather subtraction instead of addition ?
Mr F says: That's because the derivative of ${\color{red}e^{-z}}$ is ${\color{red}- e^{-z}}$.

[snip]
..

8. Originally Posted by matheagle
I would not use the chain rule on the second and there's no reason to do a quotient rule on the first.
He/she left out the derivative of -z.
How would you do the second without the chain rule?

$\sqrt{(x^{2}*5^{x})^{3}}$

9. Using $(ab)^c=a^cb^c$ we can simplify this to

$x^3(5^{3/2})^x$ and I would just use the product rule...

and get $x^3(5^{3/2})^x\ln(5^{3/2})+3x^2(5^{3/2})^x$.

This is esier to simplifier than the one you get via the chain rule...

$x^2(5^{3/2})^x\biggl( x\ln(5^{3/2})+3\biggr)$.

10. Originally Posted by matheagle
Using $(ab)^c=a^cb^c$ we can simplify this to

$x^3(5^{3/2})^x$ and I would just use the product rule...

and get $x^3(5^{3/2})^x\ln(5^{3/2})+3x^2(5^{3/2})^x$.

This is esier to simplifier than the one you get via the chain rule.
Ah, yes. I like making more work for myself

11. Not many people would notice that.
I bet the solution manual used the chain rule.
You're way too hard on yourself.
I like solving problems several ways, that way I can check my answers.

12. Originally Posted by mr fantastic
..

care to further elaborate how it is negative?

13. Originally Posted by sk8erboyla2004
care to further elaborate how it is negative?
The derivative of $e^{g(x)}$ is $g'(x)e^{g(x)}$.

So the derivative of $e^{-x}$ is $(-1)e^{-x}=-e^{-x}$.

14. Originally Posted by matheagle
The derivative of $e^{g(x)}$ is $g'(x)e^{g(x)}$.
Or, more specifically generally, the derivative of $e^{kz}$ is $k e^{kz}$.

15. It seem to me that

$f(x)=e^{x}$

$f\prime(x)=e^{x}$

When does that rule come into play?

Page 1 of 2 12 Last