For example ,

y = 17(22+x)^(41-x)^30

I tried to use math code but it didn't show the exponential numbers/function properly

$\displaystyle y = 17(22+x) ^(41-x) ^30$

- May 6th 2010, 02:00 AMNoraHow to diffrentiate an exponential function to the function base ?
For example ,

y = 17(22+x)^(41-x)^30

I tried to use math code but it didn't show the exponential numbers/function properly

$\displaystyle y = 17(22+x) ^(41-x) ^30$ - May 6th 2010, 02:07 AMFailure
- May 6th 2010, 02:12 AMNora
Thank you could you please tell me why use e^ln(y) ?(i mean the last two steps )

- May 6th 2010, 02:36 AMFailure
Why? - Well, because you do not have a rule to differentiate something like $\displaystyle f(x):= u(x)^{v(x)}$ directly (analogous to the product rule or the quotient rule, that is). Instead you do it like this:

$\displaystyle f'(x)=\left(u(x)^{v(x)}\right)'=\left(e^{v(x)\ln u(x)}\right)'=e^{v(x)\ln u(x)}\cdot\left(v(x)\ln u(x)\right)'$

where the last equality holds because of the chain rule. Now you can apply the product rule (and for $\displaystyle \ln u(x)$ the chain rule again) and are done:

$\displaystyle =u(x)^{v(x)}\cdot \left(v'(x)\ln u(x)+v(x)\cdot\frac{1}{u(x)}\cdot u'(x)\right)$ - May 6th 2010, 02:51 AMNora
aha I got it .. So it's a rule like other rules (Exponential , product , log ....from where you can get directly the derivative

Thank you so much :)

I'll solve the equation right now (Happy) - May 6th 2010, 02:57 AMFailure
Um, well, yes you might take it as a rule like the other rules, but you would not want to actually

*memorize*it, like you did (or should do) with the other, much more important rules: first, because it is not used that often, and, second, because it is easier to remember that you can replace $\displaystyle y=u(x)^{v(x)}$ by $\displaystyle y=e^{v(x)\ln u(x)}$ and then apply the rules that you already know about.

P.S: Please note that I have made a silly mistake in my first reply to your question that I have now corrected: the constant factor 17 in front of the power-term can (and should) be left as a factor in front of the remaining exponential mess... - May 6th 2010, 04:53 AMNora
well , It's already written in my book .. when you've said it's a rule I grabbed my book to look for it then I found it under (exponential function to the base of constant ) and compare it to your explanation it was so helpful :)

Thank you :) :)

and yes the number 17 i knew it was mistake because it's constant and can't be included in derivative :) thank you for clarifying

you made my calculus life easier (Rofl) - May 6th 2010, 06:56 AMSoroban
Hello, Nora!

Use logarithmic differentiation.

Quote:

$\displaystyle y \;=\; 17(22+x)^{(41-x)^{30}} $

Take logs: .$\displaystyle \ln(y) \;=\;\ln\bigg[17(22+x)^{(41-x)^{30}}\bigg] \;=\;(41-x)^{30}\cdot\ln[17(22+x)]

$

Differentiate implicitly: .$\displaystyle \frac{1}{y}\!\cdot\!y' \;=\;(41-x)^{30}\cdot\frac{17}{17(22+x)} + 30(41-x)^{29}(-1)\cdot\ln[17(22+x)] $

. . $\displaystyle \frac{y'}{y} \;=\;\frac{(41-x)^{30}}{x+22} - 30(41-x)^{29}\cdot\ln[17(22+x)] \;=\;$ .$\displaystyle \frac{(41-x)^{29}}{x+22}\bigg[(41-x) - 30(22+x)\ln[17(22+x)]\bigg]$

. . $\displaystyle y' \;=\;y\cdot\frac{(41-x)^{29}}{22+x}\bigg[(41-x) - 30(22+x)\ln[17(22+x)]\bigg] $

. . $\displaystyle y' \;=\;17(22+x)^{(41-x)^{30}}\cdot\frac{(41-x)^{29}}{22+x} \bigg[(41-x) - 30(22+x)\ln[17(22+x)]\bigg]

$

- May 6th 2010, 06:59 AMgrgrsanjaythank you
i was looking for this problem

- May 6th 2010, 07:08 AMHallsofIvy
With something like "$\displaystyle y= f(x)^{g(x)}$", there are two simple

**mistakes**we could make:

1) Treat the exponent, g(x), as a constant and use the "power rule" to get

"$\displaystyle y'= g(x)f(x)^{g(x)-1}f'(x)$".

2) Treat the base, f(x), as a constant and use the "exponential rule" to get "$\displaystyle y'= ln(f(x)) f(x)^{g(x)}g'(x)$.

The interesting thing is that the correct derivative is the**sum**of those two errors!

Taking the logarithm of both sides of $\displaystyle y= f(x)^{g(x)}$, $\displaystyle ln(y)= g(x) ln(f(x))$. On the left side, $\displaystyle (ln(y))'= \frac{1}{y} y'$. On the right, using the product rule, $\displaystyle (g(x)ln(f(x))'= g'(x)ln(f(x)+ \frac{g(x)}{f(x)}f'(x)$.

Multiplying on both sides of $\displaystyle \frac{1}{y}y'= g'(x)ln(f(x))+ \frac{g(x)}{f(x)}f'(x)$ by $\displaystyle y= f(x)^{g(x)}$, we get

$\displaystyle y'= ln(f(x))f(x)^{g(x)}g'(x)+ g(x)f(x)^{g(x)- 1} g'(x)$. - May 6th 2010, 07:42 AMgrgrsanjay
- May 9th 2010, 08:55 AMNoraThank you very much for helping me (Clapping)

**Soroban**

__HallsofIvy__