# How to diffrentiate an exponential function to the function base ?

• May 6th 2010, 03:00 AM
Nora
How 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

$y = 17(22+x) ^(41-x) ^30$
• May 6th 2010, 03:07 AM
Failure
Quote:

Originally Posted by Nora
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

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

Consider that
$y = 17(22+x)^{(41-x)^{30}}=17\cdot e^{\ln\left((22+x)^{(41-x)^{30}}\right)}=17\cdot e^{(41-x)^{30}\cdot\ln\big(22+x\big)}$
Now use the chain rule to get the derivative of that last term.
• May 6th 2010, 03:12 AM
Nora
Thank you could you please tell me why use e^ln(y) ?(i mean the last two steps )
• May 6th 2010, 03:36 AM
Failure
Quote:

Originally Posted by Nora
Thank you could you please tell me why use e^ln(y) ?(i mean the last two steps )

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

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

where the last equality holds because of the chain rule. Now you can apply the product rule (and for $\ln u(x)$ the chain rule again) and are done:
$=u(x)^{v(x)}\cdot \left(v'(x)\ln u(x)+v(x)\cdot\frac{1}{u(x)}\cdot u'(x)\right)$
• May 6th 2010, 03:51 AM
Nora
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, 03:57 AM
Failure
Quote:

Originally Posted by Nora
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)

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 $y=u(x)^{v(x)}$ by $y=e^{v(x)\ln u(x)}$ 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, 05:53 AM
Nora
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, 07:56 AM
Soroban
Hello, Nora!

Use logarithmic differentiation.

Quote:

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

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

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

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

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

. . $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, 07:59 AM
grgrsanjay
thank you
i was looking for this problem
• May 6th 2010, 08:08 AM
HallsofIvy
With something like " $y= f(x)^{g(x)}$", there are two simple mistakes we could make:

1) Treat the exponent, g(x), as a constant and use the "power rule" to get
" $y'= g(x)f(x)^{g(x)-1}f'(x)$".

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

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

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

Multiplying on both sides of $\frac{1}{y}y'= g'(x)ln(f(x))+ \frac{g(x)}{f(x)}f'(x)$ by $y= f(x)^{g(x)}$, we get
$y'= ln(f(x))f(x)^{g(x)}g'(x)+ g(x)f(x)^{g(x)- 1} g'(x)$.
• May 6th 2010, 08:42 AM
grgrsanjay
Quote:

Originally Posted by HallsofIvy
With something like " $y= f(x)^{g(x)}$", there are two simple mistakes we could make:

1) Treat the exponent, g(x), as a constant and use the "power rule" to get
" $y'= g(x)f(x)^{g(x)-1}f'(x)$".

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

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

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

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

excellent
• May 9th 2010, 09:55 AM
Nora
Thank you very much for helping me (Clapping)
Soroban

HallsofIvy