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
$\displaystyle y = 17(22+x) ^(41-x) ^30$
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:
$\displaystyle =u(x)^{v(x)}\cdot \left(v'(x)\ln u(x)+v(x)\cdot\frac{1}{u(x)}\cdot u'(x)\right)$
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...
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
Hello, Nora!
Use logarithmic differentiation.
$\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]
$
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
$\displaystyle y'= ln(f(x))f(x)^{g(x)}g'(x)+ g(x)f(x)^{g(x)- 1} g'(x)$.