Eng.msallam
New member
- Joined
- Nov 23, 2014
- Messages
- 4
hello guys , some help here please
Partial diferentiation of z=(1+siny)^1+cosx can any one help ? thanks?
Partial diferentiation of z=(1+siny)^1+cosx can any one help ? thanks?
With respect to which variable? Also, we can't "help" you when we can't yet see where you're needing that help, so: What did you get when you tried doing the differentiation?Partial diferentiation of z=(1+siny)^1+cosx can any one help ? thanks?
With respect to which variable? Also, we can't "help" you when we can't yet see where you're needing that help, so: What did you get when you tried doing the differentiation?
Please be complete. Thank you!
Any expression to the power 1 is just itself, so the power can be ignored. The answer would have been different, of course, if the power on 1 + sin(y) had been something like "2" or "3 + x". :wink:
Ah. So, after inserting the missing grouping symbols, you meant that the equation is as follows:the power is not 1 it's (1+cosx ) :mrgreen:
Those parentheses would have helped at the start!the power is not 1 it's (1+cosx ) :mrgreen:
Those parentheses would have helped at the start!
In order to differentiate when you have a power that is not a constant, first take the logarithm: If f(x)= (u(x))^(v(x) then ln(f)= ln(u(x)^v(x))= v(x)ln(u(x)).
Now, differentiate both sides of that. The derivative of \(\displaystyle ln(f(x))\) is \(\displaystyle \frac{1}{f(x)}\frac{df}{dx}\). To differentiate the right side, use the product rule.
The derivative of ln(u(x)) is, as above, \(\displaystyle \frac{1}{u(x)}\frac{du}{dx}\) so, by the product rule the derivative of the right side is \(\displaystyle \frac{dv}{dx} ln(u(x))+ v(x)\frac{1}{u(x)}\frac{du}{dx}\) so that \(\displaystyle \frac{1}{f}\frac{df}{dx}= \frac{dv}{dx}ln(u)+ \frac{u(x)}{v(x)}\frac{du}{dx}\).
In this problem u(x)= 1+ sin(x) and v(x)= 1+ cos(x) so \(\displaystyle \frac{du}{dx}= cos(x)\) and \(\displaystyle \frac{dv}{dx}= 1- sin(x)\).