Partial differentiation

[Eng.msallam]

hello guys , some help here please

Partial diferentiation of z=(1+siny)^1+cosx can any one help ? thanks?

 

[stapel]

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!

 

[Eng.msallam]

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!

we need to differentiate with respect to x and with respect to y

the problem with the power , i want to make sure of it's differentiation .

Thank you :grin:

 

[stapel]

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:

 

[Eng.msallam]

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:

the power is not 1 it's (1+cosx ) :mrgreen:

 

[stapel]

the power is not 1 it's (1+cosx ) :mrgreen:

Ah. So, after inserting the missing grouping symbols, you meant that the equation is as follows:

. . . . .z = (1 + sin(y)^(1 + cos(x))

Would you know how to differentiate (1 + sin(x))^(1 + cos(x))? If so, start by using those methods, and reply with your steps and your results. Thank you! ;-)

 

[HallsofIvy]

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 \(\displaystyle \frac{dv}{dx}= - sin(x)\).
(EDITED)

 

[Eng.msallam]

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 \(\displaystyle \frac{dv}{dx}= 1- sin(x)\).

Thank you very much

 

[HallsofIvy]

Note that I have edited my response because there was a silly error in it. I don't know why I thought that the derivative of "1+ cos(x)" would be "1- sin(x)"! Yes, I know perfectly well that the derivative of a constant, like "1", is 0 so the correct derivative is just "- sin(x)".