1. ## Implicit differetiation

Please show me how you work this out. Thanks

2. Hello,

$\displaystyle \boxed{e^{xy} \sin(yz)-e^{yz}\cos(xz)+z=0}$

--> $\displaystyle ?=\frac{\partial z}{\partial y}=z'(y)$ (for more comfort)

---> $\displaystyle \boxed{e^{x{\color{red}y}} \sin({\color{red}y}{\color{blue}z})-e^{{\color{red}y}{\color{blue}z}}\cos(x{\color{blu e}z})+{\color{blue}z}=0}$

~~~~~~~~~~
The first term will use the product rule since there are y and z in it, considering x as a constant and y as the variable.

Thus the derivative of the first term will be :

$\displaystyle xe^{xy} \sin(yz)+e^{xy} {\color{red}\dots} \cos(yz)$

The red dots correspond to the derivative of yz. We have to use the product rule.
---> $\displaystyle {\color{red}\dots}=z+yz'$

---------> $\displaystyle \boxed{xe^{xy} \sin(yz)+e^{xy} (z+yz') \cos(yz)} \ (1)$

~~~~~~~~~~
Now considering the second term $\displaystyle e^{{\color{red}y}{\color{blue}z}}\cos(x{\color{blu e}z})$

We have to use the product rule too.

$\displaystyle \underbrace{\color{red}\dots}_{\text{derivative of yz, again}} e^{yz} \cos(xz) \ \ + \ \ e^{yz} (-(xz') \sin(xz))$

---------> $\displaystyle \boxed{(z+yz') e^{yz} \cos(xz) - xz' e^{yz} \sin(xz)} \ (2)$

~~~~~~~~~~
Now, the third term, z.
Its derivative is z' (3)

------------------------------

The derivative of the whole thing will be $\displaystyle (1)-(2)+(3)=0$

And... I won't copy it, you can do it

I hope there is no mistake, check it again, it will make a good exercise

3. this is also what I need, thax

4. Originally Posted by vincent8818
this is also what I need, thax
What didn't you get ?

5. z+yz'??

6. Originally Posted by vincent8818
z+yz'??
Ok,

This is the derivative of yz, with respect to y.

It's like having y*z(y) (z : function of y).

Thus applying the product rule, we get :

1*z(y)+y*z'(y)

1 is the derivative of y, with respect to y.

So it's z(y)+yz'(y), which I simplified into z+yz'

7. now i got it, thax

8. Originally Posted by Moo
Hello,

$\displaystyle \boxed{e^{xy} \sin(yz)-e^{yz}\cos(xz)+z=0}$

--> $\displaystyle ?=\frac{\partial z}{\partial y}=z'(y)$ (for more comfort)

---> $\displaystyle \boxed{e^{x{\color{red}y}} \sin({\color{red}y}{\color{blue}z})-e^{{\color{red}y}{\color{blue}z}}\cos(x{\color{blu e}z})+{\color{blue}z}=0}$

~~~~~~~~~~
The first term will use the product rule since there are y and z in it, considering x as a constant and y as the variable.

Thus the derivative of the first term will be :

$\displaystyle xe^{xy} \sin(yz)+e^{xy} {\color{red}\dots} \cos(xy)$

The red dots correspond to the derivative of yz. We have to use the product rule.
---> $\displaystyle {\color{red}\dots}=z+yz'$

---------> $\displaystyle \boxed{xe^{xy} \sin(yz)+e^{xy} (z+yz') \cos(xy)} \ (1)$

~~~~~~~~~~
Now considering the second term $\displaystyle e^{{\color{red}y}{\color{blue}z}}\cos(x{\color{blu e}z})$

We have to use the product rule too.

$\displaystyle \underbrace{\color{red}\dots}_{\text{derivative of yz, again}} e^{yz} \cos(xz) \ \ + \ \ e^{yz} (-(xz') \sin(xz))$

---------> $\displaystyle \boxed{(z+yz') e^{yz} \cos(xz) - xz' e^{yz} \sin(xz)} \ (2)$

~~~~~~~~~~
Now, the third term, z.
Its derivative is z' (3)

------------------------------

The derivative of the whole thing will be $\displaystyle (1)-(2)+(3)=0$

And... I won't copy it, you can do it

I hope there is no mistake, check it again, it will make a good exercise
impressive!