Implicit differetiation

**al1850** · May 1st 2008, 11:47 PM

Please show me how you work this out. Thanks

**Moo** · May 2nd 2008, 12:05 AM

Hello,

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

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

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

~~~~~~~~~~
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 :

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

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

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

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

We have to use the product rule too.

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

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

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

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

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

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

**vincent8818** · May 2nd 2008, 12:16 AM

this is also what I need, thax

**Moo** · May 2nd 2008, 12:19 AM

Originally Posted by vincent8818

this is also what I need, thax

What didn't you get ?

**vincent8818** · May 2nd 2008, 12:25 AM

z+yz'??

**Moo** · May 2nd 2008, 12:33 AM

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'

**vincent8818** · May 2nd 2008, 12:41 AM

now i got it, thax

**al1850** · May 2nd 2008, 12:45 AM

Originally Posted by Moo

Hello,

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

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

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

~~~~~~~~~~
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 :

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

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

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

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

We have to use the product rule too.

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

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

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

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

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

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!

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