# Thread: Implicit Differentiation with 3 Variables

1. ## Implicit Differentiation with 3 Variables

Hi all. I've been having problems with the question below.

$\displaystyle z = sin(xyz) + xln(yz)$

find

$\displaystyle \frac{\partial z}{\partial x}$

I've tried two differentiation methods:

First

$\displaystyle \frac{\frac{\partial T} {\partial x}}{\frac{\partial T}{\partial z}}$

which generally gives me:

$\displaystyle \frac{yzcos(xyz) + ln(yz)}{xycos(xyz) + \frac{xy}{yz}}$

I'm not 100% sure about that though. Especially differentiating $\displaystyle \frac {\partial T}{\partial z}$ (the denominator of $\displaystyle \frac{\frac{\partial T} {\partial x}}{\frac{\partial T}{\partial z}}$)

Secondly
I dont know the name of this differentiation method or how to describe it, but essentially you differentiate as normal - with respect to x, treating y as a constant.
This time, however, also add $\displaystyle \frac{\partial z}{\partial x}$ whenever you encounter a $\displaystyle z$. Or is it whenever you differentiate $\displaystyle z$? Or is it when $\displaystyle z$ is left from differentiating $\displaystyle x$ - as in $\displaystyle z$ was a coefficient of $\displaystyle x$? As you can see this is where my confusion comes in.

If anyone can confirm or help me with this I would appreciate it. Thanks.

2. Originally Posted by isp_of_doom
Hi all. I've been having problems with the question below.

$\displaystyle z = sin(xyz) + xln(yz)$

find

$\displaystyle \frac{\partial z}{\partial x}$

I've tried two differentiation methods:

First

$\displaystyle \frac{\frac{\partial T} {\partial x}}{\frac{\partial T}{\partial z}}$

which generally gives me:

$\displaystyle \frac{yzcos(xyz) + ln(yz)}{xycos(xyz) + \frac{xy}{yz}}$

I'm not 100% sure about that though. Especially differentiating $\displaystyle \frac {\partial T}{\partial z}$ (the denominator of $\displaystyle \frac{\frac{\partial T} {\partial x}}{\frac{\partial T}{\partial z}}$)

Secondly
I dont know the name of this differentiation method or how to describe it, but essentially you differentiate as normal - with respect to x, treating y as a constant.
This time, however, also add $\displaystyle \frac{\partial z}{\partial x}$ whenever you encounter a $\displaystyle z$. Or is it whenever you differentiate $\displaystyle z$? Or is it when $\displaystyle z$ is left from differentiating $\displaystyle x$ - as in $\displaystyle z$ was a coefficient of $\displaystyle x$? As you can see this is where my confusion comes in.

If anyone can confirm or help me with this I would appreciate it. Thanks.
$\displaystyle z = sin(xyz) + xln(yz)$

The implicit function theorem would give the partial derivative as $\displaystyle \frac{\partial z}{\partial x}=-\frac{\frac{\partial T}{\partial x}}{\frac{\partial T}{\partial x}}$. I used this formula, and obtained the same expression you did. You just need a negative sign.

3. WAIT!!!

We're using the theorem wrong, lol. I'm surprised no one has said anything. I just realized this.

$\displaystyle z=sin(xyz)+xln(xy)$

$\displaystyle F(x,y,z)=sin(xyz)+xln(yz)-z=0$

NOW you can apply the implicit function theorem. Always move everything over to one side. The expression we obtained earlier would be true only if $\displaystyle 0=sin(xyz)+xln(yz)$.

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Your second method would work too. It is just a little more involved.

$\displaystyle z=sin(xyz)+xln(yz)$

I will write $\displaystyle z=z_x$ since z must be treated as a function of x. So we could go through the following.

$\displaystyle \frac{\partial z_x}{\partial x}=\frac{\partial}{\partial x}sin(xyz_x)+\frac{\partial}{\partial x}\left(xln(yz_x)\right)$

Treat $\displaystyle y$ as a constant, and remember that z is a function. I would write the first term as $\displaystyle \frac{\partial}{\partial x}sin(xyz_x)=\frac{d}{du}sin(u)\frac{\partial u}{\partial x}$ where $\displaystyle u=xyz_x$. Then calculate $\displaystyle \frac{\partial u}{\partial x}$ remembering that $\displaystyle z=z_x$ is a function, so apply the product rule to get $\displaystyle yz_x+xy\frac{\partial z_x}{\partial x}$. So the term is $\displaystyle \frac{\partial}{\partial x}sin(xyz_x)=cos(xyz_x)\left(yz_x+xy\frac{\partial z_x}{\partial x}\right)$. You could do something similar with the second term.

$\displaystyle \frac{\partial }{\partial x}xln(yz_x)=ln(yz_x)+\frac{xy}{yz_x}\frac{\partial z_x}{\partial x}$

$\displaystyle \frac{\partial z}{\partial x}=ln(yz)+\frac{xy}{yz}\frac{\partial z}{\partial x}+cos(xyz)\left(yz+xy\frac{\partial z}{\partial x}\right)$.

$\displaystyle \frac{\partial z}{\partial x}=\frac{ln(yz)+yzcos(xyz)}{1-xycos(xyz)+\frac{xy}{yz}}$

double checked the question and it actually states:

A function $\displaystyle z = f(x,y)$ is defined implicitly by $\displaystyle sin(xyz) + xln(yz) = 1$ Find $\displaystyle \frac{\partial z}{\partial x}$ and $\displaystyle \frac{\partial z}{\partial y}$.

In this case your first post would apply, correct?

5. Originally Posted by isp_of_doom

double checked the question and it actually states:

A function $\displaystyle z = f(x,y)$ is defined implicitly by $\displaystyle sin(xyz) + xln(yz) = 1$ Find $\displaystyle \frac{\partial z}{\partial x}$ and $\displaystyle \frac{\partial z}{\partial y}$.

In this case your first post would apply, correct?
Yes

6. Originally Posted by isp_of_doom
Hi all. I've been having problems with the question below.

$\displaystyle z = sin(xyz) + xln(yz)$

find

$\displaystyle \frac{\partial z}{\partial x}$

I've tried two differentiation methods:

First

$\displaystyle \frac{\frac{\partial T} {\partial x}}{\frac{\partial T}{\partial z}}$

which generally gives me:

$\displaystyle \frac{yzcos(xyz) + ln(yz)}{xycos(xyz) + \frac{xy}{yz}}$

I'm not 100% sure about that though. Especially differentiating $\displaystyle \frac {\partial T}{\partial z}$ (the denominator of $\displaystyle \frac{\frac{\partial T} {\partial x}}{\frac{\partial T}{\partial z}}$)

Secondly
I dont know the name of this differentiation method or how to describe it, but essentially you differentiate as normal - with respect to x, treating y as a constant.
This time, however, also add $\displaystyle \frac{\partial z}{\partial x}$ whenever you encounter a $\displaystyle z$. Or is it whenever you differentiate $\displaystyle z$? Or is it when $\displaystyle z$ is left from differentiating $\displaystyle x$ - as in $\displaystyle z$ was a coefficient of $\displaystyle x$? As you can see this is where my confusion comes in.

If anyone can confirm or help me with this I would appreciate it. Thanks.
If z= sin(xyz)+ x ln(yz) then
$\displaystyle \frac{\partial z}{\partial x}= cos(xyz) \left(yz\frac{\partial x}{\partial x}+ xz\frac{\partial y}{\partial x}+ xy\frac{\partial z}{\partial x}\right)+ ln(yz)\frac{\partial x}{\partial x}+ \frac{x}{yz}\left(z\frac{\partial y}{\partial x}+ y\frac{\partial z}{\partial x}\right)$

Of course, $\displaystyle \frac{\partial x}{\partial x}= 1$ and $\displaystyle \frac{\partial y}{\partial x}= 0$ so that becomes
$\displaystyle \frac{\partial z}{\partial x}= cos(xyz)\left(yz+ xy\frac{\partial z}{\partial x}\right)+ ln(yz)+ \frac{x}{yz}\left(y\frac{\partial z}{\partial x}\right)$.
Solve that for $\displaystyle \frac{\partial z}{\partial x}$.

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