Chain rule

asked 2014-07-09 20:54:50 -0600

Tiffany
656 ●3 ●18 ●34

So I've just now been working on 14.4 the chain rule (late, I know). And I'm having some issues with 3 and 4. I understand the chain rule and how it works, and got 1 and 2 with no issues. My issue I'm running into now has to do with it being with respect to s and t now instead of x and y.

So the problem states to use the chain rule to solve for $\frac {\partial z} { \partial s}$ and $\frac{\partial z}{\partial t}$

$$z = x^2y \quad x=\sin(st) \quad y= t^2+s^2$$

So I guess here is where I get confused. Do you replace x and y with their respective equations, then take the partial derivatives from there? Or is it some sort of combination of the original z equation plus the derivitavies of x and y separrately. The book gets the answer..$$ 2xyt\cos(st) + 2x2s, \quad 2xys\cos(st) +2x2t$$

So I'm lost, any help would be greatly appreciated!

Thanks!

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answered 2014-07-09 21:30:23 -0600

Christina
807 ●18 ●35 https://sites.google.com/...

It helped me to look at page 365, example 14.4.2. It shows the following: $$\frac{\partial f}{\partial x} = f_x\frac{\partial x}{\partial x} + f_y\frac{\partial y}{\partial x} + f_z\frac{\partial z}{\partial x}$$ So starting with problem 3, I began with the following: $$\frac{\partial z}{\partial s} = z_x \frac{\partial x}{\partial s} + z_y \frac{\partial y}{\partial s}$$

Then I calculated $z_x$, $z_y$, $ \frac{\partial x}{\partial s}$ and $\frac{\partial y}{\partial s}$ as follows: $$z_x = 2yx$$$$z_y = x^2$$ $$ \frac{\partial x}{\partial s} = tcos(st)$$$$\frac{\partial y}{\partial s} = 2s$$

Plugging those values in gives us $$\frac{\partial z}{\partial s} = 2xytcos(st) + 2sx^2$$

Which is what the answer is! Hope this helps in solving the $\frac{\partial z}{\partial t}$

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answered 2014-07-09 21:13:34 -0600

Kyouko
231 ●4 ●12

updated 2014-07-09 21:49:41 -0600

-Comparing these answers back to the ones in the book, instead of $2s2x$ i have calculated $2sx^2$. I am not sure if I am wrong or if the book is wrong in this instance, but i calculated the partial derivative on a calculator and i got the answer i have provided-

The way the problem is state, I believe you simply swap out $x$ and $y$ for their $x(s,t)$ and $y(x,t)$ counterparts thus generating the equation:

$$ z = x^2y $$

Substituting here for $x$ and $y$:

$$ z = sin^2(st)(t^2 + s^2) $$

Now it should be a simple matter of taking the partial derivatives (with quite a bit of chain rule in here)

$$ \frac {\partial z} {\partial t} = 2tsin^2(st) + (t^2 + s^2)(2ssin(st)cost(st)) $$

And then with respect to s $$ \frac {\partial z} {\partial s} = 2ssin^2(st) + (t^2 + s^2)(2tsin(st)cos(st)) $$

When we simplify this back to $x$ and $y$ terms we get the equations:

$$ \frac {\partial z} {\partial s} = 2sx^2 + 2txycos(st) $$

and

$$ \frac {\partial z} {\partial t} = 2tx^2 + 2sxycos(st) $$

The trick to this question is recognizing that $x$ and $y$ are their own functions of 's' and 't' and plugging this into the equation given to create the new function $z(s, t)$.

Hope this helps

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Chain rule

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