Multivariable Calculus - Chain Rule

**BlackBlaze** · March 16th 2010, 03:12 PM

I might have to keep coming back here, this assignment is killing me...

Given w = xy + yz, y = sin x and z = e^x, find dw/dx
$\frac{dw}{dx} = \frac{\partial w}{\partial y}\frac{dy}{dx} + \frac{\partial w}{\partial z}\frac{dz}{dx}$

$= (x+z) \cos x + ye^x$
$= x \cos x + e^x \cos x + e^x \sin x$

But if you sub in the values for y and z before taking the derivative, you get:
$x \sin x + \sin x e^x$
$\frac{dw}{dx} = (\sin x + x \cos x) + (e^x \sin x + e^x \cos x)$
$= x \cos x + e^x \cos x + e^x \sin x + \sin x$

Different from the top answer. Can anyone see where I went wrong?

Also:
Given u = x + ay and v = x + by, for what values a and b can you change:
$9(\frac{\partial ^2 f}{\partial x^2}) - 9(\frac{\partial ^2 f}{\partial x \partial y}) + 2(\frac{\partial ^2 f}{\partial y^2}) = 0$
to:
$\frac{\partial ^2 f}{\partial u \partial v} = 0$

At first I tried to factor it into (3n - 2m)(3n - m), but then I remembered that the second derivative doesn't equal the first derivative squared. Also, the middle term isn't necessarily both partial derivatives multiplied together. I found:
$\frac{\partial f}{\partial x} = \frac{\partial f}{\partial u} + \frac{\partial f}{\partial v}$
and:
$\frac{\partial f}{\partial y} = \frac{\partial f}{\partial u}a + \frac{\partial f}{\partial v}b$

But I can't think of a way to sub them in, or take the next derivative without knowing what f is.

**tom@ballooncalculus** · March 16th 2010, 03:27 PM

$<br /> \frac{dw}{dx} = \frac{\partial w}{\partial y}\frac{dy}{dx} + \frac{\partial w}{\partial z}\frac{dz}{dx}<br />$

wants to be

$<br /> \frac{dw}{dx} = \frac{\partial w}{\partial y}\frac{dy}{dx} + \frac{\partial w}{\partial z}\frac{dz}{dx} + \frac{\partial w}{\partial x}<br />$

**Soroban** · March 16th 2010, 03:45 PM

Hello, BlackBlaze!

You left out a term in your equation . . .

Given: . $w \:=\: xy + yz,\;\; y \:=\: \sin x,\;\;z \:=\: e^x$
Find $\frac{dw}{dx}$

The equation is: . $\frac{dw}{dx} \;=\; {\color{blue}\frac{\partial w}{\partial x}\!\cdot\!\frac{dx}{dx}} + \frac{\partial w}{\partial y}\!\cdot\!\frac{dy}{dx} + \frac{\partial w}{\partial z}\!\cdot\!\frac{dz}{dx}$

. . . . . . . . . . . . . . . $=\;\;\;y\cdot 1 + (x+z)\cos x + y\cdot e^x$

. . . . . . . . . . . . . . . $=\;\; \sin x + (x+e^x)\cos x + \sin x\cdot e^x$

. . . . . . . . . . . . . . . $=\;\;\sin x + x\cos x + e^x\cos x + e^x\sin x$

. . which matches your other answer . . .

**tom@ballooncalculus** · March 16th 2010, 05:11 PM

Originally Posted by BlackBlaze

I found:
$\frac{\partial f}{\partial x} = \frac{\partial f}{\partial u} + \frac{\partial f}{\partial v}$
and:
$\frac{\partial f}{\partial y} = \frac{\partial f}{\partial u}a + \frac{\partial f}{\partial v}b$

But I can't think of a way to sub them in, or take the next derivative without knowing what f is.

Carrying on from there, but writing it

$f_x = f_u + f_v$

and:

$f_y = a f_u + b f_v$

(if you've no objection), the same reasoning gets

$f_{xx} = f_{uu} + 2f_{uv} + f_{vv}$

and

$f_{yy} = a^2 f_{uu} + 2ab f_{uv} + b^2 f_{vv}$

and

$f_{xy} = a f_{uu} + (a+b) f_{uv} + b f_{vv}$

so that

$9 f_{xx} - 9 f_{xy} + 2 f_{yy} = 0$

means

$(9 - 9a + 2a^2) f_{uu} + (18 - 9(a+b) + 4ab) f_{uv} + (9 - 9b + 2b^2) f_{vv} = 0$

Then you can choose a and b to make the f_{uu} and f_{vv} drop out.

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