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

$\displaystyle \frac{dw}{dx} = \frac{\partial w}{\partial y}\frac{dy}{dx} + \frac{\partial w}{\partial z}\frac{dz}{dx}$

$\displaystyle = (x+z) \cos x + ye^x$

$\displaystyle = x \cos x + e^x \cos x + e^x \sin x$

But if you sub in the values for y and z before taking the derivative, you get:

$\displaystyle x \sin x + \sin x e^x$

$\displaystyle \frac{dw}{dx} = (\sin x + x \cos x) + (e^x \sin x + e^x \cos x)$

$\displaystyle = x \cos x + e^x \cos x + e^x \sin x + \sin x$

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:

$\displaystyle 9(\frac{\partial ^2 f}{\partial x^2}) - 9(\frac{\partial ^2 f}{\partial x \partial y}) + 2(\frac{\partial ^2 f}{\partial y^2}) = 0$

to:

$\displaystyle \frac{\partial ^2 f}{\partial u \partial v} = 0$

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:

$\displaystyle \frac{\partial f}{\partial x} = \frac{\partial f}{\partial u} + \frac{\partial f}{\partial v}$

and:

$\displaystyle \frac{\partial f}{\partial y} = \frac{\partial f}{\partial u}a + \frac{\partial f}{\partial v}b$

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