Partial differential coefficient

[jonjacson]

Homework Statement

The equation is z= e ^(x*y), the interesting thing is y is function of x too, y = ψ(x)

Calculate the partial derivative respect to x, and the total derivative.

Homework Equations

Total differential:

dz= ∂z/∂x dx + ∂z/∂y dy

The Attempt at a Solution

[/B]
Well, according to the book:

∂z/∂x= y * e^xy

But I don't agree with this result, because if y is also function of x this term will be different.

For example if y=x³, z would be = e^x4, and the partial derivative is:

∂z/∂x = 4 x³ * e ^x4

and this is not the same as:

y * e ^xy = x³e^x4

What do you think?

I just want to check if the book is right, this topic (multivariate differential calculus) is so important that I want to understand it correctly.

 

[Orodruin]

The book is right. When they talk about partial derivative they talk about the derivative considering $e^{yx}$ as a function of $x$ and $y$ as independent parameters. What you are computing is the total derivative.

 

[Ray Vickson]

Homework Statement

The equation is z= e ^(x*y), the interesting thing is y is function of x too, y = ψ(x)

Calculate the partial derivative respect to x, and the total derivative.

Homework Equations

Total differential:

dz= ∂z/∂x dx + ∂z/∂y dy

The Attempt at a Solution

[/B]
Well, according to the book:

∂z/∂x= y * e^xy

But I don't agree with this result, because if y is also function of x this term will be different.

For example if y=x³, z would be = e^x4, and the partial derivative is:

∂z/∂x = 4 x³ * e ^x4

and this is not the same as:

y * e ^xy = x³e^x4

What do you think?

I just want to check if the book is right, this topic (multivariate differential calculus) is so important that I want to understand it correctly.

If $z = e^{x\, \psi(x)}$ then there really is no such thing as a "partial derivative" of $z$; there is only an "ordinary" (1-variable) derivative $dz/dx$.
However, that being said, we do have that $z'(x)$ is expressed in terms of the partial derivatives of the function $e^{xy}$ evaluated at $y = \psi(x)$ (and, of course, the derivative $d \psi(x)/dx$ is involved as well).

To fix your difficulties when $y = x^3$ you need to use the full force of the complete chain rule:
$$\frac{d}{dx} \left( \left. e^{x y} \right|_{y = x^3} \right) = \left. \frac{\partial e^{x y}}{\partial x} \right|_{y = x^3}
+ \left. \frac{ \partial e^{xy}}{\partial y} \right|_{y = x^3} \: \cdot \frac{d \, x^3}{dx}$$