Multivariable Chain-Rule Problem

[S.R]

Homework Statement

Let g(x, y) = f(sin(y), cos(x)). Find the second partial derivative of g with respect to x (g_xx).

Homework Equations

The Attempt at a Solution

I attempted to find g_x, but I'm not entirely sure how chain rule applies in this situation.

Is this correct?

g_x = f_x(sin(y), cos(x)) * (-sin(x))

 

[tommyxu3]

To get the derivative with respect to ##x,## by chain rule: $$\frac{\partial g}{\partial x}=\frac{\partial f}{\partial x}\frac{\partial \sin y}{\partial x}+\frac{\partial f}{\partial y}\frac{\partial \cos x}{\partial x}$$
and see every term if one can be cancelled.

 

[S.R]

To get the derivative with respect to ##x,## by chain rule: $$\frac{\partial g}{\partial x}=\frac{\partial f}{\partial x}\frac{\partial \sin y}{\partial x}+\frac{\partial f}{\partial y}\frac{\partial \cos x}{\partial x}$$
and see every term if one can be cancelled.

Ah, thank-you. The first term cancels, right?

 

[tommyxu3]

Ah, thank-you. The first term cancels, right?

Yes, for the ##\sin y## doesn't depend on ##x.##

 

[tommyxu3]

However, forgetting to remind, in the calculation, mind the terms regarding ##f## you have to plug in its pair ##(\sin y,\cos x).##

 

[S.R]

However, forgetting to remind, in the calculation, mind the terms regarding ##f## you have to plug in its pair ##(\sin y,\cos x).##

Should it be d/dx(siny) as in the single var. case?

 

[tommyxu3]

To be precise, you are right, but using partial derivative doesn't make no sense based on the definition also haha (just for my laziness)

 

[S.R]

To be precise, you are right, but using partial derivative doesn't make no sense based on the definition also haha (just for my laziness)

How would I obtain g_y though? I'm also not sure how to derive the first formula.

 

[tommyxu3]

Well...then maybe you don't understand the chain rule totally...For reference: https://en.wikipedia.org/wiki/Chain_rule
To make yourself really acquire the information, I suggest getting the full picture of it including the proof.

 

[S.R]

Well...then maybe you don't understand the chain rule totally...For reference: https://en.wikipedia.org/wiki/Chain_rule
To make yourself really acquire the information, I suggest getting the full picture of it including the proof.

The way I learned the chain rule (in the context of multivariable functions) was to draw a dependency diagram. In this case, however, the dependency diagram is not clear.

 

[Ray Vickson]

Homework Statement

Let g(x, y) = f(sin(y), cos(x)). Find the second partial derivative of g with respect to x (g_xx).

Homework Equations

The Attempt at a Solution

I attempted to find g_x, but I'm not entirely sure how chain rule applies in this situation.

Is this correct?

g_x = f_x(sin(y), cos(x)) * (-sin(x))

When I learned this stuff we were advised to use notation like ##f_1##, which is "partial derivative of ##f## with respect to the first variable" and ##f_2## for "... with respect to the second variable". That way, when you swap the positions of ##x## and ##y## (as done in this problem) you avoid getting yourself hopelessly confused.