Partial derivative of tan(x + y)

[username12345]

Homework Statement

f(x, y) = \tan(x + y) \\<br /> f_x = ?<br />

Homework Equations

\frac{dy}{dx}\tan(x)= \sec^2 x

The Attempt at a Solution

I set y as constant, so I said derivative of y = 0 then took derivative of tan as above. However the answer should be
f_x = \sec^2(x + y)

Why is y included?

 

[tiny-tim]

Hi username12345!

f(x, y) = \tan(x + y) \\<br /> f_x = ?
…
However the answer should be
f_x = \sec^2(x + y)

Why is y included?

Because it's in g(x,y) in the chain rule … ∂f(g(x,y))/∂x = ∂f(g(x,y))/∂g ∂g(x,y))/∂x

 

[HallsofIvy]

The fact that the derivative of y is 0 does not mean that y itself is 0 and replacing tan(x+y) with tan(x) is saying y= 0.

What is the derivative of tan(x+ a) for constant a?

 

[username12345]

What is the derivative of tan(x+ a) for constant a?

Well, what if you asked, what is derivative of tan(2x + a)... I would say 2 \sec^2(2x + a). So if that is correct then the derivative of \tan(x + a) would be \sec^2(x + a). Is this correct?

Now considering the partial derivative with respect to x, we want y as a constant so replace a with y and we get f_x = \sec^2(x + y)