# Derivative of tan(f(x,y))

I was looking at the derivation of the derivative of `tan(x^2 + 5x + y)` on the Wolfram|Alpha. I don't understand the last step, which says the derivative of `y` is zero. Shouldn't it be `y`?

Here is the link to the derivation: http://www3.wolframalpha.com/Calculate/MSP/MSP4281a2116fh62dgcef100002h2gfc72529d6a10?MSPStoreType=image/gif&s=11&w=478&h=546

Please note that I couldn't post an image. The link above will display the image from the Wolfram|Alpha website.

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That link returns an error message: Sorry, this page does not exist on the Wolfram|Alpha site. – Alex Jun 16 '12 at 20:01

As you can see the derivative of `tan(x^2 + 5x + y)` is `(5+2 x) sec^2(5 x+x^2+y)` as you can see that you have not explicitly stated that `y` is a function of `w`. So wofram alpha assumes `y` to be constant, and the derivative of constant is zero.
This is what we get for just `y`
If you specify `y` as function of `x` (by y(x)), you get the following result