Differentiation of Trig Functions

[Hootenanny]

The question is as follows:
Find $$\frac{dy}{dx} \; \; \; xtanx \; \; \; dx$$
The standard differential is given in the formula book as
$$f(x) = \tan kx \Rightarrow f'(x) = k\sec^2 kx$$
Therefore, I got:
$$\frac{dy}{dx} = x \sec^2 x$$
However, the answer given is
$$\tan x + x \sec^2 x$$
I can't see where I've gone wrong, it seems like such a simple differential. Any help would be much appreciated.

 

[assyrian_77]

Have you heard of differentiation of products? Namely,

$$\frac{d}{dx}f(x)g(x)=f'(x)g(x)+f(x)g'(x)$$

 

[Jeff Ford]

The $x$ isn't a constant that can be left out when you differentiate. You need to use the product rule on the two terms $x$ and $\tan(x)$

 

[becz-]

what the book gave has k as a constant, in your problem x is not a constant. you need to use the product rule.

 

[Hootenanny]

Oh yes. I release what I've done now. Thank's.