0

I have the following problem:

$$ y = x\sec kx \\ y' = x\ \sec{kx}\ \tan{kx} + \sec{kx} \\ = \sec{kx} (x\ \tan{kx} + 1) $$

I'm confused about the kx is it $$\sec(k)$$ or $$\sec(kx)$$

So if I were to take the derivative would it be $$\sec(k)\tan(k)(x)$$ or $$\sec(kx)\tan(kx)$$

Lutz Lehmann
  • 126,666
  • As in the other problem you asked, with trigonometric functions, it is common to use letters to represent unspecified constants (often, constant multipliers). Traditionally (which is to say, someone made the choice a couple centuries back and it stuck...), the letters used here are from the middle of the alphabet, such as $ \ k \ , \ m \ , \text{or} \ n \ . $ Your differentiation will involve "Product Rule" and a chain-rule substitution $ \ u \ = \ kx \ . $ – colormegone Feb 21 '14 at 07:25
  • The way it's written, it's probably $\sec(kx)$. It is impossible to know for sure.

    However, if it was $\sec(k) x$, the derivative would be $2x\sec k$, not what you wrote.

     – 5xum Feb 21 '14 at 07:25
  • @5xum I rather suspect it is $ \ \sec(kx) \ . $ – colormegone Feb 21 '14 at 07:27
  • @RecklessReckoner i added my solution to the bottom of my post is that correct? –  Feb 21 '14 at 07:35

1 Answers1

1

First, I think we can comfortably assume that $y = x \sec (kx)$, not $y = x (\sec k) x$, because if the latter were the case, then presumably one would instead have written $y = x^2 \sec k$.

Next, we have to know whether $y$ is being regarded as a function of $k$, or of $x$, or of both. If we are interested in the instantaneous rate of change of $y$ with respect to $x$ (and treating $k$ as constant), then that is $$\frac{\partial y}{\partial x} = \sec (kx) \left(1 + kx \tan (kx)\right).$$ If we want the derivative of $y$ with respect to $k$ (and treating $x$ as constant), then that is $$\frac{\partial y}{\partial k} = x^2 \sec (kx) \tan (kx).$$ If $k$ is itself a function of $x$ but is unspecified, then $$\frac{dy}{dx} = \sec (kx)\left(1 + x {\textstyle \left(k + x \frac{dk}{dx}\right)} \tan (kx) \right),$$ where $k = k(x)$. This last expression is called the total derivative of $y$ with respect to $x$.

heropup
  • 135,869