# Derivatives of Trigonometric Functions

1. Oct 8, 2007

### Heat

1. The problem statement, all variables and given/known data

Find d^2y/dx^2.

y = x cos x

3. The attempt at a solution

I've been doing derivatives recently and now got into doing them with trig functions.

I thought it was,

y = x cos x = -xsinx = -xcosx

but that is the derivative of the derivative.

Can someone explain what to do in this situation?

What is the difference of dx/dy, d^2y/dx^2, dy^2/d^2x, etc....

2. Oct 8, 2007

### cristo

Staff Emeritus
You need to use the product rule to differentiate x.cosx; i.e. d/dx(uv)=v(du/dx)+u(dv/dx). To find the second derivative you simply differentiate the result of d/dx(x.cosx).

3. Oct 8, 2007

### CompuChip

First find the first derivative. You get two terms (product rule!).

dx/dy is the derivative of x with respect to y. Writing this down sort of implies that x is a function of y, like $x(y) = 3 \sqrt{1 + y^2} / y$, though it's a bit against conventions to call the function x and the variable y.

d^2 y / dx^2 is the second derivative of y with respect to x, of which y is a function. It can be viewed as shorthand for
$$\frac{d^2 y}{dx^2} = \frac{d\left( \frac{ dy }{ dx } \right) }{ dx }$$
Of course this is possible, since dy/dx is again a function of x, for example:
$$y(x) = x^2, \frac{dy}{dx} = 2 x, \frac{d^2 y}{dx} = \frac{d (2x) }{dx} = 2.$$

Oh, and dy^2/dx^2 doesn't really mean anything, it's usually an error if someone writes that down. (Though it is possible to make sense of it, by viewing x^2 as a variable, e.g. $y = 2 x^2$ s.t. $y^2 = 4 (x^2)^2 = 4 x^4$. Then $\frac{ d y^2 }{ d x^2 } = 8 x^2$ as you can see by temporarily writing $z = x^2$ everywhere, deriving, and writing it back in x^2).

Last edited: Oct 8, 2007