Derivatives of Trigonometric Functions

**Bashyboy** · April 30th 2012, 11:39 AM

Why is it that when we take derivatives of trig functions we also have to take the derivative of its angle? It is because we want to know how also the angle changes?

**Plato** · April 30th 2012, 11:58 AM

Please restate this question. It has no answer as written.

**HallsofIvy** · April 30th 2012, 12:40 PM

I believe you are referring to the "chain rule". That does not apply just to trig functions, but to any functions: if x is a function of t, x(t), then $\frac{df(x)}{dt}= \frac{df}{dx}\frac{dx}{dt}$ .

In particular, if the variable is an "angle" (actually, most applications of trig functions do not involve angles), $\theta(t)$ that depends on t, the the derivative of $sin(\theta(t))$ , with respect to t, is
$\frac{dsin(\theta)}{dt}= \frac{dsin(\theta)}{d\theta}\frac{d\theta}{dt}= cos(\theta)\frac{d\theta}{dt}$ .

Thread: Derivatives of Trigonometric Functions

LinkBack

Thread Tools

Display

Derivatives of Trigonometric Functions

Re: Derivatives of Trigonometric Functions

Re: Derivatives of Trigonometric Functions

Similar Math Help Forum Discussions

Applications of Derivatives of Trigonometric and Inverse Trigonometric Functions

Derivatives of Inverse Trigonometric Functions

derivatives with trigonometric functions

Derivatives of inverse trigonometric functions

derivatives of trigonometric functions

Search Tags