A.3 Derivatives of sin⁡𝑥 and cos⁡𝑥

On this screen we're going to develop formally the derivatives of sin⁡𝑥 and cos⁡𝑥, which we will need again and again. We'll use Desmos to help make sense of the results, and of course also practice using them in various Practice Problems below.

In case you'd simply like to memorize these results so you can move on:

Derivatives of sin x and cos x

𝑑𝑑𝑥sin⁡𝑥=cos⁡𝑥and𝑑𝑑𝑥cos⁡𝑥=−sin⁡𝑥

Yep: the derivative of sin is cosine, and the derivative of cosine is negative sine. You just gotta remember those.
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Many students find it helpful to remember that, as it turns out, the trig functions that start with "co," like "cosine," have a negative sign in their derivatives, while the trig functions without "co" at their start do not. Hence we can quickly remember that the derivative of cosine is negative sine, while the derivative of sine is (positive) cosine.

You can see a full list of the trig function derivatives on our Handy Table of Trig Function Derivatives, quickly accessible from the "Key Formulas" drop-down menu toward the upper right of every screen. We'll develop the results for sin and cos below, and for the other trig functions later in this Chapter once we have more tools.

I. Derivative of sin⁡𝑥: 𝑑𝑑𝑥sin⁡𝑥 =cos⁡𝑥

To find the derivative of sin⁡𝑥, we start, as always, with the definition of the derivative applied to the function. Recall the definition of the derivative:

𝑑𝑑𝑥𝑓(𝑥)=limℎ→0𝑓(𝑥+ℎ)−𝑓(𝑥)ℎ

We have 𝑓(𝑥) =sin⁡(𝑥). Hence 𝑓(𝑥+ℎ)=sin⁡(𝑥+ℎ)=sin⁡𝑥cos⁡ℎ+sin⁡ℎcos⁡𝑥 where we used the sine addition formula sin⁡(𝑥 +𝑦) =sin⁡𝑥cos⁡𝑦 +sin⁡𝑦cos⁡𝑥. (All such trig formulas are available in our Handy Table of Trig Formulas and Identities, also available from the Key Formulas drop-down menu in the upper right of every screen.)

Making these substitutions in the definition of the derivative for 𝑓(𝑥) =sin⁡𝑥 gives us

𝑑𝑑𝑥sin⁡𝑥=limℎ→0sin⁡(𝑥+ℎ)−sin⁡𝑥ℎ=limℎ→0(sin⁡𝑥cos⁡ℎ+sin⁡ℎcos⁡𝑥)−sin⁡𝑥ℎ=limℎ→0sin⁡𝑥cos⁡ℎ−sin⁡𝑥+sin⁡ℎcos⁡𝑥ℎ=limℎ→0sin⁡𝑥cos⁡ℎ−sin⁡𝑥ℎ+limℎ→0sin⁡ℎcos⁡𝑥ℎ=limℎ→0sin⁡𝑥(cos⁡ℎ−1)ℎ+limℎ→0cos⁡𝑥sin⁡ℎℎ[sin⁡𝑥 and cos⁡𝑥 are unaffected by the limit]=sin⁡𝑥[limℎ→0cos⁡ℎ−1ℎ]+cos⁡𝑥[limℎ→0sin⁡ℎℎ](∗)

You might recognize the "Special Trig Limits" we learned when we were first exploring limits:

lim𝜃→0cos⁡𝜃−1𝜃=0andlim𝜃→0sin⁡𝜃𝜃=1

The actual reason we introduced those then is because we need them now to complete our proof, continuing from the line marked (*):

limℎ→0sin⁡𝑥cos⁡ℎ−sin⁡𝑥ℎ+limℎ→0sin⁡ℎcos⁡𝑥ℎ[sin⁡𝑥 and cos⁡𝑥 are unaffected by the limit]𝑑𝑑𝑥sin⁡𝑥=sin⁡𝑥[limℎ→0cos⁡ℎ−1ℎ]+cos⁡𝑥[limℎ→0sin⁡ℎℎ](∗)=sin⁡𝑥[limℎ→0cos⁡ℎ−1ℎ]0+cos⁡𝑥[limℎ→0sin⁡ℎℎ]1=cos⁡𝑥

That is, we have the key result:

Derivative of sin x

𝑑𝑑𝑥sin⁡𝑥=cos⁡𝑥

The following Exploration let's you explore this relationship graphically.

EXPLORATION 1: Moveable tangent line for 𝑦 =sin⁡𝑥

The top graph below shows 𝑦 =sin⁡𝑥.

The lower graph automatically updates to show the value of the function's derivative at the same point 𝑥 =𝑎 that's in the upper graph. That is, for each value of 𝑥 =𝑎, we automatically plot the point (𝑎,𝑓′(𝑎)) on the lower graph.

Verify for yourself:

  • Are the zeros of 𝑓′(𝑥) where you expect them to be? (That is, where the tangent line to the graph of 𝑦 =sin⁡𝑥 is horizontal?)
  • Are the maximum and minimum values of 𝑓′(𝑥) where you expect them to be? (Where is the slope of the curve 𝑦 =sin⁡𝑥 steepest in the positive and negative directions?)

We hope that you can see for yourself how the way the slope of the tangent line to the curve 𝑦 =sin⁡𝑥 leads directly to the derivative 𝑓′(𝑥) =cos⁡𝑥.

II. Derivative of cos⁡𝑥: 𝑑𝑑𝑥cos⁡𝑥 = −sin⁡𝑥

The approach to finding the derivative of cos⁡𝑥 is very much the same as that for sin⁡𝑥 above; we'll just make use of a different trig addition formula.

We start of course with the definition of the derivative applied to 𝑔(𝑥) =cos⁡𝑥. Then 𝑔(𝑥+ℎ)=cos⁡(𝑥+ℎ)=cos⁡𝑥cos⁡ℎ−sin⁡𝑥sin⁡ℎ where we have used the trig addition formula cos⁡(𝑥 +𝑦) =cos⁡𝑥cos⁡𝑦 −sin⁡𝑥sin⁡𝑦.

The definition of the derivative for 𝑔(𝑥) =cos⁡𝑥 then gives us

𝑑𝑑𝑥cos⁡𝑥=limℎ→0cos⁡(𝑥+ℎ)−cos⁡𝑥ℎ=limℎ→0(cos⁡𝑥cos⁡ℎ−sin⁡𝑥sin⁡ℎ)−cos⁡𝑥ℎ=limℎ→0cos⁡𝑥cos⁡ℎ−cos⁡𝑥−sin⁡𝑥sin⁡ℎℎ=limℎ→0cos⁡𝑥cos⁡ℎ−cos⁡𝑥ℎ−limℎ→0sin⁡𝑥sin⁡ℎℎ=limℎ→0cos⁡𝑥(cos⁡ℎ−1)ℎ−limℎ→0sin⁡𝑥sin⁡ℎℎ[sin⁡𝑥 and cos⁡𝑥 are unaffected by the limit]=cos⁡𝑥[limℎ→0cos⁡ℎ−1ℎ]−sin⁡𝑥[limℎ→0sin⁡ℎℎ][Same two "special trig limits" as above]=cos⁡𝑥[limℎ→0cos⁡ℎ−1ℎ]0−sin⁡𝑥[limℎ→0sin⁡ℎℎ]1=−sin⁡𝑥

That is, we have another key result:

Derivative of cos x

𝑑𝑑𝑥cos⁡𝑥=−sin⁡𝑥

The following Exploration lets you see the graphical relationship between the function 𝑔(𝑥) =cos⁡𝑥 and its derivative 𝑔′(𝑥) = −sin⁡𝑥.

EXPLORATION 2: Moveable tangent line for 𝑦 =cos⁡𝑥

The top graph below shows 𝑦 =cos⁡𝑥.

The lower graph automatically updates to show the value of the function's derivative at the same point 𝑥 =𝑎 that's in the upper graph. That is, for each value of 𝑥 =𝑎, we automatically plot the point (𝑎,𝑓′(𝑎)) on the lower graph.

Verify for yourself:

  • Are the zeros of 𝑔′(𝑥) where you expect them to be? (That is, where the tangent line to the graph of 𝑦 =cos⁡𝑥 is horizontal?)
  • Are the maximum and minimum values of 𝑔′(𝑥) where you expect them to be? (Where is the slope of the curve 𝑦 =cos⁡𝑥 steepest in the positive and negative directions?)

We hope that you can see for yourself how the way the slope of the tangent line to the curve 𝑦 =cos⁡𝑥 leads directly to the derivative 𝑔′(𝑥) = −sin⁡𝑥.