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Derivatives of Trigonometric Functions

Find the derivatives of the sine and cosine function.

Derivatives of the Sine and Cosine Functions

We begin our exploration of the derivative for the sine function by using the formula to make a reasonable guess at its derivative. Recall that for a function \(f(x),\)

\[{f}^{'}(x)=\underset{h\to 0}{\text{lim}}\frac{f(x+h)-f(x)}{h}.\]

Consequently, for values of \(h\) very close to 0, \({f}^{'}(x)\approx \frac{f(x+h)-f(x)}{h}.\) We see that by using \(h=0.01,\)

\[\frac{d}{dx}(\text{sin}\ x)\approx \frac{\text{sin}\ (x+0.01)-\text{sin}\ x}{0.01}\]

By setting \(D(x)=\frac{\text{sin}\ (x+0.01)-\text{sin}\ x}{0.01}\) and using a graphing utility, we can get a graph of an approximation to the derivative of \(\text{sin}\ x\) ().

Upon inspection, the graph of \(D(x)\) appears to be very close to the graph of the cosine function. Indeed, we will show that

\[\frac{d}{dx}(\text{sin}\ x)=\text{cos}\ x.\]

If we were to follow the same steps to approximate the derivative of the cosine function, we would find that

\[\frac{d}{dx}(\text{cos}\ x)=\text{-}\text{sin}\ x.\]

Condensed — the full section is in OpenStax Calculus Volume 1.

Derivatives of Other Trigonometric Functions

Since the remaining four trigonometric functions may be expressed as quotients involving sine, cosine, or both, we can use the quotient rule to find formulas for their derivatives.

Example

Try it.

Find the derivative of \(f(x)=\text{tan}\ x.\)

Solution

Start by expressing \(\text{tan}\ x\) as the quotient of \(\text{sin}\ x\) and \(\text{cos}\ x:\)

\[f(x)=\text{tan}\ x=\frac{\text{sin}\ x}{\text{cos}\ x}.\]

Now apply the quotient rule to obtain

\[{f}^{'}(x)=\frac{\text{cos}\ x\ \text{cos}\ x-(\text{-}\text{sin}\ x)\text{sin}\ x}{{(\text{cos}\ x)}^{2}}.\]

Simplifying, we obtain

\[{f}^{'}(x)=\frac{{\text{cos}}^{2}x+{\ \text{sin}}^{2}x}{{\text{cos}}^{2}x}.\]

Recognizing that \({\text{cos}}^{2}x+{\text{sin}}^{2}x=1,\) by the Pythagorean theorem, we now have

\[{f}^{'}(x)=\frac{1}{{\text{cos}}^{2}x}.\]

Finally, use the identity \(\text{sec}\ x=\frac{1}{\text{cos}\ x}\) to obtain

\[{f}^{'}(x)={\text{sec}}^{2}x.\]

The derivatives of the remaining trigonometric functions may be obtained by using similar techniques. We provide these formulas in the following theorem.

Example

Try it.

Find the equation of a line tangent to the graph of \(f(x)=\text{cot}\ x\) at \(x=\frac{\text{\pi }}{4}.\)

Solution

To find an equation of the tangent line, we need a point and a slope at that point. To find the point, compute

\[f(\frac{\pi }{4})=\text{cot}\ \frac{\pi }{4}=1.\]

Thus the tangent line passes through the point \((\frac{\pi }{4},1).\) Next, find the slope by finding the derivative of \(f(x)=\text{cot}\ x\) and evaluating it at \(\frac{\pi }{4}\text{:}\)

\[{f}^{'}(x)=\text{-}{\text{csc}}^{2}x\ \text{and}\ {f}^{'}(\frac{\pi }{4})=\text{-}{\text{csc}}^{2}(\frac{\pi }{4})=-2.\]

Using the point-slope equation of the line, we obtain

\[y-1=-2(x-\frac{\pi }{4})\]

or equivalently,

\[y=-2x+1+\frac{\pi }{2}.\]

Condensed — the full section is in OpenStax Calculus Volume 1.

Higher-Order Derivatives

The higher-order derivatives of \(\text{sin}\ x\) and \(\text{cos}\ x\) follow a repeating pattern. By following the pattern, we can find any higher-order derivative of \(\text{sin}\ x\) and \(\text{cos}\ x.\)

Example

Try it.

Find the first four derivatives of \(y=\text{sin}\ x.\)

Solution

Each step in the chain is straightforward:

\[\begin{array}{lll}y & = & \text{sin}\ x \\ \frac{dy}{dx} & = & \text{cos}\ x \\ \frac{{d}^{2}y}{d{x}^{2}} & = & \text{-}\text{sin}\ x \\ \frac{{d}^{3}y}{d{x}^{3}} & = & \text{-}\text{cos}\ x \\ \frac{{d}^{4}y}{d{x}^{4}} & = & \text{sin}\ x.\end{array}\]

Example

Try it.

Find \(\frac{{d}^{74}}{d{x}^{74}}(\text{sin}\ x).\)

Solution

We can see right away that for the 74th derivative of \(\text{sin}\ x,74=4(18)+2,\) so

\[\frac{{d}^{74}}{d{x}^{74}}(\text{sin}\ x)=\frac{{d}^{72+2}}{d{x}^{72+2}}(\text{sin}\ x)=\frac{{d}^{2}}{d{x}^{2}}(\text{sin}\ x)=\text{-}\text{sin}\ x.\]

Example

Try it.

A particle moves along a coordinate axis in such a way that its position at time \(t\) is given by \(s(t)=2-\text{sin}\ t.\) Find \(v(\pi \text{/}4)\) and \(a(\pi \text{/}4).\) Compare these values and decide whether the particle is speeding up or slowing down.

Solution

First find \(v(t)={s}^{'}(t)\text{:}\)

\[v(t)={s}^{'}(t)=\text{-}\text{cos}\ t.\]

Thus,

\[v(\frac{\pi }{4})=-\frac{1}{\sqrt{2}}.\]

Next, find \(a(t)={v}^{'}(t).\) Thus, \(a(t)={v}^{'}(t)=\text{sin}\ t\) and we have

\[a(\frac{\pi }{4})=\frac{1}{\sqrt{2}}.\]

Since \(v(\frac{\pi }{4})=-\frac{1}{\sqrt{2}}<0\) and \(a(\frac{\pi }{4})=\frac{1}{\sqrt{2}}>0,\) we see that velocity and acceleration are acting in opposite directions; that is, the object is being accelerated in the direction opposite to the direction in which it is travelling. Consequently, the particle is slowing down.

Key Concepts

We can find the derivatives of sin x and cos x by using the definition of derivative and the limit formulas found earlier. The results are
\[\frac{d}{dx}\ \text{sin}\ x=\text{cos}\ x\ \frac{d}{dx}\ \text{cos}\ x=\text{-}\text{sin}\ x.\]
With these two formulas, we can determine the derivatives of all six basic trigonometric functions.

Key Equations

Derivative of sine function	\(\frac{d}{dx}(\text{sin}\ x)=\text{cos}\ x\)
Derivative of cosine function	\(\frac{d}{dx}(\text{cos}\ x)=\text{-}\text{sin}\ x\)
Derivative of tangent function	\(\frac{d}{dx}(\text{tan}\ x)={\text{sec}}^{2}x\)
Derivative of cotangent function	\(\frac{d}{dx}(\text{cot}\ x)=\text{-}{\text{csc}}^{2}x\)
Derivative of secant function	\(\frac{d}{dx}(\text{sec}\ x)=\text{sec}\ x\ \text{tan}\ x\)
Derivative of cosecant function	\(\frac{d}{dx}(\text{csc}\ x)=\text{-}\text{csc}\ x\ \text{cot}\ x\)

Derivatives of Trigonometric Functions

For the following exercises, find \(\frac{dy}{dx}\) for the given functions.

For the following exercises, find an equation of the tangent line to each of the given functions at the indicated values of \(x.\) Then use a calculator to graph both the function and the tangent line to ensure the equation for the tangent line is correct.

For the following exercises, find \(\frac{{d}^{2}y}{d{x}^{2}}\) for the given functions.

For the following exercises, use the quotient rule to derive the given equations.

For the following exercises, find the requested higher-order derivative for the given functions.

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

Find the derivative of \(f(x)=5{x}^{3}\text{sin}\ x.\)

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Using the product rule, we have
\[\begin{array}{ll}f'(x) & =\frac{d}{dx}(5{x}^{3})\cdot \text{sin}\ x+\frac{d}{dx}(\text{sin}\ x)\cdot 5{x}^{3} \\ & =15{x}^{2}\cdot \text{sin}\ x+\text{cos}\ x\cdot 5{x}^{3}.\end{array}\]
After simplifying, we obtain
\[{f}^{'}(x)=15{x}^{2}\text{sin}\ x+5{x}^{3}\text{cos}\ x.\]
Find the derivative of \(f(x)=\text{sin}\ x\ \text{cos}\ x.\)

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\({f}^{'}(x)={\text{cos}}^{2}x-{\text{sin}}^{2}x\)
Find the derivative of \(g(x)=\frac{\text{cos}\ x}{4{x}^{2}}.\)

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By applying the quotient rule, we have
\[{g}^{'}(x)=\frac{(\text{-}\text{sin}\ x)4{x}^{2}-8x(\text{cos}\ x)}{{(4{x}^{2})}^{2}}.\]
Simplifying, we obtain
\[\begin{array}{ll}{g}^{'}(x) & =\frac{-4{x}^{2}\text{sin}\ x-8x\ \text{cos}\ x}{16{x}^{4}} \\ & =\frac{\text{-}x\ \text{sin}\ x-2\ \text{cos}\ x}{4{x}^{3}}.\end{array}\]
Find the derivative of \(f(x)=\frac{x}{\text{cos}\ x}.\)

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\(\frac{\text{cos}\ x+x\ \text{sin}\ x}{{\text{cos}}^{2}x}\)
A particle moves along a coordinate axis in such a way that its position at time \(t\) is given by \(s(t)=2\ \text{sin}\ t-t\) for \(0\le t\le 2\pi .\) At what times is the particle at rest?

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To determine when the particle is at rest, set \({s}^{'}(t)=v(t)=0.\) Begin by finding \({s}^{'}(t).\) We obtain
\[{s}^{'}(t)=2\ \text{cos}\ t-1,\]
so we must solve
\[2\ \text{cos}\ t-1=0\ \text{for}\ 0\le t\le 2\pi .\]
The solutions to this equation are \(t=\frac{\pi }{3}\) and \(t=\frac{5\pi }{3}.\) Thus the particle is at rest at times \(t=\frac{\pi }{3}\) and \(t=\frac{5\pi }{3}.\)
A particle moves along a coordinate axis. Its position at time \(t\) is given by \(s(t)=\sqrt{3}t+2\ \text{cos}\ t\) for \(0\le t\le 2\pi .\) At what times is the particle at rest?

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\(t=\frac{\pi }{3},t=\frac{2\pi }{3}\)
Find the derivative of \(f(x)=\text{tan}\ x.\)

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Start by expressing \(\text{tan}\ x\) as the quotient of \(\text{sin}\ x\) and \(\text{cos}\ x:\)
\[f(x)=\text{tan}\ x=\frac{\text{sin}\ x}{\text{cos}\ x}.\]
Now apply the quotient rule to obtain
\[{f}^{'}(x)=\frac{\text{cos}\ x\ \text{cos}\ x-(\text{-}\text{sin}\ x)\text{sin}\ x}{{(\text{cos}\ x)}^{2}}.\]
Simplifying, we obtain
\[{f}^{'}(x)=\frac{{\text{cos}}^{2}x+{\ \text{sin}}^{2}x}{{\text{cos}}^{2}x}.\]
Recognizing that \({\text{cos}}^{2}x+{\text{sin}}^{2}x=1,\) by the Pythagorean theorem, we now have
\[{f}^{'}(x)=\frac{1}{{\text{cos}}^{2}x}.\]
Finally, use the identity \(\text{sec}\ x=\frac{1}{\text{cos}\ x}\) to obtain
\[{f}^{'}(x)={\text{sec}}^{2}x.\]
Find the derivative of \(f(x)=\text{cot}\ x.\)

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\({f}^{'}(x)=\text{-}{\text{csc}}^{2}x\)
Find the equation of a line tangent to the graph of \(f(x)=\text{cot}\ x\) at \(x=\frac{\text{\pi }}{4}.\)

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To find an equation of the tangent line, we need a point and a slope at that point. To find the point, compute
\[f(\frac{\pi }{4})=\text{cot}\ \frac{\pi }{4}=1.\]
Thus the tangent line passes through the point \((\frac{\pi }{4},1).\) Next, find the slope by finding the derivative of \(f(x)=\text{cot}\ x\) and evaluating it at \(\frac{\pi }{4}\text{:}\)
\[{f}^{'}(x)=\text{-}{\text{csc}}^{2}x\ \text{and}\ {f}^{'}(\frac{\pi }{4})=\text{-}{\text{csc}}^{2}(\frac{\pi }{4})=-2.\]
Using the point-slope equation of the line, we obtain
\[y-1=-2(x-\frac{\pi }{4})\]
or equivalently,
\[y=-2x+1+\frac{\pi }{2}.\]
Find the derivative of \(f(x)=\text{csc}\ x+x\ \text{tan}\ x.\)

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To find this derivative, we must use both the sum rule and the product rule. Using the sum rule, we find
\[{f}^{'}(x)=\frac{d}{dx}(\text{csc}\ x)+\frac{d}{dx}(x\ \text{tan}\ x).\]
In the first term, \(\frac{d}{dx}(\text{csc}\ x)=\text{-}\text{csc}\ x\ \text{cot}\ x,\) and by applying the product rule to the second term we obtain
\[\frac{d}{dx}(x\ \text{tan}\ x)=(1)(\text{tan}\ x)+({\text{sec}}^{2}x)(x).\]
Therefore, we have
\[{f}^{'}(x)=\text{-}\text{csc}\ x\ \text{cot}\ x+\text{tan}\ x+x\ {\text{sec}}^{2}x.\]
Find the derivative of \(f(x)=2\ \text{tan}\ x-3\ \text{cot}\ x.\)

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\({f}^{'}(x)=2\ {\text{sec}}^{2}x+3\ {\text{csc}}^{2}x\)
Find the slope of the line tangent to the graph of \(f(x)=\text{tan}\ x\) at \(x=\frac{\pi }{6}.\)

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\(\frac{4}{3}\)
Find the first four derivatives of \(y=\text{sin}\ x.\)

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Each step in the chain is straightforward:
\[\begin{array}{lll}y & = & \text{sin}\ x \\ \frac{dy}{dx} & = & \text{cos}\ x \\ \frac{{d}^{2}y}{d{x}^{2}} & = & \text{-}\text{sin}\ x \\ \frac{{d}^{3}y}{d{x}^{3}} & = & \text{-}\text{cos}\ x \\ \frac{{d}^{4}y}{d{x}^{4}} & = & \text{sin}\ x.\end{array}\]
For \(y=\text{cos}\ x,\) find \(\frac{{d}^{4}y}{d{x}^{4}}.\)

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\(\text{cos}\ x\)
Find \(\frac{{d}^{74}}{d{x}^{74}}(\text{sin}\ x).\)

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We can see right away that for the 74th derivative of \(\text{sin}\ x,74=4(18)+2,\) so
\[\frac{{d}^{74}}{d{x}^{74}}(\text{sin}\ x)=\frac{{d}^{72+2}}{d{x}^{72+2}}(\text{sin}\ x)=\frac{{d}^{2}}{d{x}^{2}}(\text{sin}\ x)=\text{-}\text{sin}\ x.\]
For \(y=\text{sin}\ x,\) find \(\frac{{d}^{59}}{d{x}^{59}}(\text{sin}\ x).\)

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\(\text{-}\text{cos}\ x\)
A particle moves along a coordinate axis in such a way that its position at time \(t\) is given by \(s(t)=2-\text{sin}\ t.\) Find \(v(\pi \text{/}4)\) and \(a(\pi \text{/}4).\) Compare these values and decide whether the particle is speeding up or slowing down.

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First find \(v(t)={s}^{'}(t)\text{:}\)
\[v(t)={s}^{'}(t)=\text{-}\text{cos}\ t.\]
Thus,
\[v(\frac{\pi }{4})=-\frac{1}{\sqrt{2}}.\]
Next, find \(a(t)={v}^{'}(t).\) Thus, \(a(t)={v}^{'}(t)=\text{sin}\ t\) and we have
\[a(\frac{\pi }{4})=\frac{1}{\sqrt{2}}.\]
Since \(v(\frac{\pi }{4})=-\frac{1}{\sqrt{2}}<0\) and \(a(\frac{\pi }{4})=\frac{1}{\sqrt{2}}>0,\) we see that velocity and acceleration are acting in opposite directions; that is, the object is being accelerated in the direction opposite to the direction in which it is travelling. Consequently, the particle is slowing down.
A block attached to a spring is moving vertically. Its position at time \(t\) is given by \(s(t)=2\ \text{sin}\ t.\) Find \(v(\frac{5\pi }{6})\) and \(a(\frac{5\pi }{6}).\) Compare these values and decide whether the block is speeding up or slowing down.

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\(v(\frac{5\pi }{6})=\text{-}\sqrt{3}<0\) and \(a(\frac{5\pi }{6})=-1<0.\) The block is speeding up.
\(y={x}^{2}-\text{sec}\ x+1\)

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\(\frac{dy}{dx}=2x-\text{sec}\ x\ \text{tan}\ x\)
\(y=3\ \text{csc}\ x+\frac{5}{x}\)
\(y={x}^{2}\text{cot}\ x\)

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\(\frac{dy}{dx}=2x\ \text{cot}\ x-{x}^{2}{\text{csc}}^{2}x\)
\(y=x-{x}^{3}\text{sin}\ x\)
\(y=\frac{\text{sec}\ x}{x}\)

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\(\frac{dy}{dx}=\frac{x\ \text{sec}\ x\ \text{tan}\ x-\text{sec}\ x}{{x}^{2}}\)
\(y=\text{sin}\ x\ \text{tan}\ x\)
\(y=(x+\text{cos}\ x)(1-\text{sin}\ x)\)

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\(\frac{dy}{dx}=(1-\text{sin}\ x)(1-\text{sin}\ x)-\text{cos}\ x(x+\text{cos}\ x)\)
\(y=\frac{\text{tan}\ x}{1-\text{sec}\ x}\)
\(y=\frac{1-\text{cot}\ x}{1+\text{cot}\ x}\)

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\(\frac{dy}{dx}=\frac{2\ {\text{csc}}^{2}x}{{(1+\text{cot}\ x)}^{2}}\)
\(y=\text{cos}\ x(1+\text{csc}\ x)\)
[T] \(f(x)=\text{-}\text{sin}\ x,x=0\)

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\(y=\text{-}x\)
[T] \(f(x)=\text{csc}\ x,x=\frac{\pi }{2}\)
[T] \(f(x)=1+\text{cos}\ x,x=\frac{3\pi }{2}\)

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\(y=x+\frac{2-3\pi }{2}\)
[T] \(f(x)=\text{sec}\ x,x=\frac{\pi }{4}\)
[T] \(f(x)={x}^{2}-\text{tan}\ x,\ x=0\)

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\(y=\text{-}x\)
[T] \(f(x)=5\ \text{cot}\ x,\ x=\frac{\pi }{4}\)
\(y=x\ \text{sin}\ x-\text{cos}\ x\)

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\(3\ \text{cos}\ x-x\ \text{sin}\ x\)
\(y=\text{sin}\ x\ \text{cos}\ x\)
\(y=x-\frac{1}{2}\ \text{sin}\ x\)

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\(\frac{1}{2}\ \text{sin}\ x\)
\(y=\frac{1}{x}+\text{tan}\ x\)
\(y=2\ \text{csc}\ x\)

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\(2\text{csc}\ x({\text{csc}}^{2}x+{\text{cot}}^{2}x)\)
\(y={\text{sec}}^{2}x\)

Symbols used here

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

\pi

pi
Ratio of a circle's circumference to its diameter, 3.14159…

f'(x),\ \frac{dy}{dx}

derivative
Instantaneous rate of change; slope of the graph.

\approx

approximately equal
Equal to the precision shown, not exactly.

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

Euler's number
2.71828…, the base whose exponential is its own derivative.

\infty

infinity
Not a number: "grows without bound" in limits and intervals.

\sin,\ \cos,\ \tan

sine, cosine, tangent
Ratios of sides in a right triangle; coordinates on the unit circle.

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\sum_{k=1}^{n} a_k

summation
Add a_k for k = 1 up to n.

\lim_{x \to a} f(x)

limit
The value f(x) approaches as x approaches a.

\int f(x)\,dx,\ \int_a^b

integral
Antiderivative (indefinite) or signed area from a to b (definite).

y',\ y''

first and second derivative of y
Prime notation for derivatives with respect to x (or t).

C,\ C_1,\ C_2

arbitrary constants
Constants of integration fixed by initial conditions.

How to: Derivatives of Trigonometric Functions

Find the derivatives of the sine and cosine function.
Find the derivatives of the standard trigonometric functions.
Calculate the higher-order derivatives of the sine and cosine.
We can find the derivatives of sin
With these two formulas, we can determine the derivatives of all six basic trigonometric functions.
Sketch one period of the position function for
Find the velocity function.
Sketch one period of the velocity function for

Questions people ask

What is a derivative in one sentence?

The slope of the graph at a point — the rate at which the output is changing there. Speed is the derivative of position.

What is an integral in one sentence?

The accumulated total of a rate — the area under the curve. Distance travelled is the integral of speed.

Why are derivatives and integrals opposites?

That is the fundamental theorem of calculus: accumulating a rate and then measuring how fast the accumulation grows gets you back the rate. Integration undoes differentiation up to a constant.

When do I use substitution and when integration by parts?

Substitution when part of the integrand is the derivative of another part (u and du both present). Parts when the integrand is a product of two unrelated kinds of function — a polynomial times an exponential, log or trig function.

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Parts of this page are adapted from OpenStax Calculus Volume 1 (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

@ action Calculus

Limits Derivatives Integrals Definite integrals Taylor series Series and sums Maxima and minima The chain rule Implicit differentiation Related rates and optimisation Integration techniques: substitution, parts, partial fractions Applications of integration: area, volume, arc length Infinite series and convergence tests