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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 \(x=\frac{\pi }{4}.\)

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.

Agora tu Nenhuma calculadora resolve esta, mas as peças dela são computáveis. Tente um abaixo, ou digite seu próprio.

Prática (40)

Experimente cada um no papel primeiro. Revele a resposta para conferir; os verificados podem ser abertos no resolvedor com todos os passos.

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

Revelar a resposta

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 \(x=\frac{\pi }{4}.\)

Revelar a resposta

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 \(f(x)=\text{csc}\ x+x\ \text{tan}\ x.\)

Revelar a resposta

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 \(y=\text{sin}\ x.\)

Revelar a resposta

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

Revelar a resposta

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.

Revelar a resposta

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.

Revelar a resposta

\(v(\frac{5\pi }{6})=\text{-}\sqrt{3}<0\) 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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\(2\text{csc}\ x({\text{csc}}^{2}x+{\text{cot}}^{2}x)\)
\(y={\text{sec}}^{2}x\)

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Como: 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

Perguntas que as pessoas fazem

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.

Partes desta página são adaptadas a partir de OpenStax Calculus Volume 1 (CC BY-NC-SA 4.0). Condensado e reexplicado aqui; erros são nossos. Como a sua fonte é licenciada para uso não comercial, esta lição é gratuita, não vende nada e é compartilhada sob a mesma licença.

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