3.5: Derivatives of Trigonometric Functions (2024)

The Derivatives of \(\sin x\) and \(\cos x\)

The derivative of the sine function is the cosine and the derivative of the cosine function is the negative sine.

\[\dfrac{d}{dx}(\sin x)=\cos x \nonumber \]

\[\dfrac{d}{dx}(\cos x)=−\sin x \nonumber \]

Proof

Because the proofs for \(\dfrac{d}{dx}(\sin x)=\cos x\) and \(\dfrac{d}{dx}(\cos x)=−\sin x\) use similar techniques, we provide only the proof for \(\dfrac{d}{dx}(\sin x)=\cos x\). Before beginning, recall two important trigonometric limits:

\(\displaystyle \lim_{h→0}\dfrac{\sin h}{h}=1\) and \(\displaystyle \lim_{h→0}\dfrac{\cos h−1}{h}=0\).

The graphs of \(y=\dfrac{\sin h}{h}\) and \(y=\dfrac{\cos h−1}{h}\) are shown in Figure \(\PageIndex{2}\).

We also recall the following trigonometric identity for the sine of the sum of two angles:

\[\sin (x+h)=\sin x\cos h+\cos x\sin h. \nonumber \]

Now that we have gathered all the necessary equations and identities, we proceed with the proof.

\[\begin{align*} \dfrac{d}{dx}(\sin x) &=\lim_{h→0}\dfrac{\sin(x+h)−\sin x}{h} & & \text{Apply the definition of the derivative.}\\[4pt]
&=\lim_{h→0}\dfrac{\sin x\cos h+\cos x\sin h−\sin x}{h} & & \text{Use trig identity for the sine of the sum of two angles.}\\[4pt]
&=\lim_{h→0}\left(\dfrac{\sin x\cos h−\sin x}{h}+\dfrac{\cos x\sin h}{h}\right) & & \text{Regroup.}\\[4pt]
&=\lim_{h→0}\left(\sin x\left(\dfrac{\cos h−1}{h}\right)+(\cos x)\left(\dfrac{\sin h}{h}\right)\right) & & \text{Factor out }\sin x\text{ and }\cos x \\[4pt]
&=(\sin x)\lim_{h→0}\left(\dfrac{\cos h−1}{h}\right)+(\cos x)\lim_{h→0}\left(\dfrac{\sin h}{h}\right) & & \text{Factor }\sin x\text{ and }\cos x \text{ out of limits.} \\[4pt]
&=(\sin x)(0)+(\cos x)(1) & & \text{Apply trig limit formulas.}\\[4pt]
&=\cos x & & \text{Simplify.} \end{align*} \nonumber \]

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Example \(\PageIndex{1}\): Differentiating a Function Containing \(\sin x\)

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

Solution

Using the product rule, we have

\[ \begin{align*} f'(x) &=\dfrac{d}{dx}(5x^3)⋅\sin x+\dfrac{d}{dx}(\sin x)⋅5x^3 \\[4pt] &=15x^2⋅\sin x+\cos x⋅5x^3. \end{align*}\]

After simplifying, we obtain

\[f′(x)=15x^2\sin x+5x^3\cos x. \nonumber \]

Exercise \(\PageIndex{1}\)

Find the derivative of \(f(x)=\sin x\cos x.\)

Hint

Don’t forget to use the product rule.

Answer

\[f′(x)=\cos^2x−\sin^2x \nonumber \]

Example \(\PageIndex{2}\): Finding the Derivative of a Function Containing cos x

Find the derivative of \(g(x)=\dfrac{\cos x}{4x^2}\).

Solution

By applying the quotient rule, we have

\[g′(x)=\dfrac{(−\sin x)4x^2−8x(\cos x)}{(4x^2)^2}. \nonumber \]

Simplifying, we obtain

\[g′(x)=\dfrac{−4x^2\sin x−8x\cos x}{16x^4}=\dfrac{−x\sin x−2\cos x}{4x^3}. \nonumber \]

Exercise \(\PageIndex{2}\)

Find the derivative of \(f(x)=\dfrac{x}{\cos x}\).

Hint

Use the quotient rule.

Answer

\(f'(x) = \dfrac{\cos x+x\sin x}{\cos^2x}\)

Example \(\PageIndex{3}\): An Application to Velocity

A particle moves along a coordinate axis in such a way that its position at time \(t\) is given by \(s(t)=2\sin t−t\) for \(0≤t≤2π.\) At what times is the particle at rest?

Solution

To determine when the particle is at rest, set \(s′(t)=v(t)=0.\) Begin by finding \(s′(t).\) We obtain

\[s′(t)=2 \cos t−1, \nonumber \]

so we must solve

\[2 \cos t−1=0\text{ for }0≤t≤2π. \nonumber \]

The solutions to this equation are \(t=\dfrac{π}{3}\) and \(t=\dfrac{5π}{3}\). Thus the particle is at rest at times \(t=\dfrac{π}{3}\) and \(t=\dfrac{5π}{3}\).

Exercise \(\PageIndex{3}\)

A particle moves along a coordinate axis. Its position at time \(t\) is given by \(s(t)=\sqrt{3}t+2\cos t\) for \(0≤t≤2π.\) At what times is the particle at rest?

Hint

Use the previous example as a guide.

Answer

\(t=\dfrac{π}{3},\quad t=\dfrac{2π}{3}\)

Example \(\PageIndex{4}\): The Derivative of the Tangent Function

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

Solution

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

\(f(x)=\tan x =\dfrac{\sin x}{\cos x}\).

Now apply the quotient rule to obtain

\(f′(x)=\dfrac{\cos x\cos x−(−\sin x)\sin x}{(\cos x)^2}\).

Simplifying, we obtain

\[f′(x)=\dfrac{\cos^2x+\sin^2 x}{\cos^2x}. \nonumber \]

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

\[f′(x)=\dfrac{1}{\cos^2x} \nonumber \]

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

\(f′(x)=\text{sec}^2 x\).

Exercise \(\PageIndex{4}\)

Find the derivative of \(f(x)=\cot x .\)

Hint

Rewrite \(\cot x \) as \(\dfrac{\cos x}{\sin x}\) and use the quotient rule.

Answer

\(f′(x)=−\csc^2 x\)

Derivatives of \(\tan x\), \(\cot x\), \(\sec x\), and \(\csc x\)

The derivatives of the remaining trigonometric functions are as follows:

\[\begin{align} \dfrac{d}{dx}(\tan x )&=\sec^2x\\[4pt]
\dfrac{d}{dx}(\cot x )&=−\csc^2x\\[4pt]
\dfrac{d}{dx}(\sec x)&=\sec x \tan x\\[4pt]
\dfrac{d}{dx}(\csc x)&=−\csc x \cot x. \end{align} \nonumber \]

Example \(\PageIndex{5}\): Finding the Equation of a Tangent Line

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

Solution

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

\(f\left(\frac{π}{4}\right)=\cot\frac{π}{4}=1\).

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

\(f′(x)=−\csc^2 x\) and \(f′\left(\frac{π}{4}\right)=−\csc^2\left(\frac{π}{4}\right)=−2\).

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

\(y−1=−2\left(x−\frac{π}{4}\right)\)

or equivalently,

\(y=−2x+1+\frac{π}{2}\).

Example \(\PageIndex{6}\): Finding the Derivative of Trigonometric Functions

Find the derivative of \(f(x)=\csc x+x\tan x .\)

Solution

To find this derivative, we must use both the sum rule and the product rule. Using the sum rule, we find

\(f′(x)=\dfrac{d}{dx}(\csc x)+\dfrac{d}{dx}(x\tan x )\).

In the first term, \(\dfrac{d}{dx}(\csc x)=−\csc x\cot x ,\) and by applying the product rule to the second term we obtain

\(\dfrac{d}{dx}(x\tan x )=(1)(\tan x )+(\sec^2 x)(x)\).

Therefore, we have

\(f′(x)=−\csc x\cot x +\tan x +x\sec^2 x\).

Exercise \(\PageIndex{5}\)

Find the derivative of \(f(x)=2\tan x −3\cot x .\)

Hint

Use the rule for differentiating a constant multiple and the rule for differentiating a difference of two functions.

Answer

\(f′(x)=2\sec^2 x+3\csc^2 x\)

Exercise \(\PageIndex{6}\)

Find the slope of the line tangent to the graph of \(f(x)=\tan x \) at \(x=\dfrac{π}{6}\).

Hint

Evaluate the derivative at \(x=\dfrac{π}{6}\).

Answer

\(\dfrac{4}{3}\)

Example \(\PageIndex{7}\): Finding Higher-Order Derivatives of \(y=\sin x\)

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

Solution

Each step in the chain is straightforward:

\[\begin{align*} y&=\sin x \\[4pt]
\dfrac{dy}{dx}&=\cos x \\[4pt]
\dfrac{d^2y}{dx^2}&=−\sin x \\[4pt]
\dfrac{d^3y}{dx^3}&=−\cos x \\[4pt]
\dfrac{d^4y}{dx^4}&=\sin x \end{align*}\]

Analysis

Once we recognize the pattern of derivatives, we can find any higher-order derivative by determining the step in the pattern to which it corresponds. For example, every fourth derivative of \(\sin x\) equals \(\sin x\), so

\[\dfrac{d^4}{dx^4}(\sin x)=\dfrac{d^8}{dx^8}(\sin x)=\dfrac{d^{12}}{dx^{12}}(\sin x)=…=\dfrac{d^{4n}}{dx^{4n}}(\sin x)=\sin x \nonumber \]

\[\dfrac{d^5}{dx^5}(\sin x)=\dfrac{d^9}{dx^9}(\sin x)=\dfrac{d^{13}}{dx^{13}}(\sin x)=…=\dfrac{d^{4n+1}}{dx^{4n+1}}(\sin x)=\cos x. \nonumber \]

Exercise \(\PageIndex{7}\)

For \(y=\cos x\), find \(\dfrac{d^4y}{dx^4}\).

Hint

See the previous example.

Answer

\(\cos x\)

Example \(\PageIndex{8}\): Using the Pattern for Higher-Order Derivatives of \(y=\sin x\)

Find \(\dfrac{d^{74}}{dx^{74}}(\sin x)\).

Solution

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

\[\dfrac{d^{74}}{dx^{74}}(\sin x)=\dfrac{d^{72+2}}{dx^{72+2}}(\sin x)=\dfrac{d^2}{dx^2}(\sin x)=−\sin x. \nonumber \]

Exercise \(\PageIndex{8}\)

For \(y=\sin x\), find \(\dfrac{d^{59}}{dx^{59}}(\sin x).\)

Hint

\(\dfrac{d^{59}}{dx^{59}}(\sin x)=\dfrac{d^{4⋅14+3}}{dx^{4⋅14+3}}(\sin x)\)

Answer

\(−\cos x\)

Example \(\PageIndex{9}\): An Application to Acceleration

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

Solution

First find \(v(t)=s′(t)\)

\[v(t)=s′(t)=−\cos t . \nonumber \]

Thus,

\(v\left(\frac{π}{4}\right)=−\dfrac{1}{\sqrt{2}}=-\dfrac{\sqrt{2}}{2}\).

Next, find \(a(t)=v′(t)\). Thus, \(a(t)=v′(t)=\sin t\) and we have

\(a\left(\frac{π}{4}\right)=\dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}}{2}\).

Since \(v\left(\frac{π}{4}\right)=−\dfrac{\sqrt{2}}{2}<0\) and \(a\left(\frac{π}{4}\right)=\dfrac{\sqrt{2}}{2}>0\), 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 traveling. Consequently, the particle is slowing down.

Exercise \(\PageIndex{9}\)

A block attached to a spring is moving vertically. Its position at time t is given by \(s(t)=2\sin t\). Find \(v\left(\frac{5π}{6}\right)\) and \(a\left(\frac{5π}{6}\right)\). Compare these values and decide whether the block is speeding up or slowing down.

Hint

Use Example \(\PageIndex{9}\) as a guide.

Answer

\(v\left(\frac{5π}{6}\right)=−\sqrt{3}<0\) and \(a\left(\frac{5π}{6}\right)=−1<0\). The block is speeding up.