4.4 Variations on Derivatives of Trigonometric Functions

Key points

 

In this section we compute the derivatives of the trigonometric functions via the definition
\[f'(x) = \lim_{x\to a}\frac{f(x)-f(a)}{x-a}.\]
In this context, there will be a clear difference between the technique to compute the difference quotient for sine, cosine, secant, and cosecant, versus those for tangent and cotangent:

• To compute the derivatives of \(\sin x\) and \(\cos x\) we will rewrite the numerators of their difference quotients as products of trigonometric functions.

  Here is a good point for you to review the technique to rewrite differences of sines or differences of cosines as product (see 4.13).

• For tangent and cotangent, the computation will be relatively straightforward application of the subtraction identities for sine and cosine.

4.4.1 Derivatives Sine, Cosine, Cosecant, Secant

 

Proposition 1. 1. \(\displaystyle \frac{d}{dx}\sin x = \lim_{x\to a} \frac{\sin x-\sin a}{x-a}= \cos a\) for all \(x\in \mathbb{R}\);

2. \(\displaystyle \frac{d}{dx}\cos x = \lim_{x\to a} \frac{\cos x-\cos a}{x-a}= -\sin a\) for all \(x\in \mathbb{R}\)

Proof. 1. Recall that (see 4.1.4)
\[\sin x – \sin a = 2\sin \frac{x – a}{2}\cos \frac{x + a}{2}, \]
so that,
\begin{align*}
\lim_{x\to a} \frac{\sin x-\sin a}{x-a}
&= \lim_{x\to a} \frac{ 2\sin \frac{x – a}{2}\cos \frac{x + a}{2}}{x-a}
\underset{?}{=}\left(\lim_{x\to a} \frac{ 2\sin \frac{x – a}{2} }{2\frac{x – a}{2}}\right)
\left(\lim_{x\to a} \cos \frac{x + a}{2}\right)=\\
&=\left(\lim_{u\to 0} \frac{\sin u }{u}\right)\left(\lim_{x\to a} \cos \frac{x + a}{2}\right)
\underset{?}{=}\cos a, \; \text{with} u = \frac{x-a}{2}.
\end{align*}

2. Again (see 4.1.4)
\[\cos x – \cos a = -2\sin \frac{x – a}{2}\sin \frac{x + a}{2}, \]
so that,
\begin{align*}
\lim_{x\to a} \frac{\cos x-\cos a}{x-a}
&= -\lim_{x\to a} \frac{ 2\sin \frac{x – a}{2}\sin \frac{x + a}{2}}{x-a}
\underset{?}{=}-\left(\lim_{x\to a} \frac{ 2\sin \frac{x – a}{2} }{2\frac{x – a}{2}}\right)
\left(\lim_{x\to a} \sin \frac{x + a}{2}\right)=\\
&=-\left(\lim_{u\to 0} \frac{\sin u }{u}\right)\left(\lim_{x\to a} \sin \frac{x + a}{2}\right)
\underset{?}{=}-\sin a, \; \text{with} u = \frac{x-a}{2}.
\end{align*}

□

We leave it to the reader to prove the following corollary, by providing all the details of the proof.
Corollary 1. 1. \(\displaystyle \frac{d}{dx}\csc x =\lim_{x\to a} \frac{\csc x-\csc a}{x-a}= -\csc a \cot a\) for all \(x\in D_{\csc}\);

2. \(\displaystyle \frac{d}{dx}\sec x = \lim_{x\to a} \frac{\sec x-\sec a}{x-a}= \sec a \tan a\) for all \(x\in D_{\sec}\)

4.4.2 Derivatives of Tangent and Cotangent

 
Proposition 2. 1. \(\displaystyle \frac{d}{dx}\tan x = \lim_{x\to a} \frac{\tan x-\tan a}{x-a}= \sec^2a\) for all \(x\in D_{\tan}\).

2. \(\displaystyle \frac{d}{dx}\cot x = \lim_{x\to a} \frac{\cot x-\cot a}{x-a}= -\csc^2a\) for all \(x\in D_{\cot}\).

Proof. 1. Firs we rewrite the difference quotient:
\begin{align*}
\frac{\tan x-\tan a}{x-a} &= \frac{\displaystyle \frac{\sin x}{\cos x}-\frac{\sin a}{\cos a}}{x-a}
= \frac{\sin x \cos a -\sin a \cos x}{(x-a)\cos x \cos a}
=\frac{\sin (x-a)}{(x-a)\cos x \cos a} =\\
&=\frac{\sin (x-a)}{(x-a)}\sec x \sec a
=\frac{\sin u}{u}\sec x \sec a, u=x-a.
\end{align*}
It now follows that
\[\lim_{x\to a} \frac{\tan x-\tan a}{x-a} = \lim_{x\to a} \frac{\sin (x-a)}{(x-a)}\sec x \sec a = \lim_{u\to 0}\frac{\sin u}{u}\left (\lim_{x\to a}\sec x \sec a\right ) \underset{?}{=}\sec^2a.\]
2. The proof is very similar to the one above for tangent, and is left to the reader.

□

4.4.3 Compositions with Powers and Roots

 

Example 1. Let \(h(x)=\csc x^5\), show that \(\displaystyle \lim_{x\to a} \frac{\csc x^5-\csc a^5}{x-a} = -5a^4\cot a^5 \csc a^5\).

Solution. (1) First of all, we rewrite \(h\) in terms of sine.
\begin{align*}
\frac{\csc x^5-\csc a^5}{x-a}
&=\frac{\displaystyle \frac{1}{\sin x^5}-\frac{1}{\sin a^5}}{x-a}
=\frac{\sin a^5-\sin x^5}{(x-a)\sin x^5\sin a^5}=\\
&=-\frac{\sin x^5-\sin a^5}{x-a}
\csc x^5\csc a^5.
\end{align*}

(2) We convert the difference \(\sin x^5 – \sin a^5\) into a product (see 4.1.4):
\begin{align*}
\sin x^5 – \sin a^5 =2\sin \frac{x^5-a^5}{2} \cos \frac{x^5+a^5}{2} .
\end{align*}

Hence,
\begin{align*}
-\frac{\sin x^5-\sin a^5}{(x-a)}\csc x^5 \csc a^5
&=-\frac{2 \sin \frac{x^5-a^5}{2} \cos \frac{x^5+a^5}{2}}{x-a}\csc x^5\csc a^5 =\\
&=-\frac{\sin \frac{x^5-a^5}{2}}{\frac{x-a}{2}}\cos \frac{x^5+a^5}{2} \csc x^5 \csc a^5.
\end{align*}

(3) To apply \(\displaystyle \lim_{u\to 0}\frac{\sin u}{u} = 1\), we need to make the substitution \(\displaystyle u = \frac{x^5-a^5}{2}\). To accomplish this note that
\[\frac{1}{\frac{x-a}{2}} = \frac{x^4+x^3a+x^2a^2+xa^3+a^4}{{\frac{x-a}{2}\left(x^4+x^3a+x^2a^2+xa^3+a^4\right)}} = \frac{x^4+x^3a+x^2a^2+xa^3+a^4}{\frac{x^5-a^5}{2}}.\]
Thus
\[-\frac{\sin \frac{x^5-a^5}{2}}{\frac{x-a}{2}}\cos \frac{x^5+a^5}{2} \csc x^5 \csc a^5 =\]
\[
\underset{?}{=}-\frac{\sin \frac{x^5-a^5}{2}}{\frac{x^5-a^5}{2}}\left(x^4+x^3a+x^2a^2+xa^3+a^4\right)
\cos \frac{x^5+a^5}{2} \csc x^5 \csc a^5.
\]

(4) We can now compute the limit:
\begin{align*}
h'(x)&= -\lim_{x\to a} \frac{\sin \frac{x^5-a^5}{2}}{\frac{x^5-a^5}{2}}\left(x^4+x^3a+x^2a^2+xa^3+a^4\right)
\cos \frac{x^5+a^5}{2} \csc x^5 \csc a^5=\\
&=-\left(\lim_{u\to 0} \frac{\sin u}{u}\right)
\left(\lim_{x\to a}\left(x^4+x^3a+x^2a^2+xa^3+a^4\right)
\cos \frac{x^5+a^5}{2} \csc x^5 \csc a^5\right)=\\
&=-(1)(a^4+a^3a+a^2a^2+aa^3+a^4)\cos a^5\csc a^5 \csc a^5 =-5a^4\cot a^5 \csc a^5.
\end{align*}

Example 2. (a) Given \(h(x) = \sec \sqrt[3]{x}\) show that
\[ h'(x)= \lim_{x\to a} \frac{\sec \sqrt[3]{x}-\sec \sqrt[3]{a}}{x-a}=\frac{1}{3\sqrt[3]{a^2}} \tan\sqrt[3]{a} \sec \sqrt[3]{a};\]
(b) find an equation for the line tangent to the graph of \(h\) at the point with \(x\)-coordinate \(x=(\pi/3)^3\).

Solution. (a) We will compute the limit in four steps.

 
(1) First of all, we rewrite \(h\) in terms of cosine.
\begin{align*}
\frac{\sec \sqrt[3]{x}-\sec \sqrt[3]{a}}{x-a}
&=\frac{\displaystyle \frac{1}{\cos \sqrt[3]{x}}-\frac{1}{\cos \sqrt[3]{a}}}{x-a}
=\frac{\cos \sqrt[3]{a}-\cos \sqrt[3]{x}}{(x-a)\cos \sqrt[3]{x}\cos \sqrt[3]{a}}=\\
&=\frac{\cos \sqrt[3]{a}-\cos \sqrt[3]{x}}{x-a}
\sec \sqrt[3]{x} \sec \sqrt[3]{a}.
\end{align*}

(2) Again, using 4.1.3:
\begin{align*}
\cos \sqrt[3]{a} – \cos \sqrt[3]{x}
= 2 \sin \frac{\sqrt[3]{x}+\sqrt[3]{a}}{2} \sin \frac{\sqrt[3]{x}-\sqrt[3]{a}}{2}.
\end{align*}
\((*)\) Here we have used the fact that sine is an odd function, that is \(\sin(-x)=-\sin x\). Hence,
\[
\frac{\sec \sqrt[3]{x}-\sec \sqrt[3]{a}}{x-a}
=2\frac{\sin \frac{\sqrt[3]{x}-\sqrt[3]{a}}{2}}{x-a}\sin \frac{\sqrt[3]{x}+\sqrt[3]{a}}{2} \sec \sqrt[3]{x}\sec \sqrt[3]{a}.
\]

(3) To make the substitution
\(\displaystyle u = \frac{\sqrt[3]{x}-\sqrt[3]{a}}{2}\),
rewrite the denominator as follows:
\[
x-a = \left(\sqrt[3]{x}\right)^3-\left(\sqrt[3]{a}\right)^3
\underset{?}{=} \left(\sqrt[3]{x}-\sqrt[3]{x}\right)
\left(\sqrt[3]{x^2}+ \sqrt[3]{xa} + \sqrt[3]{a^2}\right).
\]
Thus
\begin{align*}
\frac{\sec \sqrt[3]{x}-\sec \sqrt[3]{a}}{x-a} &=
2\frac{\sin \frac{\sqrt[3]{x}-\sqrt[3]{a}}{2}}{x-a}\sin \frac{\sqrt[3]{x}+\sqrt[3]{a}}{2} \sec \sqrt[3]{x}\sec \sqrt[3]{a} = \\
&= 2\frac{\sin \frac{\sqrt[3]{x}-\sqrt[3]{a}}{2}}{ \left(\sqrt[3]{x}-\sqrt[3]{x}\right)
\left(\sqrt[3]{x^2}+ \sqrt[3]{xa} + \sqrt[3]{a^2}\right)}
\sin \frac{\sqrt[3]{x}+\sqrt[3]{a}}{2} \sec \sqrt[3]{x}\sec \sqrt[3]{a} = \\
&= \frac{\sin u}{u}\frac{1}{
\sqrt[3]{x^2}+ \sqrt[3]{xa} + \sqrt[3]{a^2}}
\sin \frac{\sqrt[3]{x}+\sqrt[3]{a}}{2} \sec \sqrt[3]{x}\sec \sqrt[3]{a}, u=\frac{\sqrt[3]{x}-\sqrt[3]{a}}{2}.
\end{align*}

(4) Finally, to compute the limit, given that by the substitution
\[u=\frac{\sqrt[3]{x}-\sqrt[3]{a}}{2} \Longrightarrow x\to a \Longleftrightarrow u\to 0,\]
then
\begin{align*}
\lim_{x\to a}\frac{\sec \sqrt[3]{x}-\sec \sqrt[3]{a}}{x-a}
&=
\lim_{x\to a} 2\frac{\sin \frac{\sqrt[3]{x}-\sqrt[3]{a}}{2}}{\sqrt[3]{x}-\sqrt[3]{x}}\frac{1}
{\sqrt[3]{x^2}+ \sqrt[3]{xa} + \sqrt[3]{a^2}}
\sin \frac{\sqrt[3]{x}+\sqrt[3]{a}}{2} \sec \sqrt[3]{x} \, \sec \sqrt[3]{a} = \\
&= \left(\lim_{u\to 0}\frac{\sin u}{u}\right)
\left(\lim_{x\to a}\frac{1}{\sqrt[3]{x^2}+ \sqrt[3]{xa} + \sqrt[3]{a^2}}
\sin \frac{\sqrt[3]{x}+\sqrt[3]{a}}{2} \sec \sqrt[3]{x}\, \sec \sqrt[3]{a}\right) =\\
&= 1\cdot
\frac{1}{\sqrt[3]{a^2}+ \sqrt[3]{aa} + \sqrt[3]{a^2}}
\sin \frac{\sqrt[3]{a}+\sqrt[3]{a}}{2} \, \sec \sqrt[3]{a}\, \sec \sqrt[3]{a} =\\
&=
\frac{1}{3\sqrt[3]{a^2}} \sin\sqrt[3]{a} \, \sec \sqrt[3]{a}\, \sec \sqrt[3]{a}
=\frac{1}{3\sqrt[3]{a^2}} \, \tan\sqrt[3]{a} \, \sec \sqrt[3]{a}.
\end{align*}

 
(b) Since the slope of the tangent line to the graph of \(h\) at \(x=(\pi/3)^3\) is given precisely by
\(m=h’\left ( (\pi/3)^3\right )\), we have
\[
m=\frac{1}{3\sqrt[3]{((\pi/3)^3)^2}} \tan\sqrt[3]{(\pi/3)^3} \sec \sqrt[3]{(\pi/3)^3}
=\frac{1}{3((\pi/3)^2} \tan(\pi/3) \sec (\pi/3)=\frac{6\sqrt{3}}{\pi^2}.\]
Since the point with \(x\) coordinate \(x=(\pi/3)^3\) has \(y\)-coordinate given by
\(y=h\left((\pi/3)^3\right) = \sec \sqrt[3]{(\pi/3)^3}=2\), using the point-slope equation for a line, we have
\[y-2=\frac{6\sqrt{3}}{\pi^2}\left(x-\left(\frac{\pi}{3}\right)^3\right) \Longleftrightarrow
y=\frac{6\sqrt{3}}{\pi^2}x -\frac{2\sqrt{3}\pi}{9}+2.\]

Example 3. Given \(h(x) = \cot \sqrt{x}\),

(a) show that \(\displaystyle \frac{\cot \sqrt{x}-\cot \sqrt{a}}{x-a} =-\frac{\sin u}{u}\frac{1}{\sqrt{x}+\sqrt{a}} \csc \sqrt{x} \csc \sqrt{a}\), with \( u =\sqrt{x}-\sqrt{a}\);

(b) use (a) to conclude that \(\displaystyle h'(x)=-\frac{1}{2}\frac{1}{\sqrt{x}}\csc^2\sqrt{x}\);

(c) find the points where the graph of \(h\) has a horizontal tangent line.

Solution. (a) Rewrite the difference quotient in terms and sine and cosine:
\begin{align*}
\frac{\cot \sqrt{x}-\cot \sqrt{a}}{x-a}
&= \frac{\displaystyle \frac{\cos \sqrt{x}}{\sin \sqrt{x}}-\frac{\cos \sqrt{a}}{\sin \sqrt{a}}}{x-a}
= \frac{\sin \sqrt{a} \cos \sqrt{x} -\sin \sqrt{x} \cos \sqrt{a}}{(x-a)\sin \sqrt{x} \sin \sqrt{a}} =\\
&=\frac{\sin (\sqrt{a}-\sqrt{x})}{(x-a)\sin \sqrt{x} \sin \sqrt{a}}
\underset{?}{=}-\frac{\sin (\sqrt{x}-\sqrt{a})}{(x-a)} \csc \sqrt{x} \csc \sqrt{a} =\\
&=-\frac{\sin (\sqrt{x}-\sqrt{a})}{\left(\sqrt{x}-\sqrt{a}\right)\left(\sqrt{x}+\sqrt{a}\right)}
\csc \sqrt{x} \csc \sqrt{a} = \\
&=-\frac{\sin u}{u}\frac{1}{\sqrt{x}+\sqrt{a}} \csc \sqrt{x} \csc \sqrt{a}, u =\sqrt{x}-\sqrt{a}
.
\end{align*}

(b) We need to compute the limit (using \(x\to a \Longleftrightarrow u =\sqrt{x}-\sqrt{a}\to 0\)):
\[\lim_{x\to a} \frac{\cot \sqrt{x}-\cot \sqrt{a}}{x-a} =-\left(\lim_{u\to 0} \frac{\sin u}{u}\right)\left(\lim_{x\to a} \frac{1}{\sqrt{x}+\sqrt{a}} \csc \sqrt{x} \csc \sqrt{a}\right)
\underset{?}{=} -\frac{1}{2}\frac{1}{\sqrt{a}}\csc^2\sqrt{a}.\]

(c) For a horizontal tangent line \(h'(x)=0\). Thus, we have
\[h'(x)=0 \Longleftrightarrow -\frac{1}{2}\frac{1}{\sqrt{x}}\csc^2\sqrt{x}=0
\underset{?}{\Longleftrightarrow } \frac{1}{\sqrt{x}}=0 \mathrm{or} \csc^2\sqrt{x}=0
,\]
but neither of these terms can be zero. Thus, the graph of \(h\) does not have any horizontal tangent lines.

Example 4. Compute the derivative of the function. (a) \(f(x) = \cos^4 x\); (b) \(g(x) =\sqrt[3]{\tan x}\); (c) \( h(x) =\sqrt{\csc x^5}\).

Solution. (a)
\begin{align*}
\lim_{x\to a}\frac{\cos^4x-\cos^4 a}{x-a} & = \lim_{x\to a}\frac{\cos x-\cos a}{x-a}\left (\cos^3 x+ \cos^2x\cos a+\cos x\cos^2a+\cos^3a\right )=\\
&\underset{?}{=}-\sin a\cdot 4\cos^3a.
\end{align*}

(b)
\begin{align*}
\lim_{x\to a}\frac{\sqrt[3]{\tan x} – \sqrt[3]{\tan a}}{x-a} &
= \lim_{x\to a}\frac{\sqrt[3]{\tan x} – \sqrt[3]{\tan a}}{x-a} \cdot
\frac
{\sqrt[3]{\tan^2 x} + \sqrt[3]{\tan x} \sqrt[3]{\tan a} + \sqrt[3]{\tan^2 a}}
{\sqrt[3]{\tan^2 x} + \sqrt[3]{\tan x} \sqrt[3]{\tan a} + \sqrt[3]{\tan^2 a}}
=\\
&\underset{?}{=} \lim_{x\to a}\frac{ \tan x – \tan a}{x-a}\cdot \frac{1}{\sqrt[3]{\tan^2 x} + \sqrt[3]{\tan x} \sqrt[3]{\tan a} + \sqrt[3]{\tan^2 a}} =\\
&\underset{?}{=} \sec^2a \cdot \frac{1}{3}\frac{1}{\sqrt[3]{\tan^2 a}}.
\end{align*}

(c)
\begin{align*}
\lim_{x\to a}\frac{\sqrt {\csc x^5} – \sqrt{\csc a^5}}{x-a} &
\underset{?}{=} \lim_{x\to a}
\frac{ \csc x^5 – \csc a^5}{x-a}\frac{1}{\sqrt {\csc x^5} + \sqrt{\csc a^5}}
=\\
&\underset{?}{=} -5a^4\cot a^5 \csc a^5 \cdot \frac{1}{2\sqrt{\csc a^5}},
\end{align*}
see Example 1.