Derivatives of Trigonometric Functions

In our previous lesson, we have reviewed some of the fundamental concepts in trigonometric functions. Since before we can differentiate such functions, we must first be proficient in transforming functions from one form to another. Below are some of the different formulas for differentiating trigonometric functions. For the following formulas, let $u$ be a function of $x$ .

$\begin{align} \ddu\br{\sin u} &= \cos u \label{eq:derivative of sin}\\ \ddu\br{\cos u} &= -\sin u \label{eq:derivative of cos}\\ \ddu\br{\tan u} &= \sec^2 u \label{eq:derivative of tan}\\ \ddu\br{\cot u} &= -\csc^2 u \label{eq:derivative of cot}\\ \ddu\br{\sec u} &= \tan u \sec u \label{eq:derivative of sec}\\ \ddu\br{\csc u} &= -\cot u \csc u \label{eq:derivative of csc} \end{align}$

Differential Formulas for Trigonometric Functions

As you may have realized, there is only a small difference between the formulas for differentiation and its differentials.

$\begin{align} d\br{\sin u} &= \cos u\,du \label{eq:differential of sin}\\ d\br{\cos u} &= -\sin u\,du \label{eq:differential of cos}\\ d\br{\tan u} &= \sec^2 u\,du \label{eq:differential of tan}\\ d\br{\cot u} &= -\csc^2 u\,du \label{eq:differential of cot}\\ d\br{\sec u} &= \tan u \sec u\,du \label{eq:differential of sec}\\ d\br{\csc u} &= -\cot u \csc u\,du \label{eq:differential of csc} \end{align}$

$\example{1}$ Find the differential of the function $y = \sin 5x$

$\solution$

To solve for the derivative of the function, use $\eref{eq:derivative of sin}$ ,

$\begin{align*} \ddx\br{y} &= \ddx\br{\sin 5x} \\ \dydx &= \cos{5x}\cdot \ddx\br{5x} \\ &= \cos{5x}\cdot 5 \\ &= 5\cos{5x} \tagans \end{align*}$

$\example{2}$ Determine the second derivative of the function $y = \cos(3-2x)$

$\solution$

In this example, use $\eref{eq:derivative of cos}$ ,

$\begin{align*} \ddx\br{y} &= \ddx\br{\cos(3-2x)} \\ \dydx &= -\sin(3-2x)\cdot \ddx\br{3-2x} \\ &= -\sin(3-2x) (-2) \\ &= 2\sin(3-2x) \\ &= 2\sin(3-2x) \end{align*}$

Then differentiate again,

$\begin{align*} \ddx\br{\dydx} &= d\brk{2\sin(3-2x)} \\ \frac{d^2y}{dx^2} &= 2\cos(3-2x) \cdot \ddx(3-2x) \\ &= 2\cos(3-2x) \cdot (-2) \\ &= -4\cos(3-2x) \\ \frac{d^2y}{dx^2} &= -4\cos(3-2x) \tagans \end{align*}$

$\example{3}$ Differentiate the function with respect to $\theta$ : $x = \sec{2\theta}+\tan{3\theta}$

$\solution$

To solve for the derivative of the function, use $\tcAal{Equations \ref{eq:differential of sec} and \ref{eq:differential of tan}}$ ,

$\begin{align*} d\br{x} &= d\br{\sec{2\theta}+\tan{3\theta}} \\ dx &= \tan{2\theta}\sec{2\theta}\cdot d(2\theta) + \sec^2{3\theta}\cdot d(3\theta) \\ &= \tan{2\theta}\sec{2\theta}\cdot 2\,d\theta + \sec^2{3\theta}\cdot 3\,d\theta \\ &= \br{2\tan{2\theta}\sec{2\theta} + 3\sec^2{3\theta}}\,d\theta \\ \frac{dx}{d\theta}&= 2\tan{2\theta}\sec{2\theta} + 3\sec^2{3\theta} \tagans \end{align*}$

$\example{4}$ Find the derivative of $x$ with respect to $y$ of the function $y = \tan x-x$

$\solution$

To solve for the derivative of the function, use $\eref{eq:differential of tan}$ ,

$\begin{align*} d(y) &= d(\tan x-x) \\ dy &= \sec^2x\,dx-dx \\ &= (\sec^2x-1)\,dx \\ \end{align*}$

From the Pythagorean Identities,

$\begin{align*} \tan^2{x}+1 &= \sec^2{x} \\ \sec^2{x}-1 &= \tan^2{x} \end{align*}$

Substituting this to the soultion,

$\begin{align*} dy &= \tan^2{x}\,dx \\ \end{align*}$

Hence,

$\begin{align*} \frac{dx}{dy} &= \frac{1}{\tan^2x} \\ &= \cot^2x \tagans \end{align*}$

$\example{5}$ Differentiate the function with respect to $t$ . $w = t^5\csc 4t$

$\solution$

In this example, before we can use a differential formula for trigonometric function, we first need to apply the Product Rule.

$\begin{align*} d(w) &= d(t^5 \cdot \csc{4t}) \\ dw &= t^5 \cdot d(\csc{4t}) + \csc{4t} \cdot d(t^5) \\ &= t^5\brk{-\cot{4t}\csc{4t}\,(4\,dt)}+\csc{4t}\,(5t^4\,dt) \\ &= \brk{-4t^5\cot{4t}\csc{4t}+5t^4\csc{4t}}\,dt \\ \frac{dw}{dt} &= 5t^4\csc{4t}-4t^5\cot{4t}\csc{4t} \tagans \end{align*}$

$\example{6}$ Find the differential of the function $v = \sec^2\br{\frac{x}{6}}$

$\solution$

In this example, we can recall that $\sec^2\br{\frac{x}{6}}$ can also be expressed as $\brk{\sec\br{\frac{x}{6}}}^2$ . With this in mind, we can use the power rule to differentiate the function.

$\begin{align*} d(v) &= d\brk{\brk{\sec\br{\frac{x}{6}}}^2} \\ &= 2\sec\br{\frac{x}{6}} \cdot d\brk{\sec\br{\frac{x}{6}}} \\ &= 2\sec\br{\frac{x}{6}} \cdot \brk{\tan\br{\frac{x}{6}}\sec\br{\frac{x}{6}} \cdot d\br{\frac{x}{6}}} \\ &= 2\sec\br{\frac{x}{6}} \cdot \brk{\tan\br{\frac{x}{6}}\sec\br{\frac{x}{6}} \cdot \frac{1}{6}\, dx} \\ &= \frac{1}{3} \tan\br{\frac{x}{6}}\sec^2\br{\frac{x}{6}} \,dx \end{align*}$

$\example{7}$ Let $a$ be hold constant. Determine the third derivative of the function $y = \cos ax$

$\solution$

In this example, $a$ is treated as a constant. Thus,

$\begin{align*} d(y) &= d(\cos ax) \\ dy &= -\sin ax \cdot d(ax) \\ &= -\sin ax \cdot a\,dx \\ &= -a\sin ax\,dx \\ \end{align*}$

Then differentiate again,

$\begin{align*} d(dy) &= d(-a\sin ax\,dx) \\ d^2y &= -a\cos ax\,dx \cdot(a\,dx) \\ &= -a^2\cos ax\,dx^2 \end{align*}$

And differentiating for the last time,

$\begin{align*} d(d^2y) &= d(-a^2\cos ax\,dx^2) \\ d^3y &= -(-a^2\sin ax\,dx^2) \cdot(a\,dx) \\ &= a^3\sin ax\,dx^3 \\ \frac{d^3y}{dx^3} &= a^3\sin ax \tagans \end{align*}$

$\example{8}$ Determine the $y’$ of the function $s = \sec(\sin t)$

$\solution$

To evaluate this example, use the chain rule.

$\begin{align*} d(s) &= d(\sec(\sin t)) \\ ds &= \tan(\sin t)\sec(\sin t) \cdot d(\sin t) \\ &= \tan(\sin t)\sec(\sin t)\cos t \,dt \\ \frac{ds}{dt} &= \tan(\sin t)\sec(\sin t)\cos t \tagans \end{align*}$

$\example{9}$ Let $A$ , $B$ , and $k$ be constants. From the function $y = A\cos kx+B\sin kx$ , show that $\frac{d^2y}{dx^2} = -k^2x$ .

$\solution$

Differentiate the function twice.

$\begin{align*} d(y) &= d(A\cos kx+B\sin kx) \\ dy &= A(-\sin kx) (k\,dx) + B\cos kx (k\,dx) \\ &= \br{-Ak\sin kx + Bk\cos kx}\,dx \\ d(dy) &= d\brk{\br{-Ak\sin kx + Bk\cos kx}\,dx} \\ d^2y &= \brk{-Ak\cos kx (k\,dx) + Bk(-\cos kx)(k\,dx) } \,dx \\ &= \br{-Ak^2\cos kx - Bk^2\cos kx } \,dx^2 \\ \frac{d^2y}{dx^2}&= -Ak^2\cos kx - Bk^2\cos kx \\ &= -k^2\br{Ak^2\cos kx + B\cos kx} \\ \end{align*}$

Since $y = A\cos kx+B\sin kx$ ,

$\begin{align*} \frac{d^2y}{dx^2}&= -k^2y \tagans \end{align*}$

Derivatives of Trigonometric Functions

Differential Formulas for Trigonometric Functions

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