Calculus

Differentiation of Functions

Derivatives of Trigonometric Functions

Page 1
Problems 1-3
Page 2
Problems 4-17

The basic trigonometric functions include the following \(6\) functions: sine \(\left(\sin x\right),\) cosine \(\left(\cos x\right),\) tangent \(\left(\tan x\right),\) cotangent \(\left(\cot x\right),\) secant \(\left(\sec x\right)\) and cosecant \(\left(\csc x\right).\)

For each of these functions, there is an inverse trigonometric function. They are called the inverse sine \(\left(\arcsin x\right),\) inverse cosine \(\left(\arccos x\right),\) inverse tangent \(\left(\arctan x\right),\) inverse cotangent \(\left({\text{arccot }x}\right),\) inverse secant \(\left({\text{arcsec }x}\right)\) and inverse cosecant \(\left({\text{arccsc }x}\right),\) respectively.

All these functions are continuous and differentiable in their domains. Below we make a list of derivatives for these \(12\) functions.

Derivatives of Basic Trigonometric Functions

We have already derived the derivatives of sine and cosine on the Definition of the Derivative page. They are as follows:
\[{{\left( {\sin x} \right)^\prime } = \cos x,\;\;}\kern-0.3pt{{\left( {\cos x} \right)^\prime } = – \sin x.}\] Using the quotient rule it is easy to obtain an expression for the derivative of tangent:
\[
{{\left( {\tan x} \right)^\prime } }={ {\left( {\frac{{\sin x}}{{\cos x}}} \right)^\prime } }
= {\frac{{{{\left( {\sin x} \right)}^\prime }\cos x – \sin x{{\left( {\cos x} \right)}^\prime }}}{{{{\cos }^2}x}} }
= {\frac{{\cos x \cdot \cos x – \sin x \cdot \left( { – \sin x} \right)}}{{{{\cos }^2}x}} }
= {\frac{{{{\cos }^2}x + {{\sin }^2}x}}{{{{\cos }^2}x}} }
= {\frac{1}{{{{\cos }^2}x}}.}
\] The derivative of cotangent can be found in the same way. However, this can be also done using the chain rule for differentiating a composite function:
\[\require{cancel}
{{\left( {\cot x} \right)^\prime } = {\left( {\frac{1}{{\tan x}}} \right)^\prime } }
= { – \frac{1}{{{{\tan }^2}x}} \cdot {\left( {\tan x} \right)^\prime } }
= { – \frac{1}{{\frac{{{{\sin }^2}x}}{{{{\cos }^2}x}}}} \cdot \frac{1}{{{{\cos }^2}x}} }
= { – \frac{\cancel{{{\cos }^2}x}}{{{{\sin }^2}x \cdot \cancel{{{\cos }^2}x}}} }
= { – \frac{1}{{{{\sin }^2}x}}.}
\] Similarly, we find the derivatives of secant and cosecant:
\[
{{\left( {\sec x} \right)^\prime } = {\left( {\frac{1}{{\cos x}}} \right)^\prime } }
= { – \frac{1}{{{{\cos }^2}x}} \cdot {\left( {\cos x} \right)^\prime } }
= {\frac{{\sin x}}{{{{\cos }^2}x}} }
= {\frac{{\sin x}}{{\cos x}} \cdot \frac{1}{{\cos x}} }
= {\tan x\sec x,}
\] \[
{{\left( {\csc x} \right)^\prime } = {\left( {\frac{1}{{\sin x}}} \right)^\prime } }
= { – \frac{1}{{{{\sin }^2}x}} \cdot {\left( {\sin x} \right)^\prime } }
= {-\frac{{\cos x}}{{{{\sin }^2}x}} }
= {-\frac{{\cos x}}{{\sin x}} \cdot \frac{1}{{\sin x}} }
= {-\cot x\csc x.}
\]

Derivatives of Inverse Trigonometric Functions

The derivatives of the inverse trigonometric functions can be derived using the inverse function theorem. For example, the sine function \(x = \varphi \left( y \right) \) \(= \sin y\) is the inverse function for \(y = f\left( x \right) \) \(= \arcsin x.\) Then the derivative of \(y = \arcsin x\) is given by
\[
{{\left( {\arcsin x} \right)^\prime } = f’\left( x \right) = \frac{1}{{\varphi’\left( y \right)}} }
= {\frac{1}{{{{\left( {\sin y} \right)}^\prime }}} }
= {\frac{1}{{\cos y}} }
= {\frac{1}{{\sqrt {1 – {\sin^2}y} }} }
= {\frac{1}{{\sqrt {1 – {\sin^2}\left( {\arcsin x} \right)} }} }
= {\frac{1}{{\sqrt {1 – {x^2}} }}\;\;}\kern-0.3pt{\left( { – 1 < x < 1} \right).}
\] Using this technique, we can find the derivatives of the other inverse trigonometric functions:
\[
{{\left( {\arccos x} \right)^\prime } }={ \frac{1}{{{{\left( {\cos y} \right)}^\prime }}} }
= {\frac{1}{{\left( { – \sin y} \right)}} }
= {- \frac{1}{{\sqrt {1 – {{\cos }^2}y} }} }
= {- \frac{1}{{\sqrt {1 – {{\cos }^2}\left( {\arccos x} \right)} }} }
= {- \frac{1}{{\sqrt {1 – {x^2}} }}\;\;}\kern-0.3pt{\left( { – 1 < x < 1} \right),}\qquad
\] \[
{{\left( {\arctan x} \right)^\prime } }={ \frac{1}{{{{\left( {\tan y} \right)}^\prime }}} }
= {\frac{1}{{\frac{1}{{{{\cos }^2}y}}}} }
= {\frac{1}{{1 + {{\tan }^2}y}} }
= {\frac{1}{{1 + {{\tan }^2}\left( {\arctan x} \right)}} }
= {\frac{1}{{1 + {x^2}}},}
\] \[
{\left( {\text{arccot }x} \right)^\prime } = {\frac{1}{{{{\left( {\cot y} \right)}^\prime }}}}
= \frac{1}{{\left( { – \frac{1}{{{\sin^2}y}}} \right)}}
= – \frac{1}{{1 + {{\cot }^2}y}}
= – \frac{1}{{1 + {{\cot }^2}\left( {\text{arccot }x} \right)}}
= – \frac{1}{{1 + {x^2}}},
\] \[
{{\left( {\text{arcsec }x} \right)^\prime } = {\frac{1}{{{{\left( {\sec y} \right)}^\prime }}} }}
= {\frac{1}{{\tan y\sec y}} }
= {\frac{1}{{\sec y\sqrt {{{\sec }^2}y – 1} }} }
= {\frac{1}{{\left| x \right|\sqrt {{x^2} – 1} }}.}
\] In the last formula, the absolute value \(\left| x \right|\) in the denominator appears due to the fact that the product \({\tan y\sec y}\) should always be positive in the range of admissible values of \(y\), where \(y \in \left( {0,{\large\frac{\pi }{2}\normalsize}} \right) \cup \left( {{\large\frac{\pi }{2}\normalsize},\pi } \right),\) i.e. the derivative of the inverse secant is always positive.

Similarly, we can obtain an expression for the derivative of the inverse cosecant function:
\[
{{\left( {\text{arccsc }x} \right)^\prime } = {\frac{1}{{{{\left( {\csc y} \right)}^\prime }}} }}
= {-\frac{1}{{\cot y\csc y}} }
= {-\frac{1}{{\csc y\sqrt {{{\csc }^2}y – 1} }} }
= {-\frac{1}{{\left| x \right|\sqrt {{x^2} – 1} }}.}
\]

Table of Derivatives of Trigonometric Functions

The derivatives of \(6\) basic trigonometric functions and \(6\) inverse trigonometric functions considered above are presented in the following table:

Derivatives of trigonometric and inverse trigonometric functions

In the examples below, find the derivative of the given trigonometric function.

Solved Problems

Click on problem description to see solution.

 Example 1

\[y = \cos 2x – 2\sin x\]

 Example 2

\[y = \tan x + \frac{1}{3}{\tan ^3}x\]

 Example 3

\[{y = \arctan \frac{{x + 1}}{{x – 1}}\;\;}\kern-0.3pt{\left( {x \ne 1} \right)}\]

 Example 4

\[y = \arctan \left( {x – \sqrt {1 + {x^2}} } \right)\]

 Example 5

\[y = \frac{1}{{{{\cos }^n}x}}\]

 Example 6

\[y = {\cos ^2}\sin x\]

 Example 7

\[y = {\sin ^2}\sqrt x \]

 Example 8

\[y = {\sin ^3}x + {\cos ^3}x\]

 Example 9

\[y = \tan \frac{x}{2} – \cot \frac{x}{2}\]

 Example 10

Using the chain rule, derive the formula for the derivative of the inverse sine function.

 Example 11

\[y = \arcsin \sqrt {1 – {x^2}} \]

 Example 12

\[y = \frac{2}{{\sqrt 3 }}\text{arccot}\frac{{2x + 1}}{{\sqrt 3 }}\]

 Example 13

\[y = {\sin ^n}x\cos nx\]

 Example 14

Show that \({\large\frac{d}{{dx}}\normalsize} \left( {x\arcsin x + \sqrt {1 – {x^2}} } \right)\) \(= \arcsin x.\)

 Example 15

\[
{y = {\left( {\tan x} \right)^{\cos x}},\;\;\;}\kern0pt
{\text{where}\;\;0 \lt x \lt \frac{\pi }{2}.}
\]

 Example 16

\[y = \arcsin \frac{{1 – {x^2}}}{{1 + {x^2}}}\]

 Example 17

\[y = \frac{1}{x}\text{arccsc}\frac{1}{x}\]

Example 1.

\[y = \cos 2x – 2\sin x\]

Solution.

Using the linear properties of the derivative, the chain rule and the double angle formula, we obtain:
\[
{y’\left( x \right) }={ {\left( {\cos 2x – 2\sin x} \right)^\prime } }
= {{\left( {\cos 2x} \right)^\prime } – {\left( {2\sin x} \right)^\prime } }
= {\left( { – \sin 2x} \right) \cdot {\left( {2x} \right)^\prime } – 2{\left( {\sin x} \right)^\prime } }
= { – 2\sin 2x – 2\cos x }
= { – 2\sin x\cos x – 2\cos x }
= { – 2\cos x\left( {\sin x + 1} \right).}
\]

Example 2.

\[y = \tan x + \frac{1}{3}{\tan ^3}x\]

Solution.

The derivative of this function is as follows:
\[
{y’\left( x \right) }={ {\left( {\tan x + \frac{1}{3}{{\tan }^3}x} \right)^\prime } }
= {{\left( {\tan x} \right)^\prime } + {\left( {\frac{1}{3}{{\tan }^3}x} \right)^\prime } }
= {\frac{1}{{{{\cos }^2}x}} + \frac{1}{3} \cdot 3{\tan ^2}x \cdot {\left( {\tan x} \right)^\prime } }
= {\frac{1}{{{{\cos }^2}x}} + {\tan ^2}x \cdot \frac{1}{{{{\cos }^2}x}} }
= {\frac{{1 + {{\tan }^2}x}}{{{{\cos }^2}x}}.}
\] The numerator can be simplified using the trigonometric identity
\[
{1 + {\tan^2}x = {\sec ^2}x }
= {\frac{1}{{{{\cos }^2}x}}.}
\] Therefore
\[
{y’\left( x \right) }={ \frac{{1 + {{\tan }^2}x}}{{{{\cos }^2}x}} }
= {\frac{{\frac{1}{{{{\cos }^2}x}}}}{{{{\cos }^2}x}} }
= {\frac{1}{{{{\cos }^4}x}} }
= {{\sec ^4}x.}
\]

Example 3.

\[{y = \arctan \frac{{x + 1}}{{x – 1}}\;\;}\kern-0.3pt{\left( {x \ne 1} \right)}\]

Solution.

By the chain and quotient rules, we have
\[
{y’\left( x \right) }={ {\left( {\arctan \frac{{x + 1}}{{x – 1}}} \right)^\prime } }
= {\frac{1}{{1 + {{\left( {\frac{{x + 1}}{{x – 1}}} \right)}^2}}} \cdot {\left( {\frac{{x + 1}}{{x – 1}}} \right)^\prime } }
= {\frac{{1 \cdot \left( {x – 1} \right) – \left( {x + 1} \right) \cdot 1}}{{{{\left( {x – 1} \right)}^2} + {{\left( {x + 1} \right)}^2}}} }
= {\frac{{\cancel{\color{blue}{x}} – \color{red}{1} – \cancel{\color{blue}{x}} – \color{red}{1}}}{{\color{maroon}{x^2} – \cancel{\color{green}{2x}} + \color{DarkViolet}{1} + \color{maroon}{x^2} + \cancel{\color{green}{2x}} + \color{DarkViolet}{1}}} }
= {\frac{{ – \color{red}{2}}}{{\color{maroon}{2{x^2}} + \color{DarkViolet}{2}}} }
= { – \frac{1}{{1 + {x^2}}}.}
\]

Page 1
Problems 1-3
Page 2
Problems 4-17