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3. Advanced differentiation

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Inverse trig derivatives

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What you’ll learn

How to differentiate inverse trig functions using the chain rule or by simplifying

Inverse trigonometric functions

Regular trigonometric functions take an angle $θ$ as the input and output a trigonometric ratio. In a right-triangle setting, that ratio is “opposite over hypotenuse” for sine, “adjacent over hypotenuse” for cosine, and so on (SOHCAHTOA).

Inverse trigonometric functions reverse that relationship: inputting a trigonometric ratio gives the angle that produces it (within the function’s restricted range). For example,

$sin (\frac{π}{4}) = \frac{2}{2} ⟺ arcsin (\frac{2}{2}) = \frac{π}{4}$

Sidenote

Inverse notation

$arcsin (x)$ , the inverse sine function, is sometimes written as $sin^{- 1} (x)$ .

Make sure not to confuse this with the reciprocal of $sin (x)$ , or $\frac{1}{s i n ( x )} = csc (x)$ . The reciprocal is denoted $(sin (x))^{- 1}$ .

To find the derivative of $y = arcsin (x)$ , start with the relationship

$y = arcsin (x) ⟺ x = sin (y)$

Differentiate implicitly with respect to $x$ :

$\frac{d}{d x} (sin y) cos (y) \cdot y^{'} y^{'} = \frac{d}{d x} (x) = 1 = \frac{1}{cos ( y )} = \frac{1}{cos ( arcsin ( x ))}$

Now simplify the composite trig expression. The output of $sin^{- 1} (x)$ is an angle, so let

$arcsin (x) = θ ⟺ x = sin (θ)$

Think of $x$ as $\frac{x}{1}$ . In a right triangle with angle $θ$ :

opposite side $= x$
hypotenuse $= 1$

By the Pythagorean theorem, the side adjacent to $θ$ is $1 - x^{2}$ .

Using this triangle,

$y^{'} = \frac{1}{cos ( arcsin ( x ))} = \frac{1}{cos ( θ )} = \frac{1}{1 - x ^{2}}$

So, for $y = arcsin (x)$ ,

$y^{'} = \frac{1}{1 - x ^{2}}$

Repeating this same idea (implicit differentiation, then simplifying with a triangle when needed) gives the derivatives of the six inverse trig functions:

$f (x)$	$f^{'} (x)$
$sin^{- 1} (x)$	$\frac{1}{1 - x ^{2}}$
$cos^{- 1} (x)$	$\frac{- 1}{1 - x ^{2}}$
$tan^{- 1} (x)$	$\frac{1}{1 + x ^{2}}$
$cot^{- 1} (x)$	$\frac{- 1}{1 + x ^{2}}$
$sec^{- 1} (x)$	$\frac{1}{∣ x ∣ x ^{2} - 1}$
$csc^{- 1} (x)$	$\frac{- 1}{∣ x ∣ x ^{2} - 1}$

Notice that every other derivative is just the negative of the one above it.

For $sec^{- 1} (x)$ , the absolute value ensures the derivative stays positive for all $x$ in the domain (consistent with the graph’s tangent-line behavior). For $csc^{- 1} (x)$ , the absolute value similarly supports the fact that the derivative stays negative on its domain.

Examples with chain rule

If an inverse trig function has an argument that isn’t just $x$ , do two things:

Replace every $x$ in the standard derivative formula with the new argument.
Multiply by the derivative of that argument (this is the chain rule).

Find the derivative of

$y = sec^{- 1} (x^{2} + 1)$

(spoiler)

The derivative of the basic function $y = sec^{- 1} (x)$ is

$y^{'} = \frac{1}{∣ x ∣ x ^{2} - 1}$

To differentiate $y = sec^{- 1} (x^{2} + 1)$ , replace all instances of $x$ with the argument $x^{2} + 1$ and then apply the chain rule:

$y^{'} = \frac{1}{∣ x ^{2} + 1∣ ( x ^{2} + 1 ) ^{2} - 1} \cdot 2 x$

Since $x^{2} + 1 > 0$ for all real $x$ , the absolute value can be dropped:

$y^{'} = \frac{2 x}{( x ^{2} + 1 ) ( x ^{2} + 1 ) ^{2} - 1}$

Differentiate

$y = cot^{- 1} (e^{x})$

(spoiler)

The derivative of $y = cot^{- 1} (x)$ is $\frac{- 1}{1 + x ^{2}}$ .

To differentiate $y = cot^{- 1} (e^{x})$ , replace $x$ with $e^{x}$ and apply the chain rule:

$y^{'} = \frac{- 1}{1 + ( e ^{x} ) ^{2}} \cdot e^{x} = \frac{- e ^{x}}{1 + e ^{2 x}}$

Find the derivative of

$f (x) = tan (cos^{- 1} (2 x))$

in two ways:

a) Use a triangle to simplify into an algebraic function and then differentiate.

b) Differentiate and then simplify using a triangle.

Solutions

a) Simplify $\to$ differentiate

(spoiler)

Let $cos^{- 1} (2 x) = θ$ .

Then

$cos (θ) = \frac{2 x}{1}$

Using the Pythagorean theorem, the remaining side of the triangle is $1 - 4 x^{2}$ .

Then as an algebraic function,

$f (x) = tan (θ) = \frac{1 - 4 x ^{2}}{2 x}$

Differentiate with the quotient rule:

$f^{'} (x) = \frac{2 x ( \frac{1}{2 1 - 4 x ^{2}} ) ( - 8 x ) - ( 1 - 4 x ^{2} ) ( 2 )}{( 2 x ) ^{2}} = \frac{\frac{- 16 x ^{2}}{2 1 - 4 x ^{2}} - 2 1 - 4 x ^{2}}{4 x ^{2}} = \frac{\frac{- 8 x ^{2}}{1 - 4 x ^{2}} - \frac{2 ( 1 - 4 x ^{2} )}{1 - 4 x ^{2}}}{4 x ^{2}} = \frac{\frac{- 8 x ^{2} - 2 + 8 x ^{2}}{1 - 4 x ^{2}}}{4 x ^{2}} = - \frac{2}{4 x ^{2} 1 - 4 x ^{2}} = - \frac{1}{2 x ^{2} 1 - 4 x ^{2}}$

b) Differentiate $\to$ simplify

(spoiler)

Differentiate using the chain rule:

$f (x) f^{'} (x) = tan (cos^{- 1} (2 x)) = sec^{2} (cos^{- 1} (2 x)) \cdot \frac{- 2}{1 - ( 2 x ) ^{2}}$

To simplify $sec^{2} (cos^{- 1} (2 x))$ , use the same triangle as above. Since $cos (θ) = 2 x$ , we have $sec (θ) = \frac{1}{2 x}$ , so

$sec^{2} (θ) = \frac{1}{4 x ^{2}}$

Then

$f^{'} (x) = \frac{1}{4 x ^{2}} \cdot \frac{- 2}{1 - 4 x ^{2}} = - \frac{1}{2 x ^{2} 1 - 4 x ^{2}}$

Find the equation of the tangent line to $y = arccos (x^{2})$ at $x = \frac{2}{2}$ .

(spoiler)

First evaluate $y$ at $x = \frac{2}{2}$ :

$y = arccos ((\frac{2}{2})^{2}) = arccos (\frac{1}{2}) = \frac{π}{3}$

So the point of tangency is $(\frac{2}{2}, \frac{π}{3})$ .

Next find the slope by differentiating $y = arccos (x^{2})$ :

$\frac{d}{d x} arccos (x^{2})$

$= - \frac{1}{1 - ( x ^{2} ) ^{2}} \cdot 2 x$

$= - \frac{2 x}{1 - x ^{4}}$

Evaluate at $x = \frac{2}{2}$ :

$- \frac{2 \cdot \frac{2}{2}}{1 - ( \frac{2}{2} ) ^{4}} = - \frac{2}{1 - \frac{4}{16}} = - \frac{2}{\frac{3}{4}} = - \frac{2 2}{3} = - \frac{2 6}{3}$

Then the equation of tangent line at $(\frac{2}{2}, \frac{π}{3})$ is

$y - \frac{π}{3} = - \frac{2 6}{3} (x - \frac{2}{2})$

Inverse trig functions

Reverse trig functions: input a ratio, output an angle
Notation: $arcsin (x) = sin^{- 1} (x)$ ≠ $csc (x)$ (reciprocal)
Derivatives found via implicit differentiation + right-triangle simplification

Standard inverse trig derivatives

$(sin^{- 1} x)^{'} = \frac{1}{1 - x ^{2}}$ , $(cos^{- 1} x)^{'} = \frac{- 1}{1 - x ^{2}}$
$(tan^{- 1} x)^{'} = \frac{1}{1 + x ^{2}}$ , $(cot^{- 1} x)^{'} = \frac{- 1}{1 + x ^{2}}$
$(sec^{- 1} x)^{'} = \frac{1}{∣ x ∣ x ^{2} - 1}$ , $(csc^{- 1} x)^{'} = \frac{- 1}{∣ x ∣ x ^{2} - 1}$
- Co-functions (cos, cot, csc) are always the negative of their counterparts

Chain rule with inverse trig

Replace every $x$ in the formula with the inner argument
Multiply by the derivative of the inner argument
Example: $(sec^{- 1} (x^{2} + 1))^{'} = \frac{2 x}{( x ^{2} + 1 ) ( x ^{2} + 1 ) ^{2} - 1}$

Simplifying nested trig/inverse trig expressions

Use a right triangle: set inner inverse trig expression equal to $θ$ , label sides, apply Pythagorean theorem
Two valid approaches: simplify first then differentiate, or differentiate first then simplify with triangle
Both approaches yield the same result

Inverse trig derivatives

What you’ll learn

How to differentiate inverse trig functions using the chain rule or by simplifying

Inverse trigonometric functions

Inverse trigonometric functions reverse that relationship: inputting a trigonometric ratio gives the angle that produces it (within the function’s restricted range). For example,

$sin (\frac{π}{4}) = \frac{2}{2} ⟺ arcsin (\frac{2}{2}) = \frac{π}{4}$

Sidenote

Inverse notation

$arcsin (x)$ , the inverse sine function, is sometimes written as $sin^{- 1} (x)$ .

Make sure not to confuse this with the reciprocal of $sin (x)$ , or $\frac{1}{s i n ( x )} = csc (x)$ . The reciprocal is denoted $(sin (x))^{- 1}$ .

To find the derivative of $y = arcsin (x)$ , start with the relationship

$y = arcsin (x) ⟺ x = sin (y)$

Differentiate implicitly with respect to $x$ :

$\frac{d}{d x} (sin y) cos (y) \cdot y^{'} y^{'} = \frac{d}{d x} (x) = 1 = \frac{1}{cos ( y )} = \frac{1}{cos ( arcsin ( x ))}$

Now simplify the composite trig expression. The output of $sin^{- 1} (x)$ is an angle, so let

$arcsin (x) = θ ⟺ x = sin (θ)$

Think of $x$ as $\frac{x}{1}$ . In a right triangle with angle $θ$ :

opposite side $= x$
hypotenuse $= 1$

By the Pythagorean theorem, the side adjacent to $θ$ is $1 - x^{2}$ .

Using this triangle,

$y^{'} = \frac{1}{cos ( arcsin ( x ))} = \frac{1}{cos ( θ )} = \frac{1}{1 - x ^{2}}$

So, for $y = arcsin (x)$ ,

$y^{'} = \frac{1}{1 - x ^{2}}$

Repeating this same idea (implicit differentiation, then simplifying with a triangle when needed) gives the derivatives of the six inverse trig functions:

$f (x)$	$f^{'} (x)$
$sin^{- 1} (x)$	$\frac{1}{1 - x ^{2}}$
$cos^{- 1} (x)$	$\frac{- 1}{1 - x ^{2}}$
$tan^{- 1} (x)$	$\frac{1}{1 + x ^{2}}$
$cot^{- 1} (x)$	$\frac{- 1}{1 + x ^{2}}$
$sec^{- 1} (x)$	$\frac{1}{∣ x ∣ x ^{2} - 1}$
$csc^{- 1} (x)$	$\frac{- 1}{∣ x ∣ x ^{2} - 1}$

Notice that every other derivative is just the negative of the one above it.

Examples with chain rule

If an inverse trig function has an argument that isn’t just $x$ , do two things:

Replace every $x$ in the standard derivative formula with the new argument.
Multiply by the derivative of that argument (this is the chain rule).

Find the derivative of

$y = sec^{- 1} (x^{2} + 1)$

(spoiler)

The derivative of the basic function $y = sec^{- 1} (x)$ is

$y^{'} = \frac{1}{∣ x ∣ x ^{2} - 1}$

To differentiate $y = sec^{- 1} (x^{2} + 1)$ , replace all instances of $x$ with the argument $x^{2} + 1$ and then apply the chain rule:

$y^{'} = \frac{1}{∣ x ^{2} + 1∣ ( x ^{2} + 1 ) ^{2} - 1} \cdot 2 x$

Since $x^{2} + 1 > 0$ for all real $x$ , the absolute value can be dropped:

$y^{'} = \frac{2 x}{( x ^{2} + 1 ) ( x ^{2} + 1 ) ^{2} - 1}$

Differentiate

$y = cot^{- 1} (e^{x})$

(spoiler)

The derivative of $y = cot^{- 1} (x)$ is $\frac{- 1}{1 + x ^{2}}$ .

To differentiate $y = cot^{- 1} (e^{x})$ , replace $x$ with $e^{x}$ and apply the chain rule:

$y^{'} = \frac{- 1}{1 + ( e ^{x} ) ^{2}} \cdot e^{x} = \frac{- e ^{x}}{1 + e ^{2 x}}$

Find the derivative of

$f (x) = tan (cos^{- 1} (2 x))$

in two ways:

a) Use a triangle to simplify into an algebraic function and then differentiate.

b) Differentiate and then simplify using a triangle.

Solutions

a) Simplify $\to$ differentiate

(spoiler)

Let $cos^{- 1} (2 x) = θ$ .

Then

$cos (θ) = \frac{2 x}{1}$

Using the Pythagorean theorem, the remaining side of the triangle is $1 - 4 x^{2}$ .

Then as an algebraic function,

$f (x) = tan (θ) = \frac{1 - 4 x ^{2}}{2 x}$

Differentiate with the quotient rule:

b) Differentiate $\to$ simplify

(spoiler)

Differentiate using the chain rule:

$f (x) f^{'} (x) = tan (cos^{- 1} (2 x)) = sec^{2} (cos^{- 1} (2 x)) \cdot \frac{- 2}{1 - ( 2 x ) ^{2}}$

To simplify $sec^{2} (cos^{- 1} (2 x))$ , use the same triangle as above. Since $cos (θ) = 2 x$ , we have $sec (θ) = \frac{1}{2 x}$ , so

$sec^{2} (θ) = \frac{1}{4 x ^{2}}$

Then

$f^{'} (x) = \frac{1}{4 x ^{2}} \cdot \frac{- 2}{1 - 4 x ^{2}} = - \frac{1}{2 x ^{2} 1 - 4 x ^{2}}$

Find the equation of the tangent line to $y = arccos (x^{2})$ at $x = \frac{2}{2}$ .

(spoiler)

First evaluate $y$ at $x = \frac{2}{2}$ :

$y = arccos ((\frac{2}{2})^{2}) = arccos (\frac{1}{2}) = \frac{π}{3}$

So the point of tangency is $(\frac{2}{2}, \frac{π}{3})$ .

Next find the slope by differentiating $y = arccos (x^{2})$ :

$\frac{d}{d x} arccos (x^{2})$

$= - \frac{1}{1 - ( x ^{2} ) ^{2}} \cdot 2 x$

$= - \frac{2 x}{1 - x ^{4}}$

Evaluate at $x = \frac{2}{2}$ :

$- \frac{2 \cdot \frac{2}{2}}{1 - ( \frac{2}{2} ) ^{4}} = - \frac{2}{1 - \frac{4}{16}} = - \frac{2}{\frac{3}{4}} = - \frac{2 2}{3} = - \frac{2 6}{3}$

Then the equation of tangent line at $(\frac{2}{2}, \frac{π}{3})$ is

$y - \frac{π}{3} = - \frac{2 6}{3} (x - \frac{2}{2})$

Key points

Inverse trig functions

Reverse trig functions: input a ratio, output an angle
Notation: $arcsin (x) = sin^{- 1} (x)$ ≠ $csc (x)$ (reciprocal)
Derivatives found via implicit differentiation + right-triangle simplification

Standard inverse trig derivatives

$(sin^{- 1} x)^{'} = \frac{1}{1 - x ^{2}}$ , $(cos^{- 1} x)^{'} = \frac{- 1}{1 - x ^{2}}$
$(tan^{- 1} x)^{'} = \frac{1}{1 + x ^{2}}$ , $(cot^{- 1} x)^{'} = \frac{- 1}{1 + x ^{2}}$
$(sec^{- 1} x)^{'} = \frac{1}{∣ x ∣ x ^{2} - 1}$ , $(csc^{- 1} x)^{'} = \frac{- 1}{∣ x ∣ x ^{2} - 1}$
- Co-functions (cos, cot, csc) are always the negative of their counterparts

Chain rule with inverse trig

Replace every $x$ in the formula with the inner argument
Multiply by the derivative of the inner argument
Example: $(sec^{- 1} (x^{2} + 1))^{'} = \frac{2 x}{( x ^{2} + 1 ) ( x ^{2} + 1 ) ^{2} - 1}$

Simplifying nested trig/inverse trig expressions

Use a right triangle: set inner inverse trig expression equal to $θ$ , label sides, apply Pythagorean theorem
Two valid approaches: simplify first then differentiate, or differentiate first then simplify with triangle
Both approaches yield the same result