Derivatives of Inverse Trigonometric Functions

Like differentiating trigonometric functions, we can also find the derivative of their corresponding inverse functions. These are derived using implicit differentiation and trigonometric identities. Implicit differentiation refers to the rate of change of the inverse functions with respect to its argument.

Below are the derivatives of the six basic inverse trigonometric functions:

Now, we will derive each inverse function.

Inverse Sine Function (arcsin x or sin^-1 x)

The arcsine function, arcsin⁡ x, is the inverse of the sine function. 

Formula

${\dfrac{d}{dx}\left( \sin ^{-1}x\right) =\dfrac{1}{\sqrt{1-x^{2}}}}$, which has a domain of -1 < x < 1

Proof

Let us consider y = arcsin x

By the definition of inverse sine, sin y = x

Differentiating both sides with respect to x, we get

${\dfrac{d}{dx}\left( \sin y\right)}$ = 1

⇒ ${\cos y\dfrac{dy}{dx}}$ = 1

⇒ ${\dfrac{dy}{dx}=\dfrac{1}{\cos y}}$

⇒ ${\dfrac{dy}{dx}=\sec y}$ …..(i)

As we know, sin² y + cos² y = 1

⇒ cos y = ${\sqrt{1-\sin ^{2}y}}$

⇒ cos y = ${\sqrt{1-x^{2}}}$

From (i), we get

⇒ ${\dfrac{dy}{dx}=\dfrac{1}{\sqrt{1-x^{2}}}}$

⇒ ${\dfrac{d}{dx}\left( \arcsin x\right) =\dfrac{1}{\sqrt{1-x^{2}}}}$

Thus, ${\dfrac{d}{dx}\left( \sin ^{-1}x\right) =\dfrac{1}{\sqrt{1-x^{2}}}}$

Inverse Cosine Function (arccos x or cos^-1 x)

The arccosine function, arccos⁡ x, is the inverse of the cosine function. 

Formula

${\dfrac{d}{dx}\left( \cos ^{-1}x\right) =-\dfrac{1}{\sqrt{1-x^{2}}}}$, where -1 < x < 1

Proof

Let us assume y = arccos x

By the definition of inverse cosine, cos y = x

Differentiating both sides with respect to x, we get

${\dfrac{d}{dx}\left( \cos y\right)}$ = 1

⇒ ${-\sin y\dfrac{dy}{dx}}$ = 1

⇒ ${\dfrac{dy}{dx}=-\dfrac{1}{\sin y}}$

⇒ ${\dfrac{dy}{dx}=-\text{cosec}\, y}$…..(i)

As we know, sin² y + cos² y = 1

⇒ sin y = ${\sqrt{1-\cos ^{2}y}}$

⇒ sin y = ${\sqrt{1-x^{2}}}$

From (i), we get

⇒ ${\dfrac{dy}{dx}=-\dfrac{1}{\sqrt{1-x^{2}}}}$

⇒ ${\dfrac{d}{dx}\left( \arccos x\right) =-\dfrac{1}{\sqrt{1-x^{2}}}}$

Thus, ${\dfrac{d}{dx}\left( \cos ^{-1}x\right) = -\dfrac{1}{\sqrt{1-x^{2}}}}$

Inverse Tangent Function (arctan x or tan^-1 x)

The arctangent function, arctan⁡ x, is the inverse of the tangent function. 

Formula

${\dfrac{d}{dx}\left( \tan ^{-1}x\right) =\dfrac{1}{1+x^{2}}}$, where x ∈ ℝ

Proof

Let us assume y = arctan x

By the definition of inverse tangent, tan y = x

Differentiating both sides with respect to x, we get

${\dfrac{d}{dx}\left( \tan y\right)}$ = 1

⇒ ${\sec ^{2}y\dfrac{dy}{dx}=1}$

⇒ ${\dfrac{dy}{dx}=\dfrac{1}{\sec ^{2}y}}$…..(i)

As we know, sec² y – tan² y = 1

⇒ sec² y = 1 + tan² y 

⇒ sec² y = 1 + x² 

From (i), we get

⇒ ${\dfrac{dy}{dx}=\dfrac{1}{1+x^{2}}}$

⇒ ${\dfrac{d}{dx}\left( \arctan x\right) =\dfrac{1}{1+x^{2}}}$

Thus, ${\dfrac{d}{dx}\left( \tan ^{-1}x\right) =\dfrac{1}{1+x^{2}}}$

Inverse Secant Function (arcsec x or sec^-1 x)

The arcsecant function, arcsec⁡ x, is the inverse of the secant function. 

Formula

${\dfrac{d}{dx}\left( \sec ^{-1}x\right) =\dfrac{1}{\left| x\right| \sqrt{x^{2}-1}}}$, where |x| > 1

Proof

Let us assume y = arcsec x

By the definition of inverse secant, sec y = x

Differentiating both sides with respect to x, we get

${\dfrac{d}{dx}\left( \sec y\right)}$ = 1

⇒ ${\sec y\tan y\dfrac{dy}{dx}=1}$

⇒ ${\dfrac{dy}{dx}=\dfrac{1}{\sec y\tan y}}$ …..(i)

As we know, sec² y – tan² y = 1

⇒ tan² y = sec² y – 1

⇒ tan² y = x² – 1

⇒ tan y = ${\sqrt{x^{2}-1}}$

Also, sec y = x

From (i), we get

⇒ ${\dfrac{dy}{dx}=\dfrac{1}{\left| x\right| \sqrt{x^{2}-1}}}$

⇒ ${\dfrac{d}{dx}\left( \text {arcsec}\, x\right) =\dfrac{1}{\left| x\right| \sqrt{x^{2}-1}}}$

Thus, ${\dfrac{d}{dx}\left( \sec ^{-1}x\right) =\dfrac{1}{\left| x\right| \sqrt{x^{2}-1}}}$

Note: Here, the absolute value sign around x is used instead of just x because, for the graph of cosec^-1 x, the slope of the tangent is always negative. This means the derivative of cosec^-1 x remains negative regardless of the sign of x. Thus, the absolute value is necessary to ensure this consistency.

Inverse Cosecant Function (arccosec x or cosec^-1 x)

The arccosecant function, arccosec⁡ x, is the inverse of the cosecant function. 

Formula

${\dfrac{d}{dx}\left( \text{cosec} ^{-1}\, x\right) =-\dfrac{1}{\left| x\right| \sqrt{x^{2}-1}}}$, where |x| > 1

Proof

Let us consider y = arccosec x

By the definition of inverse cosecant, cosec y = x

Differentiating both sides with respect to x, we get

${\dfrac{d}{dx}\left( \text{cosec}\, y\right)}$ = 1

⇒ ${-\text{cosec}\, y\cot y\dfrac{dy}{dx}=1}$⇒ ${\dfrac{dy}{dx}=-\dfrac{1}{\text{cosec}\, y\cot y}}$ …..(i)

As we know, cosec² y – cot² y = 1

⇒ cot² y = cosec² y – 1

⇒ cot² y = x² – 1

⇒ cot y = ${\sqrt{x^{2}-1}}$

Also, cosec y = x

From (i), we get

⇒ ${\dfrac{dy}{dx}=-\dfrac{1}{\left| x\right| \sqrt{x^{2}-1}}}$

${\dfrac{d}{dx}\left( \text{arccosec}\, x\right) =-\dfrac{1}{\left| x\right| \sqrt{x^{2}-1}}}$

Thus, ${\dfrac{d}{dx}\left( \text{cosec} ^{-1}\, x\right) =-\dfrac{1}{\left| x\right| \sqrt{x^{2}-1}}}$

Note: Similar to the derivative of sec^-1 x, the absolute value is used because the graph of sec^-1 x always has tangents with positive slopes. This ensures that the derivative remains consistent regardless of the sign of x.

Inverse Cotangent Function (arccot x or cot^-1 x)

The arccotangent function, arccot⁡ x, is the inverse of the cotangent function. 

Formula

${\dfrac{d}{dx}\left( \cot ^{-1}x\right) =-\dfrac{1}{1+x^{2}}}$, where x ∈ ℝ

Proof

Let us consider y = arccot x

By the definition of inverse cotangent, cot y = x

Differentiating both sides with respect to x, we get

${\dfrac{d}{dx}\left( \cot y\right)}$ = 1

⇒ ${-\text{cosec} ^{2}\, y\dfrac{dy}{dx}=1}$

⇒ ${\dfrac{dy}{dx}=-\dfrac{1}{\text{cosec} ^{2}\, y}}$ …..(i)

As we know, cosec² y – cot² y = 1

⇒ cosec² y = 1 + cot² y 

⇒ cosec² y = 1 + x² 

From (i), we get

⇒ ${\dfrac{dy}{dx}=-\dfrac{1}{1+x^{2}}}$

⇒ ${\dfrac{d}{dx}\left( \text{arccot}\, x\right) =-\dfrac{1}{1+x^{2}}}$

Thus, ${\dfrac{d}{dx}\left( \cot ^{-1}x\right) =-\dfrac{1}{1+x^{2}}}$

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