Core Definition
The derivatives of the six inverse trigonometric functions are algebraic expressions containing no trigonometric terms:
> d d x [ sin − 1 x ] = 1 1 − x 2 d d x [ cos − 1 x ] = − 1 1 − x 2 > d d x [ tan − 1 x ] = 1 1 + x 2 d d x [ cot − 1 x ] = − 1 1 + x 2 > d d x [ sec − 1 x ] = 1 ∣ x ∣ x 2 − 1 d d x [ csc − 1 x ] = − 1 ∣ x ∣ x 2 − 1 > \begin{aligned}
> \frac{d}{dx}[\sin^{-1}x] &= \frac{1}{\sqrt{1 - x^2}} &\quad \frac{d}{dx}[\cos^{-1}x] &= -\frac{1}{\sqrt{1 - x^2}} \\
> \frac{d}{dx}[\tan^{-1}x] &= \frac{1}{1 + x^2} &\quad \frac{d}{dx}[\cot^{-1}x] &= -\frac{1}{1 + x^2} \\
> \frac{d}{dx}[\sec^{-1}x] &= \frac{1}{|x|\sqrt{x^2 - 1}} &\quad \frac{d}{dx}[\csc^{-1}x] &= -\frac{1}{|x|\sqrt{x^2 - 1}}
> \end{aligned} > d x d [ sin − 1 x ] > d x d [ tan − 1 x ] > d x d [ sec − 1 x ] = 1 − x 2 1 = 1 + x 2 1 = ∣ x ∣ x 2 − 1 1 d x d [ cos − 1 x ] d x d [ cot − 1 x ] d x d [ csc − 1 x ] = − 1 − x 2 1 = − 1 + x 2 1 = − ∣ x ∣ x 2 − 1 1 >
When layered with a differentiable inner function
u = g ( x ) u = g(x) u = g ( x ) , the
Chain Rule yields:
d d x [ sin − 1 u ] = u ′ 1 − u 2 , d d x [ tan − 1 u ] = u ′ 1 + u 2 , d d x [ sec − 1 u ] = u ′ ∣ u ∣ u 2 − 1 \frac{d}{dx}[\sin^{-1}u] = \frac{u'}{\sqrt{1 - u^2}}, \qquad \frac{d}{dx}[\tan^{-1}u] = \frac{u'}{1 + u^2}, \qquad \frac{d}{dx}[\sec^{-1}u] = \frac{u'}{|u|\sqrt{u^2 - 1}} d x d [ sin − 1 u ] = 1 − u 2 u ′ , d x d [ tan − 1 u ] = 1 + u 2 u ′ , d x d [ sec − 1 u ] = ∣ u ∣ u 2 − 1 u ′
Concept Explanation
1. Geometric Derivation of d d x [ sin − 1 ( x ) ] \frac{d}{dx}[\sin^{-1}(x)] d x d [ sin − 1 ( x )]
Let y = sin − 1 ( x ) y = \sin^{-1}(x) y = sin − 1 ( x ) x ∈ [ − 1 , 1 ] x \in [-1, 1] x ∈ [ − 1 , 1 ] y ∈ [ − π / 2 , π / 2 ] y \in [-\pi/2, \pi/2] y ∈ [ − π /2 , π /2 ] sin ( y ) = x \sin(y) = x sin ( y ) = x
Differentiating implicitly with respect to x x x cos ( y ) d y d x = 1 ⟹ d y d x = 1 cos ( y ) \cos(y) \frac{dy}{dx} = 1 \implies \frac{dy}{dx} = \frac{1}{\cos(y)} cos ( y ) d x d y = 1 ⟹ d x d y = cos ( y ) 1
In a reference right triangle with angle y y y x x x 1 1 1 1 − x 2 \sqrt{1 - x^2} 1 − x 2
Therefore, cos ( y ) = adjacent hypotenuse = 1 − x 2 \cos(y) = \frac{\text{adjacent}}{\text{hypotenuse}} = \sqrt{1 - x^2} cos ( y ) = hypotenuse adjacent = 1 − x 2
Substituting back:
d d x [ sin − 1 x ] = 1 1 − x 2 \frac{d}{dx}[\sin^{-1}x] = \frac{1}{\sqrt{1 - x^2}} d x d [ sin − 1 x ] = 1 − x 2 1
2. Geometric Derivation of d d x [ tan − 1 ( x ) ] \frac{d}{dx}[\tan^{-1}(x)] d x d [ tan − 1 ( x )]
Let y = tan − 1 ( x ) y = \tan^{-1}(x) y = tan − 1 ( x ) tan ( y ) = x \tan(y) = x tan ( y ) = x y ∈ ( − π / 2 , π / 2 ) y \in (-\pi/2, \pi/2) y ∈ ( − π /2 , π /2 )
Differentiating implicitly with respect to x x x sec 2 ( y ) d y d x = 1 ⟹ d y d x = 1 sec 2 ( y ) \sec^2(y) \frac{dy}{dx} = 1 \implies \frac{dy}{dx} = \frac{1}{\sec^2(y)} sec 2 ( y ) d x d y = 1 ⟹ d x d y = sec 2 ( y ) 1
Using the Pythagorean identity sec 2 ( y ) = 1 + tan 2 ( y ) \sec^2(y) = 1 + \tan^2(y) sec 2 ( y ) = 1 + tan 2 ( y ) sec 2 ( y ) = 1 + x 2 \sec^2(y) = 1 + x^2 sec 2 ( y ) = 1 + x 2
Substituting back yields the universally smooth derivative:
d d x [ tan − 1 x ] = 1 1 + x 2 \frac{d}{dx}[\tan^{-1}x] = \frac{1}{1 + x^2} d x d [ tan − 1 x ] = 1 + x 2 1
3. The "Co-" Mnemonic Pattern
Just as with ordinary trigonometric derivatives, every inverse trigonometric function beginning with "co-" (cos − 1 , cot − 1 , csc − 1 \cos^{-1}, \cot^{-1}, \csc^{-1} cos − 1 , cot − 1 , csc − 1
d d x [ cos − 1 x ] = − d d x [ sin − 1 x ] , d d x [ cot − 1 x ] = − d d x [ tan − 1 x ] \frac{d}{dx}[\cos^{-1}x] = -\frac{d}{dx}[\sin^{-1}x], \qquad \frac{d}{dx}[\cot^{-1}x] = -\frac{d}{dx}[\tan^{-1}x] d x d [ cos − 1 x ] = − d x d [ sin − 1 x ] , d x d [ cot − 1 x ] = − d x d [ tan − 1 x ] Engineering Applications
Robotics & Mechatronics (Joint Angular Rates): In robotic arms, inverse kinematics converts linear target coordinates into joint angles via θ = tan − 1 ( y / x ) \theta = \tan^{-1}(y/x) θ = tan − 1 ( y / x ) d θ d t \frac{d\theta}{dt} d t d θ Robotic Kinematics & Angular Velocity Mapping (Mechanical Engineering) ).Radar & Telemetry Telecommunications: Automated ground tracking stations calculate the angular sweep velocity d θ d t \frac{d\theta}{dt} d t d θ θ ( z ) = tan − 1 ( h / z ) \theta(z) = \tan^{-1}(h/z) θ ( z ) = tan − 1 ( h / z ) Tracking Actuator Angular Velocity (Robotics & Computer Engineering) ).Computer Vision & Autonomous Vehicles: Autonomous vehicle LiDAR systems translate depth matrices into optical steering angles using arc-tangent differentiation.
Example Problems
Example 1: Chain Rule with Inverse Sine
Find the derivative of f ( x ) = sin − 1 ( 5 x 3 ) f(x) = \sin^{-1}(5x^3) f ( x ) = sin − 1 ( 5 x 3 )
Solution
Step 1: Identify the outer rule and inner function
The outer function is sin − 1 ( u ) \sin^{-1}(u) sin − 1 ( u ) u = 5 x 3 u = 5x^3 u = 5 x 3 d d x [ sin − 1 ( u ) ] = u ′ 1 − u 2 \frac{d}{dx}[\sin^{-1}(u)] = \frac{u'}{\sqrt{1 - u^2}} d x d [ sin − 1 ( u )] = 1 − u 2 u ′
Step 2: Differentiate the inner function
u ′ = d d x [ 5 x 3 ] = 15 x 2 u' = \frac{d}{dx}[5x^3] = 15x^2 u ′ = d x d [ 5 x 3 ] = 15 x 2 Step 3: Substitute and simplify the radicand
f ′ ( x ) = 15 x 2 1 − ( 5 x 3 ) 2 = 15 x 2 1 − 25 x 6 f'(x) = \frac{15x^2}{\sqrt{1 - (5x^3)^2}} = \frac{15x^2}{\sqrt{1 - 25x^6}} f ′ ( x ) = 1 − ( 5 x 3 ) 2 15 x 2 = 1 − 25 x 6 15 x 2
Example 2: Operational Layering with Product Rule
Find the derivative of g ( x ) = x 2 + 1 csc − 1 ( x ) g(x) = \sqrt{x^2 + 1} \csc^{-1}(x) g ( x ) = x 2 + 1 csc − 1 ( x ) ∣ x ∣ > 1 |x| > 1 ∣ x ∣ > 1
Solution
Step 1: Apply the Product Rule
Let u = ( x 2 + 1 ) 1 2 u = (x^2 + 1)^{\frac{1}{2}} u = ( x 2 + 1 ) 2 1 v = csc − 1 ( x ) v = \csc^{-1}(x) v = csc − 1 ( x )
u ′ = 1 2 ( x 2 + 1 ) − 1 2 ( 2 x ) = x x 2 + 1 u' = \frac{1}{2}(x^2 + 1)^{-\frac{1}{2}}(2x) = \frac{x}{\sqrt{x^2 + 1}} u ′ = 2 1 ( x 2 + 1 ) − 2 1 ( 2 x ) = x 2 + 1 x v ′ = − 1 ∣ x ∣ x 2 − 1 v' = -\frac{1}{|x|\sqrt{x^2 - 1}} v ′ = − ∣ x ∣ x 2 − 1 1 Step 2: Assemble g ′ ( x ) = u ′ v + u v ′ g'(x) = u'v + uv' g ′ ( x ) = u ′ v + u v ′
g ′ ( x ) = ( x x 2 + 1 ) csc − 1 ( x ) + x 2 + 1 ( − 1 ∣ x ∣ x 2 − 1 ) g'(x) = \left(\frac{x}{\sqrt{x^2 + 1}}\right)\csc^{-1}(x) + \sqrt{x^2 + 1}\left(-\frac{1}{|x|\sqrt{x^2 - 1}}\right) g ′ ( x ) = ( x 2 + 1 x ) csc − 1 ( x ) + x 2 + 1 ( − ∣ x ∣ x 2 − 1 1 ) Step 3: Combine terms
g ′ ( x ) = x csc − 1 ( x ) x 2 + 1 − x 2 + 1 ∣ x ∣ x 2 − 1 g'(x) = \frac{x \csc^{-1}(x)}{\sqrt{x^2 + 1}} - \frac{\sqrt{x^2 + 1}}{|x|\sqrt{x^2 - 1}} g ′ ( x ) = x 2 + 1 x csc − 1 ( x ) − ∣ x ∣ x 2 − 1 x 2 + 1
Connections & References