Derivatives of Inverse Trigonometric Functions

Proofs of the formulas of the derivatives of inverse trigonometric functions are presented along with several other examples involving sums, products and quotients of functions. Another method to find the derivative of inverse functions is also included and may be used.

1 - Derivative of y=arcsin⁡(x) y = \arcsin(x)

Let y=arcsin⁡(x) y = \arcsin(x) which may be written as
x=sin⁡(y) x = \sin(y)
Differentiate both side of the above with respect to x x dxdx=d(sin⁡(y))dx \dfrac{dx}{dx} = \dfrac{d (\sin(y))} {dx} Simplify the left side and use the chain rule on the right hand side 1=cos⁡(y)dydx 1 = \cos(y) \dfrac{dy}{dx} The above gives dydx=1cos⁡y \dfrac{dy}{dx} = \dfrac{1}{\cos y} The trigonometric identity sin⁡2y+cos⁡2y=1 \sin^2 y + \cos^2 y = 1 gives cos⁡(y)=1−sin⁡2(y) \cos(y) = \sqrt{1 - \sin^2(y)} From above, we have x=sin⁡(y) x = \sin(y) , hence cos⁡(y)=1−x2 \cos(y) = \sqrt{1 - x^2}
Substitute cos⁡(y)=1−x2 \cos(y) = \sqrt{1 - x^2} in dydx=1cos⁡y \dfrac{dy}{dx} = \dfrac{1}{\cos y} to obtain dydx=11−x2 \dfrac{dy}{dx} = \dfrac{1}{\sqrt{1 - x^2}} Hence d(arcsin⁡(x))dx=11−x2 \Large \color{red}{\dfrac{d(\arcsin(x))}{dx} = \dfrac{1}{\sqrt{1 - x^2}}}

2 - Derivative of arccos⁡(x) \arccos(x)

Let y=arccos⁡(x) y = \arccos(x) which may be written as x=cos⁡(y) x = \cos(y) The differentiation of both sides of the above, with respect to x x using the chain rule on the right hand side, gives 1=−sin⁡(y)dydx 1 = - \sin(y) \dfrac{dy}{dx} The above gives dydx=−1sin⁡(y) \dfrac{dy}{dx} = - \dfrac{1}{\sin(y) } The trigonometric identity sin⁡2y+cos⁡2y=1 \sin^2 y + \cos^2 y = 1 gives sin⁡(y)=1−cos⁡2(y) \sin(y) = \sqrt{1 - \cos^2 (y)} Use x=cos⁡(y) x = \cos(y) from above to write sin⁡(y)=1−x2 \sin(y) = \sqrt{1 - x^2} Substitute sin⁡y \sin y in dydx=−1sin⁡(y) \dfrac{dy}{dx} = - \dfrac{1}{\sin(y) } to obtain d(arccos⁡(x))dx=−11−x2 \Large \color{red}{\dfrac{d(\arccos(x))}{dx} = - \dfrac{1}{\sqrt{1 - x^2}}}

3 - Derivative of arctan⁡(x) \arctan(x)

Let y=arctan⁡(x) y = \arctan(x) which may be written as x=tan⁡(y) x = \tan(y) We differentiate both side with respect to x, using the chain rule on the right hand side, to obtain 1=sec⁡2(y)dydx 1 = \sec^2(y) \dfrac{dy}{dx} The above gives dydx=1sec⁡2(y)=cos⁡2(y) \dfrac{dy}{dx} = \dfrac{1}{\sec^2(y) } = \cos^2(y) We now use the trignometric identity cos⁡2(y)=11+tan⁡2(y) \cos^2(y) = \dfrac{1}{1+\tan^2(y)} and x=tan⁡(y) x = \tan(y) from above to express sec⁡2(y) \sec^2(y) in terms of x as follows cos⁡2(y)=11+x2 \cos^2(y) = \dfrac{1}{1+x^2} Substitute in Hence dydx=cos⁡2(y) \dfrac{dy}{dx} = \cos^2(y) to obtain d(arctan⁡(x))dx=11+x2 \Large \color{red}{\dfrac{d(\arctan(x))}{dx} = \dfrac{1}{1+x^2 }}

4 - Derivative of arccot(x) \text{arccot}(x)

Let y=arccot(x) y = \text{arccot}(x) which may be written as x=cot⁡(y) x = \cot(y) Differentiate both side with respect to x, using the chain rule on the right hand side, gives 1=−csc⁡2(y)dydx 1 = - \csc^2(y) \dfrac{dy}{dx} The above gives dydx=−1csc⁡2(y)=−sin⁡2(y) \dfrac{dy}{dx} = - \dfrac{1}{\csc^2(y) } = - \sin^2(y) Use the trigonometric identity sin⁡2(y)=11+cot⁡2(y) \sin^2(y) = \dfrac{1}{1 + \cot^2(y)} and x=cot⁡(y) x = \cot(y) from above to express sin⁡2(y) \sin^2(y) in terms of x as follows sin⁡2(y)=11+cot⁡2(y)=11+x2 \sin^2(y) = \dfrac{1}{1 + \cot^2(y)} = \dfrac{1}{1 + x^2} Hence d(arccot(x))dx=−11+x2 \Large \color{red}{\dfrac{d(\text{arccot}(x))}{dx} = - \dfrac{1}{1+x^2 }}

5 - Derivative of arcsec(x) \text{arcsec}(x)

Let y=arcsec(x) y = \text{arcsec}(x) which may be written as x=sec⁡(y) x = \sec(y) The differentiation of both side with respect to x, using the chain rule on the right hand side, gives 1=sec⁡(y)tan⁡(y)dydx 1 = \sec(y) \tan(y) \dfrac{dy}{dx} The above gives dydx=1sec⁡(y)tan⁡(y) \dfrac{dy}{dx} = \dfrac{1}{\sec(y) \tan(y) } Use the identity tan⁡y=sec⁡2y−1 \tan y = \sqrt{\sec^2 y - 1} and x=sec⁡(y) x = \sec(y) from above to express sec⁡(y)tan⁡(y) \sec(y) \tan(y) in terms of x as follows sec⁡(y)tan⁡(y)=xx2−1 \sec(y) \tan(y) = x \sqrt{x^2 - 1} Hence d(arcsec(x))dx=1xx2−1 \Large \color{red}{\dfrac{d(\text{arcsec}(x))}{dx} = \dfrac{1}{x \sqrt{x^2 - 1} }}

6 - Derivative of arccsc(x) \text{arccsc}(x)

Let y=arccsc(x) y = \text{arccsc}(x) which may be written as x=csc⁡(y) x = \csc(y) The differentiation of the left and right sides of the above, using the chain rule on the right hand side, gives 1=−csc⁡(y)cot⁡(y)dydx 1 = - \csc(y) \cot(y) \dfrac{dy}{dx} The above gives dydx=−1csc⁡(y)cot⁡(y) \dfrac{dy}{dx} = - \dfrac{1}{ \csc(y) \cot(y) } Use the trigonometric identity cot⁡y=csc2y−1 \cot y = \sqrt {csc^2 y - 1} and x=csc⁡(y) x = \csc(y) from above to express csc⁡(y)cot⁡(y) \csc(y) \cot(y) in terms of x as follows csc⁡(y)cot⁡(y)=csc⁡(y)csc⁡2y−1=xx2−1 \csc(y) \cot(y) = \csc(y) \sqrt {\csc^2 y - 1} = x \sqrt{x^2 - 1} Hence d(arccsc(x))dx=−1xx2−1 \Large \color{red}{\dfrac{d(\text{arccsc}(x))}{dx} = - \dfrac{1}{x \sqrt{x^2 - 1} }}

Examples with Solutions

Example 1

Find the first derivative of f(x)=xarcsin⁡x f(x) = x \arcsin x Solution to Example 1:
Let h(x)=x h(x) = x and g(x)=arcsin⁡x g(x) = \arcsin x , function f f is considered as the product of functions h h and g g : f(x)=h(x)g(x) f(x) = h(x) g(x) .
Use the product rule of differentiation: f′(x)=h(x)g′(x)+g(x)h′(x) f '(x) = h(x) g '(x) + g(x) h '(x) , to differentiate function f f as follows
f′(x)=x(11−x2)+arcsin⁡x⋅1 f '(x) = x \left( \dfrac{1}{\sqrt{1 - x^2}} \right) + \arcsin x \cdot 1
=x1−x2+arcsin⁡x = \dfrac{x}{\sqrt{1 - x^2}} + \arcsin x

Example 2

Find the first derivative of f(x)=arctan⁡x+x2 f(x) = \arctan x + x^2
Solution to Example 2:
Let g(x)=arctan⁡x g(x) = \arctan x and h(x)=x2 h(x) = x^2 , function f f may be considered as the sum of functions g g and h h : f(x)=g(x)+h(x) f(x) = g(x) + h(x) . Hence we use the sum rule, f′(x)=g′(x)+h′(x) f '(x) = g '(x) + h '(x) , to differentiate function f f as follows
f′(x)=11+x2+2x f '(x) = \dfrac{1}{1 + x^2} + 2x
=2x3+2x+11+x2 = \dfrac{2x^3 + 2x + 1}{1 + x^2}

Example 3

Find the first derivative of f(x)=arcsin⁡(2x+2) f(x) = \arcsin (2x + 2)
Let u(x)=2x+2 u(x) = 2x + 2 , function f f may be considered as the composition f(x)=arcsin⁡(u(x)) f(x) = \arcsin(u(x)) . Hence we use the chain rule, f′(x)=(dudx)d(arcsin⁡(u))du f '(x) = \left( \dfrac{du}{dx} \right) \dfrac{d(\arcsin(u))}{du} , to differentiate function f f as follows
g′(x)=2(11−u2) g '(x) = 2 \left( \dfrac{1}{\sqrt{1 - u^2}} \right)
=21−(2x+2)2 = \dfrac{2}{\sqrt{1 - (2x + 2)^2}}

More References and links

Derivative of Inverse Function
differentiation and derivatives