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)

Let
arcsin(x)
which may be written as
x = sin(y)
we now differentiate both side of the above with respect to x using the chain rule on the right hand side
differentiate both sides
dy/dx
cos(y)
Hence
\LARGE {\dfrac{d(\arcsin(x))}{dx} = \dfrac{1}{\sqrt{1 - x^2}}}

2 - Derivative of arccos(x)

Let
y = \arccos(x)
which may be written as
x = \cos(y)
The differentiation of both side of the above with respect to x, using the chain rule on the right hand side, gives
1 = - \sin(y) \dfrac{dy}{dx}
The above gives
\dfrac{dy}{dx} = - \dfrac{1}{\sin(y) }
We now express sin(y) in terms of x
\sin(y) = \sqrt{1 - cos^2(y)} = \sqrt{1 - x^2}
Hence
\LARGE {\dfrac{d(\arccos(x))}{dx} = - \dfrac{1}{\sqrt{1 - x^2}}}

3 - Derivative of arctan(x)

Let
y = \arctan(x)
which may be written as
x = \tan(y)
We differentiate the left and right sides of the above, using the chain rule on the right hand side, to obtain
1 = \sec^2(y) \dfrac{dy}{dx}
The above gives
\dfrac{dy}{dx} = \dfrac{1}{\sec^2(y) } = cos^2(y)
We now express sec^2(y) in terms of x
\cos^2(y) = \dfrac{1}{1+\tan^2(y)} = \dfrac{1}{1+x^2}
Hence
\LARGE {\dfrac{d(\arctan(x))}{dx} = \dfrac{1}{1+x^2 }}

4 - Derivative of arccot(x)

Let
y = \text{arccot}(x)
which may be written as
x = \cot(y)
Differentiate both side with respect to x, using the chain rule on the right hand side, gives
1 = - \csc^2(y) \dfrac{dy}{dx}
The above gives
\dfrac{dy}{dx} = - \dfrac{1}{\csc^2(y) } = - sin^2(y)
We now express csc^2(y) in terms of x
\sin^2(y) = \dfrac{1}{1 + \cot^2(y)} = \dfrac{1}{1 + x^2}
Hence
\LARGE {\dfrac{d(\text{arccot}(x))}{dx} = - \dfrac{1}{1+x^2 }}

5 - Derivative of arcsec(x)

Let
y = \text{arcsec}(x)
which may be written as
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) \dfrac{dy}{dx}
The above gives
\dfrac{dy}{dx} = \dfrac{1}{\sec(y) \tan(y) }
We now express sec(y) tan(y) in terms of x
\sec(y) \tan(y) = x \sqrt{x^2 - 1}
Hence
\LARGE {\dfrac{d(\text{arcsec}(x))}{dx} = \dfrac{1}{x \sqrt{x^2 - 1} }}

6 - Derivative of arccsc(x)

Let
y = \text{arccsc}(x)
which may be written as
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) \dfrac{dy}{dx}
The above gives
\dfrac{dy}{dx} = - \dfrac{1}{ \csc(y) \cot(y) }
We now express \csc(y) \cot(y) in terms of x
\csc(y) \cot(y) = x \sqrt{x^2 - 1}
Hence
\LARGE {\dfrac{d(\text{arccsc}(x))}{dx} = - \dfrac{1}{x \sqrt{x^2 - 1} }}


Solution to Example 1:
  • Let h(x) = x and g(x) = arcsin x, function f is considered as the product of functions h and g: f(x) = h(x) g(x). Hence we use the product rule, f '(x) = h(x) g '(x) + g(x) h '(x), to differentiate function f as follows
    f '(x) = x (1 / √(1 - x 2)) + arcsin x * 1 = x / √(1 - x 2) + arcsin x

Example 2

Find the first derivative of f(x) = arctan x + x 2
Solution to Example 2:
  • Let g(x) = arctan x and h(x) = x 2, function f may be considered as the sum of functions g and h: f(x) = g(x) + h(x). Hence we use the sum rule, f '(x) = g '(x) + h '(x), to differentiate function f as follows
    f '(x) = 1 / (1 + x 2) + 2x = (2x 3 + 2x + 1) / (1 + x 2)

Example 3: Find the first derivative of f(x) = arcsin (2x + 2)

Example 3
  • Let u(x) = 2x + 2, function f may be considered as the composition f(x) = arcsin(u(x)). Hence we use the chain rule, f '(x) = (du/dx) d(arcsin(u))/du, to differentiate function f as follows
    g '(x) = (2)(1 / √(1 - u 2)
    = 2 / √(1 - (2x + 2) 2)

More References and links

Derivative of Inverse Function
differentiation and derivatives

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