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.

which may be written as

we now differentiate both side of the above with respect to x using the chain rule on the right hand side

Hence

\LARGE {\dfrac{d(\arcsin(x))}{dx} = \dfrac{1}{\sqrt{1 - x^2}}}

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}}}

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 }}

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 }}

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} }}

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} }}

- 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

- 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})

- 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})

differentiation and derivatives