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)Letwhich 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}}}
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} }}
ExamplesExample 1Find the derivative of f(x) = x arcsin xSolution to Example 1:
Example 2Find the first derivative of f(x) = arctan x + x^{ 2}Solution to Example 2:
Example 3: Find the first derivative of f(x) = arcsin (2x + 2) Example 3
More References and linksDerivative of Inverse Functiondifferentiation and derivatives 