Derivatives of Inverse Trigonometric Functions
Solved Problems
Example 7.
\[y = \arctan \left( {x - \sqrt {1 + {x^2}} } \right)\]
Solution.
We apply the chain rule twice and simplify the resulting expression:
\[\require{cancel} y'\left( x \right) = {\left[ {\arctan \left( {x - \sqrt {1 + {x^2}} } \right)} \right]^\prime } = \frac{1}{{1 + {{\left( {x - \sqrt {1 + {x^2}} } \right)}^2}}} \cdot {\left( {x - \sqrt {1 + {x^2}} } \right)^\prime } = \frac{1}{{1 + {x^2} - 2x\sqrt {1 + {x^2}} + 1 + {x^2}}} \cdot \left( {1 - \frac{{\cancel{2}x}}{{\cancel{2}\sqrt {1 + {x^2}} }}} \right) = \frac{1}{{2\left( {1 + {x^2} - x\sqrt {1 + {x^2}} } \right)}} \cdot \frac{{\sqrt {1 + {x^2}} - x}}{{\sqrt {1 + {x^2}} }} = \frac{1}{{2{{\left( {\sqrt {1 + {x^2}} } \right)}^2}}} = \frac{1}{{2\left( {1 + {x^2}} \right)}}.\]
Example 8.
Calculate the derivative of the function \(y = \arccos x\arctan x\) at \(x = 0.\)
Solution.
By the product rule,
\[y^\prime = \left( {\arccos x\arctan x} \right)^\prime = \left( {\arccos x} \right)^\prime\arctan x + \arccos x\left( {\arctan x} \right)^\prime = - \frac{1}{{\sqrt {1 - {x^2}} }} \cdot \arctan x + \arccos x \cdot \frac{1}{{1 + {x^2}}} = \frac{{\arccos x}}{{1 + {x^2}}} - \frac{{\arctan x}}{{\sqrt {1 - {x^2}} }}.\]
Substitute \(x = 0:\)
\[y^\prime\left( 0 \right) = \frac{{\arccos 0}}{{1 + {0^2}}} - \frac{{\arctan 0}}{{\sqrt {1 - {0^2}} }} = \frac{{\frac{\pi }{2}}}{1} - \frac{0}{1} = \frac{\pi }{2}.\]
Example 9.
Using the chain rule, derive the formula for the derivative of the inverse sine function.
Solution.
The function \(y\left( x \right) = \arcsin x\) is defined on the open interval \(\left( { - 1,1} \right).\) The sine of the inverse sine is equal
\[\sin \left( {\arcsin x} \right) = x.\]
We take the derivative of both sides (the left-hand side is considered as a composite function).
\[\left[ {\sin \left( {\arcsin x} \right)} \right]^\prime = x',\;\; \Rightarrow \cos \left( {\arcsin x} \right) \cdot \left( {\arcsin x} \right)^\prime = 1.\]
It follows that the derivative of inverse sine function is given by
\[\left( {\arcsin x} \right)^\prime = \frac{1}{{\cos \left( {\arcsin x} \right)}} = \frac{1}{{\sqrt {1 - {{\left[ {\cos \left( {\arcsin x} \right)} \right]}^2}} }} = \frac{1}{{\sqrt {1 - {x^2}} }},\]
where \( - 1 \lt x \lt 1.\)
Example 10.
\[y = \arcsin \sqrt {1 - {x^2}} \]
Solution.
By the chain rule, we have
\[y'\left( x \right) = \left( {\arcsin \sqrt {1 - {x^2}} } \right)^\prime = \frac{1}{{\sqrt {1 - {{\left( {\sqrt {1 - {x^2}} } \right)}^2}} }} \cdot {\left( {\sqrt {1 - {x^2}} } \right)^\prime } = \frac{1}{{\sqrt {1 - {{\left( {\sqrt {1 - {x^2}} } \right)}^2}} }} \cdot \frac{1}{{2\sqrt {1 - {x^2}} }} \cdot {\left( {1 - {x^2}} \right)^\prime } = \frac{1}{{\sqrt {\cancel{1} - \cancel{1} + {x^2}} }} \cdot \frac{1}{{2\sqrt {1 - {x^2}} }} \cdot \left( { - 2x} \right) = \frac{{ - \cancel{2}x}}{{\cancel{2}\sqrt {{x^2}} \sqrt {1 - {x^2}} }} = - \frac{{x}}{{\left| x \right|\sqrt {1 - {x^2}} }}.\]
The ratio \(\frac{x}{{\left| x \right|}}\) is just a sign of the variable \(x\) (\(\text{sign }x\)). Therefore, the final answer is written as
\[\left( {\arcsin \sqrt {1 - {x^2}} } \right)^\prime = - \frac{{\text{sign }x}}{{\sqrt {1 - {x^2}} }}.\]
Example 11.
Calculate the derivative of \(y = \arctan \sqrt {{e^x}} \) at \(x = 0.\)
Solution.
By the chain rule,
\[y^\prime = \left( {\arctan \sqrt {{e^x}} } \right)^\prime = \frac{1}{{1 + {{\left( {\sqrt {{e^x}} } \right)}^2}}} \cdot \left( {\sqrt {{e^x}} } \right)^\prime = \frac{1}{{1 + {e^x}}} \cdot \frac{1}{{2\sqrt {{e^x}} }} \cdot \left( {{e^x}} \right)^\prime = \frac{{{e^x}}}{{2\sqrt {{e^x}} \left( {1 + {e^x}} \right)}} = \frac{{\sqrt {{e^x}} }}{{2\left( {1 + {e^x}} \right)}}.\]
For \(x = 0\) we have
\[y^\prime\left( 0 \right) = \frac{{\sqrt {{e^0}} }}{{2\left( {1 + {e^0}} \right)}} = \frac{{\sqrt 1 }}{{2\left( {1 + 1} \right)}} = \frac{1}{4}.\]
Example 12.
Compute the derivative of the function \(y = \arcsin \frac{1}{{\sqrt x }}\) at \(x = 2.\)
Solution.
By the chain rule,
\[y^\prime = \left( {\arcsin \frac{1}{{\sqrt x }}} \right)^\prime = \frac{1}{{\sqrt {1 - {{\left( {\frac{1}{{\sqrt x }}} \right)}^2}} }} \cdot \left( {\frac{1}{{\sqrt x }}} \right)^\prime = \frac{1}{{\sqrt {1 - \frac{1}{x}} }} \cdot \left( {{x^{ - \frac{1}{2}}}} \right)^\prime = \sqrt {\frac{x}{{x - 1}}} \cdot \left( { - \frac{1}{2}{x^{ - \frac{3}{2}}}} \right) = \sqrt {\frac{x}{{x - 1}}} \cdot \left( { - \frac{1}{{2\sqrt {{x^3}} }}} \right) = - \frac{\cancel{\sqrt x }}{{2x\cancel{\sqrt x} \sqrt {x - 1} }} = - \frac{1}{{2x\sqrt {x - 1} }}.\]
Substitute \(x = 2:\)
\[y^\prime\left( 2 \right) = - \frac{1}{{2 \cdot 2\sqrt {2 - 1} }} = - \frac{1}{4}.\]
Example 13.
\[y = \frac{2}{{\sqrt 3 }}\text{arccot}\frac{{2x + 1}}{{\sqrt 3 }}\]
Solution.
\[y'\left( x \right) = \left( {\frac{2}{{\sqrt 3 }}\text{arccot}\frac{{2x + 1}}{{\sqrt 3 }}} \right)^\prime = \frac{2}{{\sqrt 3 }} \cdot \left( { - \frac{1}{{1 + {{\left( {\frac{{2x + 1}}{{\sqrt 3 }}} \right)}^2}}}} \right) \cdot {\left( {\frac{{2x + 1}}{{\sqrt 3 }}} \right)^\prime } = - \frac{2}{{\sqrt 3 }} \cdot \frac{1}{{1 + \frac{{4{x^2} + 4x + 1}}{3}}} \cdot \frac{2}{{\sqrt 3 }} = - \frac{4}{3} \cdot \frac{1}{{\frac{{3 + 4{x^2} + 4x + 1}}{3}}} = - \frac{4}{3} \cdot \frac{3}{{4{x^2} + 4x + 4}} = - \frac{{\cancel{4} \cdot \cancel{3}}}{{\cancel{3} \cdot \cancel{4}\left( {{x^2} + x + 1} \right)}} = - \frac{1}{{{x^2} + x + 1}}.\]
Example 14.
Show that \({\frac{d}{{dx}}} \left( {x\arcsin x + \sqrt {1 - {x^2}} } \right)\) \(= \arcsin x.\)
Solution.
Convert the left side as follows:
\[\frac{d}{{dx}}\left( {x\arcsin x} \right) + \frac{d}{{dx}}\sqrt {1 - {x^2}} = \arcsin x,\;\; \Rightarrow x' \cdot \arcsin x + x \cdot {\left( {\arcsin x} \right)^\prime } + \frac{1}{{2\sqrt {1 - {x^2}} }} \cdot \left( {1 - {x^2}} \right) = \arcsin x.\]
Since \({\left( {\arcsin x} \right)^\prime } = {\frac{1}{{\sqrt {1 - {x^2}} }}},\) we have
\[1 \cdot \arcsin x + x \cdot \frac{1}{{\sqrt {1 - {x^2}} }} + \frac{{\left( { - 2x} \right)}}{{2\sqrt {1 - {x^2}} }} = \arcsin x,\;\; \Rightarrow \arcsin x + \cancel{\frac{x}{{\sqrt {1 - {x^2}} }}} - \cancel{\frac{x}{{\sqrt {1 - {x^2}} }}} = \arcsin x,\;\; \Rightarrow \arcsin x \equiv \arcsin x.\]
Thus, the identity is proved.
Example 15.
\[y = \arcsin \frac{{1 - {x^2}}}{{1 + {x^2}}}\]
Solution.
Note that that inverse sine function is defined on the interval \(\left[ { - 1,1} \right]\). In our case, the condition that defines the allowed values of \(x\) looks as follows:
\[- 1 \le \frac{{1 – {x^2}}}{{1 + {x^2}}} \le 1,\;\; \Rightarrow - 1 – {x^2} \le 1 – {x^2} \le 1 + {x^2}, \Rightarrow \left\{ {\begin{array}{*{20}{l}} { – 1 – {x^2} \le 1 – {x^2}}\\ {1 – {x^2} \le 1 + {x^2}} \end{array}} \right.,\;\; \Rightarrow \left\{ {\begin{array}{*{20}{l}} { – 1 \le 1}\\ { – {x^2} \le {x^2}} \end{array}} \right..\]
It is seen that these inequalities are satisfied for any real \(x.\)
We compute the derivative using the chain rule:
\[y'\left( x \right) = {\left( {\arcsin \frac{{1 - {x^2}}}{{1 + {x^2}}}} \right)^\prime } = \frac{1}{{\sqrt {1 - {{\left( {\frac{{1 - {x^2}}}{{1 + {x^2}}}} \right)}^2}} }} \cdot {\left( {\frac{{1 - {x^2}}}{{1 + {x^2}}}} \right)^\prime } = \frac{{1 + {x^2}}}{{\sqrt {{{\left( {1 + {x^2}} \right)}^2} - {{\left( {1 - {x^2}} \right)}^2}} }} \cdot \frac{{\left( {2x} \right) \cdot \left( {1 + {x^2}} \right) - \left( {1 - {x^2}} \right) \cdot 2x}}{{{{\left( {1 + {x^2}} \right)}^2}}} = \frac{{1 + {x^2}}}{{\sqrt {4{x^2}} }} \cdot \frac{{ - \color{blue}{2x} - \cancel{\color{red}{2{x^3}}} - \color{blue}{2x} + \cancel{\color{red}{2{x^3}}}}}{{{{\left( {1 + {x^2}} \right)}^2}}} = \frac{{ - \color{blue}{4x}}}{{\left( {1 + {x^2}} \right)\sqrt {4{x^2}} }} = \frac{{ - 2x}}{{\left( {1 + {x^2}} \right)\left| x \right|}}.\]
Take into account that the ratio \(\frac{x}{{\left| x \right|}}\) is equal to \( \pm 1\) depending on the sign of the variable \(x,\) that is
\[\frac{x}{{\left| x \right|}} = \text{sign }x\;\;\left( {x \ne 0} \right).\]
Then the derivative can be written as
\[y'\left( x \right) = \left( {\arcsin \frac{{1 - {x^2}}}{{1 + {x^2}}}} \right)^\prime = - \frac{{2\text{sign }x}}{{1 + {x^2}}}\,\left( {x \ne 0} \right).\]
Example 16.
\[y = \frac{1}{x}\text{arccsc}\frac{1}{x}\]
Solution.
We start calculating the derivative by the product rule:
\[y'\left( x \right) = \left( {\frac{1}{x}\text{arccsc} \frac{1}{x}} \right)^\prime = {\left( {\frac{1}{x}} \right)^\prime }\text{arccsc} \frac{1}{x} + \frac{1}{x}{\left( {\text{arccsc} \frac{1}{x}} \right)^\prime }.\]
Using the formula for the derivative of the inverse cosecant function
\[\left( {\text{arccsc}\,z} \right)^\prime = - \frac{1}{{\left| z \right|\sqrt {{z^2} - 1} }}\]
and the chain rule, we have
\[y'\left( x \right) = - \frac{1}{{{x^2}}}\text{arccsc}\frac{1}{x} - \frac{1}{x} \cdot \frac{1}{{\frac{1}{{\left| x \right|}}\sqrt {{{\left( {\frac{1}{x}} \right)}^2} - 1} }} \cdot {\left( {\frac{1}{x}} \right)^\prime }.\]
Simplify this expression:
\[y'\left( x \right) = - \frac{1}{{{x^2}}}\text{arccsc} \frac{1}{x} - \frac{{\left| x \right|}}{{x\sqrt {{{\left( {\frac{1}{x}} \right)}^2} - 1} }} \cdot \left( { - \frac{1}{{{x^2}}}} \right) = \frac{{\left| x \right|}}{{{x^3}\sqrt {{{\left( {\frac{1}{x}} \right)}^2} - 1} }} - \frac{1}{{{x^2}}}\text{arccsc} \frac{1}{x} = \frac{{\left| x \right|}}{{{x^3}\sqrt {\frac{{1 - {x^2}}}{{{x^2}}}} }} - \frac{1}{{{x^2}}}\text{arccsc} \frac{1}{x} = \frac{{{{\left| x \right|}^2}}}{{{x^3}\sqrt {1 - {x^2}} }} - \frac{1}{{{x^2}}}\text{arccsc} \frac{1}{x} = \frac{1}{{x\sqrt {1 - {x^2}} }} - \frac{1}{{{x^2}}}\text{arccsc} \frac{1}{x}.\]
The domain of the given function and its derivative is of the form: \(x \in \left( { - 1,0} \right) \cup \left( {0,1} \right).\)
Example 17.
\[y = \arccos \left( {\cos x} \right)\]
Solution.
\[y^\prime = \left( {\arccos \left( {\cos x} \right)} \right)^\prime = - \frac{1}{{\sqrt {1 - {{\cos }^2}x} }} \cdot \left( {\cos x} \right)^\prime = - \frac{{\left( { - \sin x} \right)}}{{\sqrt {{{\sin }^2}x} }} = \frac{{\sin x}}{{\sqrt {{{\sin }^2}x} }} = \frac{{\sin x}}{{\left| {\sin x} \right|}}.\]
We used here the classic definition of the square root which states that
\[\sqrt {{z^2}} = \left| z \right|,\]
where \(z\) is a real number.
Thus the derivative depends on the sign of the sine function.
If \(\sin x \gt 0,\) then
\[y^\prime = \frac{{\sin x}}{{\left| {\sin x} \right|}} = \frac{{\cancel{\sin x}}}{{\cancel{\sin x}}} = 1.\]
If \(\sin x \lt 0,\) then
\[y^\prime = \frac{{\sin x}}{{\left| {\sin x} \right|}} = \frac{{\sin x}}{{\left( { - \sin x} \right)}} = - \frac{{\sin x}}{{\sin x}} = - 1.\]
Combining these results, we can write
\[y^\prime = \frac{{\sin x}}{{\left| {\sin x} \right|}} = \text{sign}\left( {\sin x} \right).\]
Example 18.
Determine the derivative of the function \(y = \arcsin \frac{{2x}}{{1 + {x^2}}}\) at \(x = 2.\)
Solution.
We differentiate the given function using the chain and quotient rules:
\[y^\prime = \left( {\arcsin \frac{{2x}}{{1 + {x^2}}}} \right)^\prime = \frac{1}{{\sqrt {1 - {{\left( {\frac{{2x}}{{1 + {x^2}}}} \right)}^2}} }} \cdot \left( {\frac{{2x}}{{1 + {x^2}}}} \right)^\prime = \frac{1}{{\sqrt {1 - \frac{{4{x^2}}}{{1 + 2{x^2} + {x^4}}}} }} \cdot \frac{{2\left( {1 + {x^2}} \right) - 2x \cdot 2x}}{{{{\left( {1 + {x^2}} \right)}^2}}} = \frac{1}{{\sqrt {\frac{{1 + 2{x^2} + {x^4} - 4{x^2}}}{{1 + 2{x^2} + {x^4}}}} }} \cdot \frac{{2 + 2{x^2} - 4{x^2}}}{{{{\left( {1 + {x^2}} \right)}^2}}} = \sqrt {\frac{{1 + 2{x^2} + {x^4}}}{{1 - 2{x^2} + {x^4}}}} \cdot \frac{{2\left( {1 - {x^2}} \right)}}{{{{\left( {1 + {x^2}} \right)}^2}}} = \sqrt {\frac{{{{\left( {1 + {x^2}} \right)}^2}}}{{{{\left( {1 - {x^2}} \right)}^2}}}} \cdot \frac{{2\left( {1 - {x^2}} \right)}}{{{{\left( {1 + {x^2}} \right)}^2}}} = \frac{{2\cancel{\left( {1 + {x^2}} \right)}\left( {1 - {x^2}} \right)}}{{\left| {1 - {x^2}} \right|{{\left( {1 + {x^2}} \right)}^\cancel{2}}}} = \frac{{2\left( {1 - {x^2}} \right)}}{{\left| {1 - {x^2}} \right|\left( {1 + {x^2}} \right)}} = \frac{{2\,\text{sign}\left( {1 - {x^2}} \right)}}{{1 + {x^2}}}.\]
For \(x = 2\) we get
\[y^\prime\left( 2 \right) = \frac{{2\,\text{sign}\left( {1 - {2^2}} \right)}}{{1 + {2^2}}} = \frac{2}{5}\text{sign}\left( { - 3} \right) = \frac{2}{5} \cdot \left( { - 1} \right) = - \frac{2}{5}.\]
