Calculus

Differentiation of Functions

Differentiation of Functions Logo

Derivatives of Inverse Functions

  • Inverse functions are functions that “reverse” each other.

    We consider a function \(f\left( x \right)\), which is strictly monotonic on an interval \(\left( {a,b} \right)\). If there exists a point \({x_0}\) in this interval such that \(f’\left( {{x_0}} \right) \ne 0\), then the inverse function \(x = \varphi \left( y \right)\) is also differentiable at \({y_0} = f\left( {{x_0}} \right)\) and its derivative is given by

    \[\varphi’\left( {{y_0}} \right) = \frac{1}{{f’\left( {{x_0}} \right)}}.\]

    Let us prove this theorem (called the inverse function theorem).

    Suppose that the variable \(y\) gets an increment \(\Delta y \ne 0\) at the point \({y_0}.\) The corresponding increment of the variable \(x\) at the point \({x_0}\) is denoted by \(\Delta x\), where \(\Delta x \ne 0\) due to the strict monotonicity of \(y = f\left( x \right)\). The ratio of the increments is written as

    \[\frac{{\Delta x}}{{\Delta y}} = \frac{1}{{\frac{{\Delta y}}{{\Delta x}}}}.\]

    Suppose that \(\Delta y \to 0\). Then \(\Delta x \to 0\), since the inverse function \(x = \varphi \left( y \right)\) is continuous at \({y_0}\). In the limit when \(\Delta x \to 0\), the right side of the relationship becomes

    \[
    {\lim\limits_{\Delta x \to 0} \frac{1}{{\frac{{\Delta y}}{{\Delta x}}}} = \frac{1}{{\lim\limits_{\Delta x \to 0} \frac{{\Delta y}}{{\Delta x}}}} }
    = {\frac{1}{{f’\left( {{x_0}} \right)}}.}
    \]

    In this case, the left hand side also approaches a limit, which by definition is equal to the derivative of the inverse function:

    \[\lim\limits_{\Delta y \to 0} \frac{{\Delta x}}{{\Delta y}} = \varphi’\left( {{y_0}} \right).\]

    Thus,

    \[\varphi’\left( {{y_0}} \right) = \frac{1}{{f’\left( {{x_0}} \right)}},\]

    that is the derivative of the inverse function is the inverse of the derivative of the original function.

    In the examples below, find the derivative of the function \(y = f\left( x \right)\) using the derivative of the inverse function \(x = \varphi \left( y \right).\)


  • Solved Problems

    Click a problem to see the solution.

    Example 1

    \[y = \sqrt[\large n\normalsize]{x}\]

    Example 2

    \[y = \arcsin x\]

    Example 3

    \[y = \ln x\]

    Example 4

    \[y = \sqrt[\large 3\normalsize]{{x + 1}}\]

    Example 5

    \[y = \arccos \left( {1 – 2x} \right)\]

    Example 6

    \[y = \sqrt {1 + \sqrt x } \]

    Example 7

    \[y = \arctan \frac{1}{x}\]

    Example 8

    \[y = \sqrt x \]

    Example 9

    \[y = 2x + 4\]

    Example 10

    Given the function \(y = {x^5} + 2{x^3} + 3x\). Find the derivative of the inverse function at \(x = 1.\)

    Example 11

    Given the function \(y = {x^2} – x\). Find the derivative of the inverse function at \(x = 1.\)

    Example 12

    Given the function \(y = {e^x} + 2x + 1\). Find the derivative of the inverse function at \(x = 0.\)

    Example 13

    Find the derivative of the inverse function at \(x = 1\) for the function \(y = \sin \left( {x – 1} \right) + {x^2}.\)

    Example 14

    Find the derivative of the inverse function of \(y = {x^2} + 2\ln x\) and calculate its value at \(x = 1.\)

    Example 15

    Find the derivative of the inverse function for \(y = {x^3} – 3x\) and calculate its value at \(x = -2.\)

    Example 16

    Find the derivative of the inverse function for \(y = 2{x^3} – 1\) and calculate its value at \(x = 2.\)

    Example 17

    \[y = {\log _2}\left( {\frac{x}{3}} \right)\]

    Example 18

    Find the derivative of \(y = \text{arcsec }x\) at \(x = \sqrt 2.\)

    Example 19

    Find the derivative of the inverse function for \(y = x \cdot {3^x}\) provided \(x \gt 0.\)

    Example 20

    Find the derivative of the inverse hyperbolic sine function \(y = \text{arcsinh } x.\)

    Example 1.

    \[y = \sqrt[\large n\normalsize]{x}\]

    Solution.

    We first determine the inverse function for the given function \(y = \sqrt[\large n\normalsize]{x}\). To do this, we express the variable \(x\) in terms of \(y:\)

    \[
    {y = f\left( x \right) = \sqrt[\large n\normalsize]{x},\;\;}\Rightarrow
    {{y^n} = {\left( {\sqrt[\large n\normalsize]{x}} \right)^n},\;\;}\Rightarrow
    {x = \varphi \left( y \right) = {y^n}.}
    \]

    By the inverse function theorem, we can write:

    \[
    {{\left( {\sqrt[\large n\normalsize]{x}} \right)^\prime } = f’\left( x \right) }
    = {\frac{1}{{\varphi’\left( y \right)}} }
    = {\frac{1}{{{{\left( {{y^n}} \right)}^\prime }}} }
    = {\frac{1}{{n{y^{n – 1}}}}.}
    \]

    Now we substitute \(y = \sqrt[\large n\normalsize]{x}\) instead of \(y.\) As a result, we obtain an expression for the derivative of the given function:

    \[
    {{\left( {\sqrt[\large n\normalsize]{x}} \right)^\prime } = \frac{1}{{n{y^{n – 1}}}} }
    = {\frac{1}{{n{{\left( {\sqrt[\large n\normalsize]{x}} \right)}^{n – 1}}}} }
    = {\frac{1}{{n\sqrt[\large n\normalsize]{{{x^{n – 1}}}}}}\;\;\;}\kern-0.3pt{\left( {x \gt 0} \right).}
    \]

    Example 2.

    \[y = \arcsin x\]

    Solution.

    The arcsine function is the is the inverse of the sine function. Therefore \(x = \varphi \left( y \right) \) \(= \sin y.\) Then the derivative of \(\arcsin x\) is

    \[
    {{\left( {\arcsin x} \right)^\prime } = f’\left( x \right) }
    = {\frac{1}{{\varphi’\left( y \right)}} }
    = {\frac{1}{{{{\left( {\sin y} \right)}^\prime }}} }
    = {\frac{1}{{\cos y}} }
    = {\frac{1}{{\sqrt {1 – {{\sin }^2}y} }} }
    = {\frac{1}{{\sqrt {1 – {{\sin }^2}\left( {\arcsin x} \right)} }} }
    = {\frac{1}{{\sqrt {1 – {x^2}} }},}
    \]

    where \(-1 \lt x \lt 1.\)

    Example 3.

    \[y = \ln x\]

    Solution.

    The natural logarithm and the exponential function are mutually inverse functions. Therefore, \(x = \varphi \left( y \right) = {e^y}\), where \(x \gt 0\), \(y \in \mathbb{R}\). The derivative of the natural logarithm is easy to calculate through the derivative of the exponential function:

    \[
    {{\left( {\ln x} \right)^\prime } = f’\left( x \right) }
    = {\frac{1}{{\varphi’\left( y \right)}} }
    = {\frac{1}{{{{\left( {{e^y}} \right)}^\prime }}} }
    = {\frac{1}{{{e^y}}} }
    = {\frac{1}{{{e^{\ln x}}}} }
    = {\frac{1}{x}}
    \]

    Here we have used a logarithmic identity, according to which

    \[{e^{\ln x}} = x.\]

    Page 1
    Problems 1-3
    Page 2
    Problems 4-20