Inverse function rule: Difference between revisions

Latest revision as of 04:51, 29 December 2025

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The thick blue curve and the thick red curve are inverse to each other. A thin curve is the derivative of the same colored thick curve. Inverse function rule:

f^{'} (x) = \frac{1}{[f^{- 1}] (f (x))}

Example for arbitrary

x_{0} \approx 5.8

:

f^{'} (x_{0}) = \frac{1}{4}

[f^{- 1}] (f (x_{0})) = 4

Script error: No such module "sidebar". In calculus, the inverse function rule is a formula that expresses the derivative of the inverse of a bijective and differentiable function Template:Mvar in terms of the derivative of Template:Mvar. More precisely, if the inverse of $f$ is denoted as $f^{- 1}$ , where $f^{- 1} (y) = x$ if and only if $f (x) = y$ , then the inverse function rule is, in Lagrange's notation,

[f^{- 1}] (y) = \frac{1}{f^{'} (f^{- 1} (y))} .

This formula holds in general whenever $f$ is continuous and injective on an interval Template:Mvar, with $f$ being differentiable at $f^{- 1} (y)$ ( $\in I$ ) and where $f^{'} (f^{- 1} (y)) \neq 0$ . The same formula is also equivalent to the expression

𝒟 [f^{- 1}] = \frac{1}{(𝒟 f) \circ (f^{- 1})},

where $𝒟$ denotes the unary derivative operator (on the space of functions) and $\circ$ denotes function composition.

Geometrically, a function and inverse function have graphs that are reflections, in the line $y = x$ . This reflection operation turns the gradient of any line into its reciprocal.^[1]

Assuming that $f$ has an inverse in a neighbourhood of $x$ and that its derivative at that point is non-zero, its inverse is guaranteed to be differentiable at $x$ and have a derivative given by the above formula.

The inverse function rule may also be expressed in Leibniz's notation. As that notation suggests,

\frac{d x}{d y} \frac{d y}{d x} = 1 .

This relation is obtained by differentiating the equation $f^{- 1} (y) = x$ in terms of Template:Mvar and applying the chain rule, yielding that:

\frac{d x}{d y} \frac{d y}{d x} = \frac{d x}{d x}

considering that the derivative of Template:Mvar with respect to Template:Mvar is 1.

Derivation

Let $f$ be an invertible (bijective) function, let $x$ be in the domain of $f$ , and let $y = f (x) .$ Let $g = f^{- 1} .$ So, $f (g (y)) = y .$ Differentiating this equation with respect to Template:Tmath, and using the chain rule, one gets

f^{'} (g (y)) \cdot g^{'} (y) = 1 .

That is,

g^{'} (y) = \frac{1}{f^{'} (g (y))}

or

{[f^{- 1}]}^{'} (y) = \frac{1}{f^{'} (f^{- 1} (y))} .

Examples

$y = x^{2}$ (for positive Template:Mvar) has inverse $x = \sqrt{y}$ .

\frac{d y}{d x} = 2 x; \frac{d x}{d y} = \frac{1}{2 \sqrt{y}} = \frac{1}{2 x}

\frac{d y}{d x} \frac{d x}{d y} = 2 x \cdot \frac{1}{2 x} = 1 .

At $x = 0$ , however, there is a problem: the graph of the square root function becomes vertical, corresponding to a horizontal tangent for the square function.

$y = e^{x}$ (for real Template:Mvar) has inverse $x = \ln y$ (for positive $y$ )

\frac{d y}{d x} = e^{x}; \frac{d x}{d y} = \frac{1}{y} = e^{- x}

\frac{d y}{d x} \frac{d x}{d y} = e^{x} e^{- x} = 1 .

Additional properties

Integrating this relationship gives

f^{- 1} (y) = \int \frac{1}{f^{'} (f^{- 1} (y))} d y + C .

This is only useful if the integral exists. In particular we need

f^{'} (x)

to be non-zero across the range of integration.

It follows that a function that has a continuous derivative has an inverse in a neighbourhood of every point where the derivative is non-zero. This need not be true if the derivative is not continuous.

Another very interesting and useful property is the following:

\int f^{- 1} (y) d y = y f^{- 1} (y) - F (f^{- 1} (y)) + C

where

F

denotes the antiderivative of

f

.

The inverse of the derivative of f(x) is also of interest, as it is used in showing the convexity of the Legendre transform.

Let $z = f^{'} (x)$ then we have, assuming $f^{″} (x) \neq 0$ : $\frac{d}{d z} {[f^{'}]}^{- 1} (z) = \frac{1}{f^{″} (x)}$ This can be shown using the previous notation $y = f (x)$ . Then we have:

f^{'} (x) = \frac{d y}{d x} = \frac{d y}{d z} \frac{d z}{d x} = \frac{d y}{d z} f^{″} (x) \Rightarrow \frac{d y}{d z} = \frac{f^{'} (x)}{f^{″} (x)}

Therefore:

\frac{d}{d z} [f^{'}]^{- 1} (z) = \frac{d x}{d z} = \frac{d y}{d z} \frac{d x}{d y} = \frac{f^{'} (x)}{f^{″} (x)} \frac{1}{f^{'} (x)} = \frac{1}{f^{″} (x)}

By induction, we can generalize this result for any integer $n \geq 1$ , with $z = f^{(n)} (x)$ , the nth derivative of f(x), and $y = f^{(n - 1)} (x)$ , assuming $f^{(i)} (x) \neq 0 for 0 < i \leq n + 1$ :

\frac{d}{d z} {[f^{(n)}]}^{- 1} (z) = \frac{1}{f^{(n + 1)} (x)}

Higher order derivatives

The chain rule given above is obtained by differentiating the identity $f^{- 1} (y) = x$ with respect to Template:Mvar, where $y = f (x)$ . One can continue the same process for higher derivatives. Differentiating the identity twice with respect to Template:Mvar, one obtains

\frac{d^{2} y}{d x^{2}} \frac{d x}{d y} + \frac{d}{d x} (\frac{d x}{d y}) (\frac{d y}{d x}) = 0,

that is simplified further by the chain rule as

\frac{d^{2} y}{d x^{2}} \frac{d x}{d y} + \frac{d^{2} x}{d y^{2}} {(\frac{d y}{d x})}^{2} = 0 .

Replacing the first derivative, using the identity obtained earlier, we get

\frac{d^{2} y}{d x^{2}} = - \frac{d^{2} x}{d y^{2}} {(\frac{d y}{d x})}^{3}

which implies

\frac{d^{2} x}{d y^{2}} = - \frac{d^{2} y / d x^{2}}{{(d y / d x)}^{3}} .

Similarly for the third derivative we have

\frac{d^{3} y}{d x^{3}} = - \frac{d^{3} x}{d y^{3}} {(\frac{d y}{d x})}^{4} - 3 \frac{d^{2} x}{d y^{2}} \frac{d^{2} y}{d x^{2}} {(\frac{d y}{d x})}^{2} .

Using the formula for the second derivative, we get

\frac{d^{3} y}{d x^{3}} = - \frac{d^{3} x}{d y^{3}} {(\frac{d y}{d x})}^{4} + 3 {(\frac{d^{2} y}{d x^{2}})}^{2} {(\frac{d y}{d x})}^{- 1}

which implies

\frac{d^{3} x}{d y^{3}} = - \frac{d^{3} y / d x^{3}}{{(d y / d x)}^{4}} + 3 \frac{{(d^{2} y / d x^{2})}^{2}}{{(d y / d x)}^{5}} .

These formulas can also be written using Lagrange's notation:

[f^{- 1}] (y) = - \frac{f^{″} (f^{- 1} (y))}{{[f^{'} (f^{- 1} (y))]}^{3}},

[f^{- 1}] (y) = - \frac{f^{‴} (f^{- 1} (y))}{{[f^{'} (f^{- 1} (y))]}^{4}} + 3 \frac{{[f^{″} (f^{- 1} (y))]}^{2}}{{[f^{'} (f^{- 1} (y))]}^{5}} .

In general, higher order derivatives of an inverse function can be expressed with Faà di Bruno's formula. Alternatively, the Template:Mvarth derivative can be written succinctly as:

{[f^{- 1}]}^{(n)} (y) = {[{(\frac{1}{f^{'} (t)} \frac{d}{d t})}^{n} t]}_{t = f^{- 1} (y)} .

From this expression, one can also derive the Template:Mvarth-integration of inverse function with base-point Template:Mvar using Cauchy formula for repeated integration whenever $f (f^{- 1} (y)) = y$ :

{[f^{- 1}]}^{(- n)} (y) = \frac{1}{n!} (f^{- 1} (a) (y - a)^{n} + \int_{f^{- 1} (a)}^{f^{- 1} (y)} {(y - f (u))}^{n} d u) .

Example

$y = e^{x}$ has the inverse $x = \ln y$ . Using the formula for the second derivative of the inverse function,

\frac{d y}{d x} = \frac{d^{2} y}{d x^{2}} = e^{x} = y; {(\frac{d y}{d x})}^{3} = y^{3};

so that

\frac{d^{2} x}{d y^{2}} \cdot y^{3} + y = 0; \frac{d^{2} x}{d y^{2}} = - \frac{1}{y^{2}},

which agrees with the direct calculation.

References

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Template:Calculus topics

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[1]

@@ Line 4: / Line 4: @@
 [[File:Umkehrregel 2.png|thumb|right|250px|The thick blue curve and the thick red curve are inverse to each other. A thin curve is the derivative of the same colored thick curve.
-Inverse function rule:<br><math>{\color{CornflowerBlue}{f'}}(x) = \frac{1}{{\color{Salmon}{(f^{-1})'}}({\color{Blue}{f}}(x))}</math><br><br>Example for arbitrary <math>x_0 \approx 5.8</math>:<br><math>{\color{CornflowerBlue}{f'}}(x_0) = \frac{1}{4}</math><br><math>{\color{Salmon}{(f^{-1})'}}({\color{Blue}{f}}(x_0)) = 4~</math>]]
+Inverse function rule:<br><math>{\color{CornflowerBlue}{f'}}(x) = \frac{1}{{\color{Salmon}{\left[f^{-1}\right]'}}({\color{Blue}{f}}(x))}</math><br><br>Example for arbitrary <math>x_0 \approx 5.8</math>:<br><math>{\color{CornflowerBlue}{f'}}(x_0) = \frac{1}{4}</math><br><math>{\color{Salmon}{\left[f^{-1}\right]'}}({\color{Blue}{f}}(x_0)) = 4~</math>]]
 {{calculus|expanded=differential}}
 In [[calculus]], the '''inverse function rule''' is a [[formula]] that expresses the [[derivative]] of the [[inverse function|inverse]] of a [[bijective]] and [[differentiable function]] {{Mvar|f}} in terms of the derivative of {{Mvar|f}}. More precisely, if the inverse of <math>f</math> is denoted as <math>f^{-1}</math>, where <math>f^{-1}(y) = x</math> if and only if <math>f(x) = y</math>, then the inverse function rule is, in [[Lagrange's notation]],
-:<math>\left[f^{-1}\right]'(y)=\frac{1}{f'\left( f^{-1}(y) \right)}</math>.
+:<math>\left[f^{-1}\right]'(y)=\frac{1}{f'\left( f^{-1}(y) \right)}.</math>
 This formula holds in general whenever <math>f</math> is [[continuous function|continuous]] and [[Injective function|injective]] on an interval {{Mvar|I}}, with <math>f</math> being differentiable at <math>f^{-1}(y)</math>(<math>\in I</math>) and where<math>f'(f^{-1}(y)) \ne 0</math>. The same formula is also equivalent to the expression
@@ Line 22: / Line 22: @@
 The inverse function rule may also be expressed in [[Leibniz's notation]]. As that notation suggests,
-:<math>\frac{dx}{dy}\,\cdot\, \frac{dy}{dx} = 1.</math>
+:<math>\frac{dx}{dy}\,\frac{dy}{dx} = 1.</math>
 This relation is obtained by differentiating the equation <math>f^{-1}(y)=x</math> in terms of {{Mvar|x}} and applying the [[chain rule]], yielding that:
-:<math>\frac{dx}{dy}\,\cdot\, \frac{dy}{dx} = \frac{dx}{dx}</math>
+:<math>\frac{dx}{dy}\,\frac{dy}{dx} = \frac{dx}{dx}</math>
 considering that the derivative of {{Mvar|x}} with respect to ''{{Mvar|x}}'' is 1.
 == Derivation ==
-Let <math>f</math> be an invertible (bijective) function, let <math>x</math> be in the domain of <math>f</math>, and let <math>y=f(x).</math> Let <math>g=f^{-1}.</math> So, <math>f(g(y))=y.</math> Derivating this equation with respect to {{tmath|y}}, and using the [[chain rule]], one gets
+Let <math>f</math> be an invertible (bijective) function, let <math>x</math> be in the domain of <math>f</math>, and let <math>y=f(x).</math> Let <math>g=f^{-1}.</math> So, <math>f(g(y))=y.</math> Differentiating this equation with respect to {{tmath|y}}, and using the [[chain rule]], one gets
 :<math>f'(g(y))\cdot g'(y)=1.</math>
 That is,
@@ Line 37: / Line 37: @@
 or
 :<math>
-  (f^{-1})^{\prime}(y) = \frac{1}{f^{\prime}(f^{-1}(y))}.
+  \left[f^{-1}\right]^{\prime}(y) = \frac{1}{f^{\prime}(f^{-1}(y))}.
 </math>
-==Examples==
+== Examples ==
 * <math>y = x^2</math> (for positive {{Mvar|x}}) has inverse <math>x = \sqrt{y}</math>.
@@ Line 49: / Line 49: @@
 \frac{dx}{dy} = \frac{1}{2\sqrt{y}}=\frac{1}{2x} </math>
-:<math> \frac{dy}{dx}\,\cdot\,\frac{dx}{dy} = 2x \cdot\frac{1}{2x} = 1. </math>
+:<math> \frac{dy}{dx}\,\frac{dx}{dy} = 2x \cdot\frac{1}{2x} = 1. </math>
 At <math>x=0</math>, however, there is a problem: the graph of the square root function becomes vertical, corresponding to a horizontal tangent for the square function.
@@ Line 60: / Line 60: @@
 \frac{dx}{dy} = \frac{1}{y} = e^{-x} </math>
-:<math> \frac{dy}{dx}\,\cdot\,\frac{dx}{dy} = e^x \cdot e^{-x} = 1. </math>
+:<math> \frac{dy}{dx}\,\frac{dx}{dy} = e^x e^{-x} = 1. </math>
 ==Additional properties==
@@ Line 66: / Line 66: @@
 * [[Integral|Integrating]] this relationship gives
-::<math>{f^{-1}}(x)=\int\frac{1}{f'({f^{-1}}(x))}\,{dx} + C.</math>
+::<math>{f^{-1}}(y)=\int\frac{1}{f'({f^{-1}}(y))}\,{dy} + C.</math>
 :This is only useful if the integral exists. In particular we need <math>f'(x)</math> to be non-zero across the range of integration.
@@ Line 74: / Line 74: @@
 * Another very interesting and useful property is the following:
-::<math> \int f^{-1}(x)\, {dx} = x f^{-1}(x) - F(f^{-1}(x)) + C </math>
+::<math> \int f^{-1}(y)\, {dy} = y f^{-1}(y) - F(f^{-1}(y)) + C </math>
-:Where <math> F </math> denotes the antiderivative of <math> f </math>.
+:where <math> F </math> denotes the antiderivative of <math> f </math>.
 * The inverse of the derivative of f(x) is also of interest, as it is used in showing the convexity of the [[Legendre transformation|Legendre transform]].
-Let  <math> z = f'(x)</math>  then we have, assuming <math> f''(x) \neq 0</math>:<math display="block"> \frac{d(f')^{-1}(z)}{dz} = \frac{1}{f''(x)}</math>This can be shown using the previous notation <math> y = f(x)</math>. Then we have:
+Let  <math> z = f'(x)</math>  then we have, assuming <math> f''(x) \neq 0</math>:<math display="block"> \frac{d}{dz}\left[f'\right]^{-1}(z) = \frac{1}{f''(x)}</math>This can be shown using the previous notation <math> y = f(x)</math>. Then we have:
 :<math display="block"> f'(x) = \frac{dy}{dx} = \frac{dy}{dz} \frac{dz}{dx} =  \frac{dy}{dz} f''(x) \Rightarrow \frac{dy}{dz} = \frac{f'(x) }{f''(x)}</math>Therefore:
-:<math> \frac{d(f')^{-1}(z)}{dz} = \frac{dx}{dz} = \frac{dy}{dz}\frac{dx}{dy} = \frac{f'(x)}{f''(x)}\frac{1}{f'(x)} = \frac{1}{f''(x)}</math>
+:<math> \frac{d}{dz}[f']^{-1}(z) = \frac{dx}{dz} = \frac{dy}{dz}\frac{dx}{dy} = \frac{f'(x)}{f''(x)}\frac{1}{f'(x)} = \frac{1}{f''(x)}</math>
 By induction, we can generalize this result for any integer <math> n \ge 1</math>, with  <math> z = f^{(n)}(x)</math>, the nth derivative of f(x), and  <math> y = f^{(n-1)}(x)</math>, assuming <math> f^{(i)}(x) \neq 0 \text{ for } 0 < i \le n+1 </math>:
-:<math> \frac{d(f^{(n)})^{-1}(z)}{dz} = \frac{1}{f^{(n+1)}(x)}</math>
+:<math> \frac{d}{dz}\left[f^{(n)}\right]^{-1}(z) = \frac{1}{f^{(n+1)}(x)}</math>
-== Higher derivatives ==
+== Higher order derivatives ==
-The [[chain rule]] given above is obtained by differentiating the identity <math>f^{-1}(f(x))=x</math> with respect to {{Mvar|x}}. One can continue the same process for higher derivatives. Differentiating the identity twice with respect to ''{{Mvar|x}}'', one obtains
+The [[chain rule]] given above is obtained by differentiating the identity <math>f^{-1}(y)=x</math> with respect to {{Mvar|y}}, where <math>y=f(x)</math>. One can continue the same process for higher derivatives. Differentiating the identity twice with respect to ''{{Mvar|x}}'', one obtains
-:<math> \frac{d^2y}{dx^2}\,\cdot\,\frac{dx}{dy} + \frac{d}{dx} \left(\frac{dx}{dy}\right)\,\cdot\,\left(\frac{dy}{dx}\right)  =  0, </math>
+:<math> \frac{d^2y}{dx^2}\,\frac{dx}{dy} + \frac{d}{dx} \left(\frac{dx}{dy}\right)\,\left(\frac{dy}{dx}\right)  =  0, </math>
 that is simplified further by the chain rule as
-:<math> \frac{d^2y}{dx^2}\,\cdot\,\frac{dx}{dy} + \frac{d^2x}{dy^2}\,\cdot\,\left(\frac{dy}{dx}\right)^2  =  0.</math>
+:<math> \frac{d^2y}{dx^2}\,\frac{dx}{dy} + \frac{d^2x}{dy^2}\,\left(\frac{dy}{dx}\right)^2  =  0.</math>
 Replacing the first derivative, using the identity obtained earlier, we get
-:<math> \frac{d^2y}{dx^2} = - \frac{d^2x}{dy^2}\,\cdot\,\left(\frac{dy}{dx}\right)^3. </math>
+:<math> \frac{d^2y}{dx^2} = - \frac{d^2x}{dy^2}\,\left(\frac{dy}{dx}\right)^3 </math>
-Similarly for the third derivative:
+which implies
-:<math> \frac{d^3y}{dx^3} = - \frac{d^3x}{dy^3}\,\cdot\,\left(\frac{dy}{dx}\right)^4 -
+:<math>  \frac{d^2x}{dy^2} = -\frac{d^2y/dx^2}{\left(dy/dx\right)^3}. </math>
-\frac{d^2x}{dy^2}\,\cdot\,\frac{d^2y}{dx^2}\,\cdot\,\left(\frac{dy}{dx}\right)^2</math>
-or using the formula for the second derivative,
+Similarly for the third derivative we have
-:<math> \frac{d^3y}{dx^3} = - \frac{d^3x}{dy^3}\,\cdot\,\left(\frac{dy}{dx}\right)^4 +
+:<math> \frac{d^3y}{dx^3} = - \frac{d^3x}{dy^3}\,\left(\frac{dy}{dx}\right)^4 -
-\left(\frac{d^2x}{dy^2}\right)^2\,\cdot\,\left(\frac{dy}{dx}\right)^5</math>
+\frac{d^2x}{dy^2}\,\frac{d^2y}{dx^2}\,\left(\frac{dy}{dx}\right)^2.</math>
-These formulas can also be written using Lagrange's notation. If ''{{Mvar|f}}'' and ''{{Mvar|g}}'' are inverses, then
+Using the formula for the second derivative, we get
-:<math> g''(x) = \frac{-f''(g(x))}{[f'(g(x))]^3}</math>
+:<math> \frac{d^3y}{dx^3} = - \frac{d^3x}{dy^3}\,\left(\frac{dy}{dx}\right)^4 +
+\left(\frac{d^2y}{dx^2}\right)^2\,\left(\frac{dy}{dx}\right)^{-1}</math>
-Higher derivatives of an inverse function can also be expressed with [[Faà di Bruno's formula]] and can be written succinctly as:
+which implies
-:<math>\left[f^{-1}\right]^{(n)}(x) = \left[\left(\frac{1}{f'(t)}\frac{d}{dt}\right)^{n} t\right]_{t = f^{-1}(x)}</math>
+:<math> \frac{d^3x}{dy^3} = - \frac{d^3y/dx^3}{\left(dy/dx\right)^4} + 3\frac{\left(d^2y/dx^2\right)^2}{\left(dy/dx\right)^5}. </math>
-From this expression, one can also derive the ''n''th-integration of inverse function with base-point ''a'' using [[Cauchy formula for repeated integration]] whenever <math>f(f^{-1}(x)) = x</math>:
+These formulas can also be written using Lagrange's notation:
-:<math>\left[f^{-1}\right]^{(-n)}(x) = \frac{1}{n!} \left(f^{-1}(a)(x-a)^n + \int_{f^{-1}(a)}^{f^{-1}(x)}\left(x-f(u)\right)^{n}\,du\right)</math>
+:<math> \left[f^{-1}\right]''(y) = -\frac{f''(f^{-1}(y))}{\left[f'(f^{-1}(y))\right]^3}, </math>
-==Example==
+:<math> \left[f^{-1}\right]'''(y) = -\frac{f'''(f^{-1}(y))}{\left[f'(f^{-1}(y))\right]^4} + 3\frac{\left[f''(f^{-1}(y))\right]^2}{\left[f'(f^{-1}(y))\right]^5}. </math>
+In general, higher order derivatives of an inverse function can be expressed with [[Faà di Bruno's formula]]. Alternatively, the {{Mvar|n}}th derivative can be written succinctly as:
+:<math> \left[f^{-1}\right]^{(n)}(y) = \left[\left(\frac{1}{f'(t)}\frac{d}{dt}\right)^{n} t\right]_{t = f^{-1}(y)}. </math>
+From this expression, one can also derive the {{Mvar|n}}th-integration of inverse function with base-point {{Mvar|a}} using [[Cauchy formula for repeated integration]] whenever <math>f(f^{-1}(y)) = y</math>:
+:<math> \left[f^{-1}\right]^{(-n)}(y) = \frac{1}{n!} \left(f^{-1}(a)(y-a)^n + \int_{f^{-1}(a)}^{f^{-1}(y)}\left(y-f(u)\right)^{n}\,du\right). </math>
+=== Example ===
 * <math>y = e^x</math> has the inverse <math>x = \ln y</math>. Using the formula for the second derivative of the inverse function,
@@ Line 140: / Line 150: @@
 \mbox{ }\mbox{ }\mbox{ }\mbox{ };
 \mbox{ }\mbox{ }\mbox{ }\mbox{ }
-\frac{d^2x}{dy^2} = -\frac{1}{y^2}
+\frac{d^2x}{dy^2} = -\frac{1}{y^2},</math>
-</math>,
 which agrees with the direct calculation.
-==See also==
+== See also ==
 {{Portal|Mathematics}}
@@ Line 159: / Line 168: @@
 * {{annotated link|Vector calculus identities}}
-==References==
+== References ==
 {{Reflist}}
 * {{Cite book|last=Marsden|first=Jerrold E.|url=https://authors.library.caltech.edu/25054/10/CalcUch8-invfunc-chainrule.pdf|title=Calculus unlimited|date=1981|publisher=Benjamin/Cummings Pub. Co|first2=Alan |last2=Weinstein|isbn=0-8053-6932-5|location=Menlo Park, Calif.|chapter=Chapter 8: Inverse Functions and the Chain Rule}}

Inverse function rule: Difference between revisions

Latest revision as of 04:51, 29 December 2025

Contents

Derivation

Examples

Additional properties

Higher order derivatives

Example

See also

References

Navigation menu

Inverse function rule: Difference between revisions

Latest revision as of 04:51, 29 December 2025

Derivation

Examples

Additional properties

Higher order derivatives

Example

See also

References

Navigation menu

Search