Reverse rule

${\displaystyle \tan \alpha =\tan(90^{\circ }-\beta )={\frac {1}{\tan \beta }}}$

${\displaystyle {\color {Periwinkle}f'}(x)={\frac {1}{{\color {Salmon}(f^{-1})'}({\color {Blue}f}(x))}}}$

${\displaystyle {\color {Periwinkle}f'}(x_{0})={\frac {1}{4}}}$
${\displaystyle {\color {Salmon}(f^{-1})'}({\color {Blue}f}(x_{0}))=4~}$

The inverse rule (sometimes called the inverse rule) is a rule of differential calculus . It says that for a reversible ( i.e. bijective ) real function , ${\displaystyle f}$

• which is differentiable at the point and ${\displaystyle x}$
• has no horizontal tangent there, d. H. for which applies ${\displaystyle f'(x)\neq 0}$

its inverse function at the point is also differentiable with derivative ${\displaystyle f^{-1}}$ ${\displaystyle y=f(x)}$

${\displaystyle (f^{-1})'(y)={\frac {1}{f'(f^{-1}(y))}}={\frac {1}{f'(x)}}.}$

The validity of this equation can be illustrated with a sketch: The formation of the inverse function corresponds to an exchange of the coordinates and . The graphs of the function and its inverse function are therefore symmetrical to one another with respect to the bisector of the I. and III. Quadrant with the equation . The derivative of a function at a certain point corresponds to the slope of the associated tangent , i.e. equal to the tangent of the angle of inclination with respect to the horizontal. This gives: ${\displaystyle x}$ ${\displaystyle y}$ ${\displaystyle f}$ ${\displaystyle f^{-1}}$ ${\displaystyle y=x}$

${\displaystyle f'(x)=\tan \alpha =\tan(90^{\circ }-\beta )={\frac {1}{\tan \beta }}={\frac {1}{(f^{-1})'(f(x))}}.}$

Evidence Sketches

The inverse rule can be shown directly by taking the difference quotient

${\displaystyle {\frac {f(x+h)-f(x)}{h}}}$

transformed to the effect that he too

${\displaystyle {\frac {1}{\frac {f^{-1}(f(x+h))-f^{-1}(f(x))}{f(x+h)-f(x)}}}}$

to then substitute with. At the limit crossing for and thus also (note that differentiable functions are especially continuous) the assertion follows. ${\displaystyle t=f(x+h)-f(x)}$ ${\displaystyle h\to 0}$ ${\displaystyle t\to 0}$

Alternatively, using the chain rule results in the property

${\displaystyle f\left(f^{-1}(y)\right)=y}$

of the inverse function when differentiating according to both sides of the equation also the inverse rule (with ): ${\displaystyle y}$ ${\displaystyle f^{-1}(y)=x}$

${\displaystyle f'\left(f^{-1}(y)\right)\cdot \left(f^{-1}\right)'(y)=1.}$

However, the differentiability of is already assumed at the point , while it is also proven in the first proof sketch. ${\displaystyle f^{-1}}$ ${\displaystyle y}$

In a very similar way, one also obtains an expression for the 2nd derivative of the inverse function by differentiating the last equation again using the product rule (again is or ): ${\displaystyle \left(f^{-1}\right)''(y)}$ ${\displaystyle y}$ ${\displaystyle f^{-1}(y)=x}$ ${\displaystyle f(x)=y}$

${\displaystyle \left(f^{-1}\right)''(y)=-{\frac {f''(x)}{\left(f'(x)\right)^{3}}}.}$

Examples

The inverse of the exponential function is the natural logarithm ${\displaystyle y=f(x)=e^{x}}$

${\displaystyle f^{-1}(y)=\ln(y)\,.}$

Because so true ${\displaystyle f'(x)=e^{x}}$

${\displaystyle \ln '(y)={\frac {1}{e^{\ln(y)}}}={\frac {1}{y}}\,.}$

Another important application of the inverse rule is the derivatives of the inverse functions of the trigonometric functions . So z. B. for the derivative of the arcsine for due ${\displaystyle -1<y<1}$ ${\displaystyle \sin '=\cos }$

${\displaystyle \arcsin '(y)={\frac {1}{\cos \left(\arcsin(y)\right)}}\,.}$

If you rearrange the trigonometric Pythagoras according to the cosine, you get

${\displaystyle \cos(y)={\sqrt {1-\sin ^{2}(y)}}}$.

Because of this it follows: ${\displaystyle f(f^{-1}(y))=y}$

${\displaystyle \arcsin '(y)={\frac {1}{\sqrt {1-\sin ^{2}\left(\arcsin(y)\right)}}}={\frac {1}{\sqrt {1-y^{2}}}}\,.}$

The same applies to the derivatives of the arccosine and the arctangent .

Alternative formulations and generalizations

Alternative conditions

If one demands the continuity of the first derivative of , then the prerequisite is sufficient , since it follows directly on a small area around and from this in turn the existence of the inverse function of on this small area follows (consider the monotony of !). This basic idea is the starting point for the multidimensional generalization of the inverse rule, the theorem of the inverse mapping. ${\displaystyle f}$ ${\displaystyle f'(x)\neq 0}$ ${\displaystyle f'\neq 0}$ ${\displaystyle x}$ ${\displaystyle f}$ ${\displaystyle f}$

Different spellings in physics and other natural sciences

In physics and other natural sciences, Leibniz's notation with differentials is sometimes used. The reverse rule then takes the following form:

${\displaystyle {\frac {\mathrm {d} y}{\mathrm {d} x}}={\frac {1}{\frac {\mathrm {d} x}{\mathrm {d} y}}}.}$

Generalizations

The inverse rule can be generalized to the derivatives of functions in several dimensions. The multidimensional equivalent of the inverse rule is the theorem of inverse mapping .

literature

• Konrad Königsberger : Analysis 1 . 6th revised edition. Springer, Berlin et al. 2004, ISBN 3-540-40371-X , ( Springer textbook ).