Implicit differentiation

The implicit differentiation (also implicit derivation ) is a possibility to derive a function, which is not explicitly given by a term, but only implicitly by an equation (also implicit curve ), with the help of the multidimensional differential calculus . It can often also be used to determine the derivative of functions that are explicitly given but difficult to derive in this form.

rule

Does the differentiable function satisfy the equation ${\displaystyle f\colon \mathbb {R} \to \mathbb {R} }$

${\displaystyle F(x,f(x))=0}$,

where also , is a differentiable function , it means that the function is constant (namely the null function). Its derivative is therefore also constant zero. With the help of the multidimensional chain rule one then obtains ${\displaystyle F\colon \mathbb {R} ^{2}\to \mathbb {R} ,\ F\colon (x,y)\mapsto F(x,y)}$ ${\displaystyle x\mapsto F(x,f(x))}$

${\displaystyle 0={\frac {\mathrm {d} }{\mathrm {d} x}}F(x,f(x))={\frac {\partial F}{\partial x}}+{\frac {\partial F}{\partial y}}\,f'=F_{x}+F_{y}\,f'\,.}$

Where and are the partial derivatives of . To simplify the notation, the function arguments have been omitted. ${\displaystyle F_{x}={\tfrac {\partial F}{\partial x}}}$ ${\displaystyle F_{y}={\tfrac {\partial F}{\partial y}}}$ ${\displaystyle F}$ ${\displaystyle (x,f(x))}$

If it applies at one point , then this also applies to all in a neighborhood of and the equation can be solved for : ${\displaystyle F_{y}(x_{0},f(x_{0}))\neq 0}$ ${\displaystyle x_{0}}$ ${\displaystyle x}$ ${\displaystyle x_{0}}$ ${\displaystyle \,f'}$

${\displaystyle f'=-{\frac {F_{x}}{F_{y}}}}$

or in detail

${\displaystyle f'(x)=-{\frac {F_{x}(x,f(x))}{F_{y}(x,f(x))}}\,.}$

Higher derivatives

By applying the product and chain rule , higher derivatives of implicit functions can also be calculated. The second derivation is thus : ${\displaystyle f''}$

${\displaystyle f''(x)=-{\frac {F_{xx}F_{y}^{2}+F_{yy}F_{x}^{2}-2F_{xy}F_{x}F_{y}}{F_{y}^{3}}}}$

with , , . ${\displaystyle F_{xx}={\tfrac {\partial ^{2}F}{\partial x^{2}}}}$ ${\displaystyle F_{yy}={\tfrac {\partial ^{2}F}{\partial y^{2}}}}$ ${\displaystyle F_{xy}={\tfrac {\partial ^{2}F}{\partial x\partial y}}}$

Examples

example 1

Find the derivative function of the natural logarithm . You can also represent this implicitly ${\displaystyle f'(x)}$ ${\displaystyle \ln(x)}$

${\displaystyle f(x)=\ln(x)\Leftrightarrow e^{f(x)}-x=0}$,

then derive the equation

${\displaystyle e^{f(x)}\cdot f'(x)-1=0}$,

again set ${\displaystyle f(x)=\ln(x)}$

${\displaystyle x\cdot f'(x)-1=0}$

and move

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

Example 2

The function , can not be derived without transformations with the conventional derivation rules, since both the exponent and the base power are variable. First, one can eliminate the exponent by taking the logarithm : ${\displaystyle f(x)=x^{x}}$ ${\displaystyle x>0}$

${\displaystyle \ln f(x)=x\ln x}$.

Now one derives implicitly by deriving both sides conventionally according to : ${\displaystyle x}$

${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}(\ln f(x))={\frac {\mathrm {d} }{\mathrm {d} x}}(x\ln x)}$

The left side can be calculated with the chain rule , the right with the product rule and the rule for the derivation of the logarithm:

${\displaystyle {\frac {1}{f(x)}}\cdot f'(x)=\ln x+x{\frac {1}{x}}}$

If you dissolve and bet , you get the solution: ${\displaystyle f'(x)}$ ${\displaystyle f(x)=x^{x}}$

${\displaystyle f'(x)=f(x)\,(\ln x+1)=x^{x}\left(\ln x+1\right)}$.

Example 3

The circle with center and radius is given by the equation . Parts of it can be written as a graph of a function . Their derivative can be calculated with the help of implicit differentiation as follows: ${\displaystyle (0,0)}$ ${\displaystyle r}$ ${\displaystyle x^{2}+y^{2}=r^{2}}$ ${\displaystyle y=f(x)}$

Insert into the defining equation : ${\displaystyle y=f(x)}$

${\displaystyle x^{2}+f(x)^{2}=r^{2}}$

Deriving this equation gives

${\displaystyle 2\,x+2\,f(x)\,f'(x)=0\,.}$

For results in dissolving after ${\displaystyle f(x)\neq 0}$ ${\displaystyle f'(x)}$

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

It follows from this that the tangent to the circle at the point with has the slope . ${\displaystyle (x,y)}$ ${\displaystyle y\neq 0}$ ${\displaystyle -{\frac {x}{y}}}$

Individual evidence

1. ^ Gerhard Marinell: Mathematics for social and economic scientists. 7th edition. Oldenbourg Wissenschaftsverlag, Munich 2001, ISBN 3-486-25567-3 , pp. 135-136 ( limited preview in the Google book search).
2. ↑ Jörg Feldvoss, Higher derivatives of implicit functions, 2000: https://www.southalabama.edu/mathstat/personal_pages/feldvoss/impldiff.pdf