Functions are set by a ratio of independent variables. In case the equation setting function is not solvable rather variable, then function is considered set implicitly. For differentiation of implicit functions there is a special algorithm.

## Instruction

1. Consider the implicit function set by some equation. At the same time it is impossible to express dependence of y (x) in an explicit form. Lead the equation to a type of F (x, y) =0. To find derivative y' (x) from implicit function, at first differentiate F equation (x, y) =0 in relation to variable x, considering that y is differentiated on x. Use rules of calculation of derivative difficult function.

2. Solve the equation of rather derivative y received after differentiation' (x). The total dependence will also be a derivative of implicitly set function on variable x.

3. Study an example for the best understanding of material. Let function be set in an implicit form as y=cos(x−y). Lead the equation to a type of y−cos(x−y) =0. Differentiate it the equations on variable x, applying rules of differentiation of difficult function. We receive, y '+sin(x−y) × (1−y') =0, i.e. y '+sin(x−y) −y '×sin(x−y) =0. Now solve the received equation concerning y': y '× (1−sin(x−y))=−sin(x−y). As a result it turns out that y' (x) =sin(x−y)÷ (sin(x−y) −1).

4. Find a derivative of implicit function of several variables as follows. Let function z (x1, x2, …, xn) in an implicit form be set by F equation (x1, x2, …, xn, z) =0. Find derivative F' | x1, including x2 variables, …, xn, z constant. Similarly calculate derivative F' | x2, …, F' | xn, F' | z. After that express private derivatives as z' | to x1=−F' | x1÷F' | z, z' | x2=−F' | x2÷F' | z, …, z' | xn=−F' | xn÷F' | z.

5. Review an example. Let function of two unknown z=z be set (x, y) formula 2x²z−2z²+yz²=6x+6z+5. Lead the equation to a type of F (x, y, z) =0: 2x²z−2z²+yz²−6x−6z−5=0. Find derivative F' | x, including y, z constant: F' | x=4xz−6. Similarly, derivative F' | y=z², F' | z=2x²-4z+2yz−6. Then z' | x=−F' | x÷F' | z=(6−4xz)÷ (2x²−4z+2yz−6), and z' | y=−F' | y÷F' | z=−z²÷ (2x²−4z+2yz−6).