Question

Suppose that x and y are both differentiable functions of t and are related by the given equation. Use implicit differentiation with respect to t to determine StartFraction dy Over dt EndFraction dy dt in terms of​ x, y, and StartFraction dx Over dt EndFraction dx dt.

Answer 1

Let z = f(x, y) where f(x, y) =0 then the implicit function is

`(dy)/(dx) =`
`(-δ f/ δ x )/(δ f/δ y )`

Example:-
`(dy)/(dx) = (-(y+2x))/((x+2y))`

Explanation:

Partial differentiation:-

• Let Z = f(x ,y) be a function of two variables x and y. Then

`\lim_(x \to 0) (f(x+dx,y)-f(x,y))/(dx)` Exists , is said to be partial derivative or Partial differentiational co-efficient of Z or f(x, y)with respective to x.

It is denoted by δ z / δ x or δ f / δ x

• Let Z = f(x ,y) be a function of two variables x and y. Then

`\lim_(x \to 0) (f(x,y+dy)-f(x,y))/(dy)` Exists , is said to be partial derivative or Partial differentiational co-efficient of Z or f(x, y)with respective to y

It is denoted by δ z / δ y or δ f / δ y

Implicit function:-

Let z = f(x, y) where f(x, y) =0 then the implicit function is

`(dy)/(dx) =`
`(-δ f/ δ x )/(δ f/δ y )`

The total differential co-efficient

d z = δ z/δ x +
`(dy)/(dx)` δ z/δ y

Implicit differentiation process

• differentiate both sides of the equation with respective to 'x'

• move all d y/dx terms to the left side, and all other terms to the right side

• factor out d y / dx from the left side
• Solve for d y/dx , by dividing

Example :
`x^2 + x y +y^2 =1`

solution:-

differentiate both sides of the equation with respective to 'x'

`2x + x (dy)/(dx) + y (1) + 2y(dy)/(dx) = 0`

move all d y/dx terms to the left side, and all other terms to the right side

`x (dy)/(dx) + 2y(dy)/(dx) = - (y+2x)`

Taking common d y/dx

`(dy)/(dx) (x+2y) = -(y+2x)`

`(dy)/(dx) = (-(y+2x))/((x+2y))`