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

Answer:

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

$\frac{dy}{dx} =$$\frac{-δ f/ δ x }{δ f/δ y }$

Example:- $\frac{dy}{dx} = \frac{-(y+2x)}{(x+2y)}$

Step-by-step explanation:

Partial differentiation:-

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

$\lim_{x \to 0} \frac{f(x+dx,y)-f(x,y)}{dx}$    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

$\lim_{x \to 0} \frac{f(x,y+dy)-f(x,y)}{dy}$    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

$\frac{dy}{dx} =$$\frac{-δ f/ δ x }{δ f/δ y }$

The total differential co-efficient

d z = δ z/δ x +  $\frac{dy}{dx}$ 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'

$2x + x \frac{dy}{dx} + y (1) + 2y\frac{dy}{dx} = 0$

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

$x \frac{dy}{dx} + 2y\frac{dy}{dx} = - (y+2x)$

Taking common d y/dx

$\frac{dy}{dx} (x+2y) = -(y+2x)$

$\frac{dy}{dx} = \frac{-(y+2x)}{(x+2y)}$