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Inga [223]
3 years ago
10

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.
Mathematics
1 answer:
stepan [7]3 years ago
7 0

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:

<u>Partial differentiation</u>:-

  • 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

<u>Implicit function</u>:-

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

<u>Implicit differentiation process</u>

  • 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)}

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