Question

find dy/dx of x^2 + xy + y^2 = 100 by first principle of derivatives ??

Answer 1

Hope it will help you .

Answer 2

it is given that x² + xy + y² = 100

we have to find dy/dx by first principle of derivatives.

d(x² + xy + y²)/dx = d(100)/dx

⇒d(x²)/dx + d(xy)/dx + d(y²)/dx = 0

as d(xⁿ)/dx = nxⁿ-¹ so, d(x²)/dx = 2x ,

⇒2x + d(xy)/dx + 2y dy/dx = 0

using product rule, d{f(x).g(x)}/dx = f(x) d{g(x)}/dx + g(x) d{f(x)}/dx

⇒2x + x dy/dx + y dx/dx + 2y dy/dx = 0

⇒2x + x dy/dx + y + 2y dy/dx = 0

⇒(2x + y) + (x + 2y)dy/dx = 0

⇒dy/dx = -(2x + y)/(x + 2y)

hence dy/dx of the given equation is -(2x + y)/(x + 2y).

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