Question

if 3x^2+4xy+2y^2+x-8=0, then the value of dy/dx at (-1,3) is​

Answer 1

this is the correct answer

Answer 2

Answer:

The correct answer of this question is $y^{1} = \frac{- (6x + 4 y +1)}{4(x+y)}$

Step-by-step explanation:

Given - $3x^2+4xy+2y^2+x-8=0,$

To Find - Write the value of dy/dx at (-1,3)

$3x^2+4xy+2y^2+x-8=0,$

$6x + 4 ( y + xy^{1} ) + 4y . y^{1} + 1 = 0$$y^{1} = \frac{- (6x + 4 y +1)}{4(x+y)}$

The change in y with respect to x is denoted by the symbol dy/dx. If y(x) is a function, the derivative is denoted by the letter y' (x). Differentiation is the process of determining a function's derivative. The derivative of a function is shown as the slope of a function.

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