Question

please help immediately ​

Answer 1

Answer:    $f'(1)=\dfrac{4}{2y+1}$

Step-by-step explanation:

To find the tangent, you need to find the derivative with respect to x.

Then substitute x = 1 into the derivative.

Given:         0 = 2x² - xy - y²

Derivative:  0 = 4x - y' - 2yy'

Solve for y': 2yy' + y' = 4x

                   y'(2y + 1) = 4x

                   $y'=\dfrac{4x}{2y+1}$

$\text{Substitute x = 1:}\quad y'(1)=\dfrac{4}{2y+1}$