Question

question 3
if ( 2x^2 + 3x + xy = 2 ) and ( y(2) = -6 ), find ( y(2) ) by implicit differentiation.
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question 4
find the slope of the tangent line to the curve
( x^3 + 3xy - y^3 = 128 )
at the point ( (-4, -4) ).
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Explanation:

Question 3

Step1: Differentiate both sides implicitly

$\frac{d}{dx}(2x^2 + 3x + xy) = \frac{d}{dx}(2)$
$4x + 3 + y + x\frac{dy}{dx} = 0$

Step2: Isolate $\frac{dy}{dx}$

$x\frac{dy}{dx} = -4x - 3 - y$
$\frac{dy}{dx} = \frac{-4x - 3 - y}{x}$

Step3: Substitute $x=2, y=-6$

$\frac{dy}{dx}\bigg|_{(2,-6)} = \frac{-4(2) - 3 - (-6)}{2}$
$=\frac{-8 - 3 + 6}{2} = \frac{-5}{2}$

Step1: Differentiate both sides implicitly

$\frac{d}{dx}(x^3 + 3xy - y^3) = \frac{d}{dx}(128)$
$3x^2 + 3y + 3x\frac{dy}{dx} - 3y^2\frac{dy}{dx} = 0$

Step2: Simplify and isolate $\frac{dy}{dx}$

Divide by 3: $x^2 + y + x\frac{dy}{dx} - y^2\frac{dy}{dx} = 0$
$\frac{dy}{dx}(x - y^2) = -x^2 - y$
$\frac{dy}{dx} = \frac{-x^2 - y}{x - y^2}$

Step3: Substitute $x=-4, y=-4$

$\frac{dy}{dx}\bigg|_{(-4,-4)} = \frac{-(-4)^2 - (-4)}{-4 - (-4)^2}$
$=\frac{-16 + 4}{-4 - 16} = \frac{-12}{-20} = \frac{3}{5}$

Answer:

$\frac{-5}{2}$

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Question 4