Respuesta :
Differentiate the function/equation with respect to x and solve for the derivative, dy/dx. The value of dy/dx at the given point is the slope of the tangent line to the curve at that point. Then use the point-slope formula to get the equation of the tangent.
1.
[tex]y = \dfrac{x^2}{x+2} \implies \dfrac{dy}{dx} = \dfrac{2x\times(x+2) - x\times1}{(x+2)^2} = \dfrac{x(x+4)}{(x+2)^2}[/tex]
When x = 2, the derivative is
[tex]\dfrac{dy}{dx}\bigg|_{x=2} = \dfrac{2(2+4)}{(2+2)^2} = \dfrac34[/tex]
Then the equation of the tangent line at (2, 1) is
[tex]y - 1 = \dfrac34 (x - 2) \implies \boxed{y = \dfrac{3x}4 - \dfrac12}[/tex]
2.
[tex]x^3 + 2y^2 = 10y \implies 3x^2 + 4y \dfrac{dy}{dx} = 10 \dfrac{dy}{dx} \implies \dfrac{dy}{dx} = \dfrac{3x^2}{10-4y}[/tex]
When x = 2 and y = 1, the derivative is
[tex]\dfrac{dy}{dx}\bigg|_{(x,y)=(2,1)} = \dfrac{3\times2^2}{10-4\times1} = 2[/tex]
Then the tangent at (2, 1) has equation
[tex]y - 1 = 2 (x - 2) \implies \boxed{y = 2x - 3}[/tex]
