Calculus question

Question

Using implicit differentiation find the equation of tangent to the line √(5x+9y)=2+xy^2+y at point (0,1)

Answer 1

Differentiate on both sides and plug in (0,1) to solve for dy/dx:

√(5x+9y)=2+xy^2+y

(5x+9y)' * 1/(2√(5x+9y)) = y^2 + 2xy(dy/dx) + dy/dx

(5 + 9(dy/dx)) * 1/(2√(5x+9y)) = y^2 + (2xy+1)(dy/dx)

(5 + 9(dy/dx)) * 1/(2√(5(0)+9(1))) = 1^2 + (2(0)(1)+1)(dy/dx)

(5 + 9(dy/dx)) * 1/(2√9) = 1 + dy/dx

(5 + 9(dy/dx)) * 1/6 = 1 + dy/dx

5/6 + (9/6)(dy/dx) = 1 + dy/dx

5/6 + (3/6)(dy/dx) = 1

(3/6)(dy/dx) = 1/6

(1/2)(dy/dx) = 1/6

dy/dx = 1/3

As this is the slope of our tangent line for the function at (0,1), we can use point-slope form to determine the final equation of the tangent line:

y-y₁ = m(x-x₁)

y-1 = (1/3)(x-0)

y-1 = (1/3)x

y = (1/3)x+1

Thus, the equation of the tangent line is y = (1/3)x+1.

I hope this helped!