CALCULUS QUESTIONS: Implicit differentiation?

Question

1.)
Consider
x^2+y^2=20.
a) Find the equation of tangent at (2, 4).
b.) Find the equation of normal at (2, 4).

2.)
Find d^2y/dx^2 for x^2-y^2=16

Can someone please show me how to do this step by step?
Thank you!

Answer 1

x² + y² = 20.

Apply 'derivative with respect to x' each side.

d/dx(x² + y²) = (d/dx)(20)

Power Rule of Derivative is (d/dx)(xⁿ) = nx^n-1

2x + 2y(dy/dx) = 0

Subtract 2x from each side.

2y(dy/dx) = - 2x

Divide each side by 2y.

(dy/dx) = - 2x/2y = -x/y

a). slope (dy/dx) = -x/y = -2/4 = -1/2

'point-slope' form is (y - y₁) = m(x - x₁)

Substitute (x₁, y₁) = (2, 4) and slope m = -1/2 in the 'point-slope' form.

(y - 4) = (-1/2)(x - 2)

y - 4 = (-x/2) + 1

Add 4 to each side.

y = (-x/2) + 5

The tangent line equation is y = (-x/2) + 5.

Answer 2

Note: a line tangent to a circle is perpendicular to the radius to the tangency.

x² + y² = 20.

Apply 'derivative with respect to x' each side.

d/dx(x² + y²) = (d/dx)(20)

Power Rule of Derivative is (d/dx)(xⁿ) = nx^n-1

2x + 2y(dy/dx) = 0

Subtract 2x from each side.

2y(dy/dx) = - 2x

Divide each side by 2y.

(dy/dx) = - 2x/2y = -x/y

Slope (dy/dx) = -x/y = -2/4 = -1/2

The perpendicular slope is (-1/m) = (-1/2)

The normula slope is 2.

Point–slope form: (y - y₁)= m(x - x₁).

Given the point is (2, 4)

(y - 4) = 2(x - 2)

y - 4 = 2x - 4

Add 4 to each side.

y = 2x.

Therefore the normal equation is y = 2x.

Answer 3

2). x² - y² = 16

Apply 'derivative with respect to x' each side.

(d/dx)(x² - y²) = (d/dx)(16)

Power Rule of Derivative is (d/dx)(xⁿ) = nx^n-1

2x - 2y(dy/dx) = 0

Subtract 2x from each side.

- 2y(dy/dx) = - 2x

Divide each side by -2y.

(dy/dx) = (-2x) / (-2y) = x/y

Again apply 'derivative with respect to x' each side.

(d²y/dx²) = (d/dx)(x/y)

The Quotient Rule.  (d/dx)(u/v) = (vu'-uv') / (v²)

(d²y/dx²) = [y(1)-xy') / (y²)

(d²y / dx²) = [y - xy' ] / y².