Answer to Question #113559 in Calculus for Appiah William

Question

Find an equation of the tangent line to the curve y = ln(x + 3) when x = -2.

Answer 1

The equation of the tangent to the curve at the point with coordinates (x₀; y₀) defines the equation

$y - y₀ = y '(x₀) \times (x - x₀)y−y₀=y$

where y '(x₀) is the derivative of the original function at the point of tangency.

Find the derivative of the function

$y'(x) = (ln(x+3))' =\frac{1}{x+3}$

The value of the derivative at x₀ = -2

$y '(-2) = \frac{1}{-2+3}=1$

The coordinates of the point of contact: x₀ = -2; y = ln (1)

We write the equation of the tangent to the curve y = ln (x+3) at x₀ = -2

     $y - ln (1) = 1 \times(x - 3)$

$y = x-3+ ln(1)$

y=x-3, ln1=0

The tangent equation is determined by the equation

$y - y₀ = - (\frac{1}{ y '(x₀))} \times (x - x₀)$

              $y - ln (1) = -1\times (x - 3)$

                        y = -x +3+ ln (1), ln1=0

y=3-x