Question

How do you find the equation of the line tangent to the graph of `f(x)=xln(2x+3)` at the point (-1,0)?

Answer 1

`y= -2x - 2`.

Explanation:

We need to find the derivative of `f(x)`. However, due to the complexity of the function, it is advisable you differentiate it in parts, as does this solution

Step 1: Differentiate `ln(2x + 3)` using the chain rule

Let `y = lnu` and `u = 2x + 3`. Then `dy/(du) = 1/u` and `(du)/dx = 2`.

`dy/dx = dy/(du) xx (du)/dx`

`dy/dx = 1/u xx 2`

`dy/dx = 1/(2x + 3) xx 2`

`dy/dx = 2/(2x+ 3)`

Step 2: Differentiate the entire function using the product rule

`f'(x) = 1(ln(2x+ 3)) + 2/(2x+ 3) xx x`

`f'(x) = ln(2x + 3) + (2x)/(2x + 3)`

Step 3: Find the slope of the tangent

We know the derivative at any point in the function represents the instantaneous rate of change, or the slope, of the function at that given point.

Thus,

`m_"tangent" = ln(2(-1) + 3) + (2(-1))/(2(-1) + 3)`

`m_"tangent" = ln(1) + -2/1`

`m_"tangent" = 0 - 2`

`m_"tangent" = -2`

Step 4: Deduce the equation of the tangent

Recall that the equation of a linear function can be written as `y - y_1 = m(x - x_1)`. Here `(x_1, y_1) = (-1, 0)` and `m = -2`.

`y - y_1 = m(x- x_1)`

`y - 0 = -2(x- (-1))`

`y = -2x - 2`

Hopefully this helps!