What is the Second Step in Feynman's Deduction of the Sound Wave Equation?

[PeSoberbo]

I am studying the sound wave equation deducted by Feynman in his lectures. In section 47-3:

P₀ + P_e = f(d₀ + d_e) = f(d₀) + d_e f'(d₀)

Where f'(d₀) stands for the derivative of f(d) evaluated at d=d₀. Also, d_e is very small.

I do not understand the second step of the equality. Can anyone help me?
Link to the lectures: http://www.feynmanlectures.caltech.edu/I_47.html#Ch47-S1

 

[Mark44]

I am studying the sound wave equation deducted by Feynman in his lectures. In section 47-3:

P₀ + P_e = f(d₀ + d_e) = f(d₀) + d_e f'(d₀)

Where f'(d₀) stands for the derivative of f(d) evaluated at d=d₀. Also, d_e is very small.

I do not understand the second step of the equality. Can anyone help me?
Link to the lectures: http://www.feynmanlectures.caltech.edu/I_47.html#Ch47-S1

The last expression in your equation is only an approximation to the second expression, so '=' should really be $\approx$. The approximation is called a linear approximation that uses a value on the tangent line at (d₀, f(d₀)) instead of the value at (d₀ + d_e, f(d₀ + d_e)).

The underlying concept is this:
$$\frac{f(d_0 + d_e) - f(d_0)}{d_e} \approx f'(d_0)$$
Multiply both sides by d_e and then add f(d₀ to both sides to get the relationship you show.

 

[PeSoberbo]

Thank you Mark!