# How is the maximum of a derivative found?

## Question:

How is the maximum of a derivative found?

## Maximum of a Derivative

The derivative is defined as the slope of the tangent line at a specific point. The tangent line represents the instantaneous rate of change of the function at a specific point.

The derivative can also help us find any local extrema if there are any. The Second Derivative Test is used to identify each critical point as either a local maximum, a local minimum, or neither (a saddle point).

In order to find a local maximum/minimum of a function, we need to use the second derivative test.

But, in this case, we will set the first derivative as the function we want to maximize.

So we want to maximize {eq}f'(x) {/eq}

Steps:

1. Calculate {eq}f''(x) {/eq} and equate it to zero.

2. The solutions to {eq}f''(x)=0 {/eq} are the critical points.

3. Calculate {eq}f'''(x) {/eq}

4. Plug-in the critical point/s to {eq}f'''(x) {/eq}

5. If {eq}f'''(x)=c {/eq}

then you have three position results, based upon the value of {eq}c{/eq}.

If {eq}c{/eq} is positive, then (c,f(c)) is a local minimum of the derivative.

If {eq}c{/eq} is negative, then (c,f(c)) is a local maximum of the derivative.

If {eq}c{/eq} is 0, then (c,f(c)) is a saddle point of the derivative.

#### Example: Find the maximum of {eq}f'(x)=2x^3+2x^2 {/eq}.

Steps using the second derivative test:

1. $$f''(x)= 6x^2+4x$$

then, $$6x^2+4x=0$$

2. The critical points are

$$6x^2+4x=0$$

$$2x(3x+2)=0$$

$$x=0, x=-\frac{2}{3}$$

3. $$f'''(x)= 12x+4$$

4.

$$\text{at }x=0\\ f'''(0)=4$$

$$\text{at }x=-\frac{2}{3}\\ f'''(-\frac{2}{3})=-\frac{12}{3}$$

Thus, the local maximum of the derivative is at (-{eq}\frac{2}{3} {/eq}, f(-{eq}\frac{2}{3} {/eq}).