7.5: Derivatives and Graphs

The derivative of a function is just the slope of the tangent line at a given point, so if we have the graph of \(y=f(x)\), we can use it to estimate the graph of \(y=f'(x)\) (even without knowing the explicit definition of \(f\))

Simply draw the tangent line at a few points on \(y=f(x)\) and estimate the slope.

Then plot those values (the slope of the tangent line) as points on the graph of \(y=f'(x)\):

The height of the graph of \(f'(x)\) should match the slope on the graph of \(y=f(x)\)
You can see that \(f(x)\) has a negative slope that becomes less negative as the value of \(x\) gets bigger.
That is why we connect these points so that the values of \(f'(x)\) on the graph are negative but becoming less negative and approaching zero as \(x\) moves toward the right.

Lets try another, this time with some asymptotes involved:

We don't need as many points, just a enough to have some points to plot, and the rest can be done by determining if the slope is getting bigger or smaller as we move around.

In \((-\infty,1)\) the slope starts off as a tiny positive, eventually reaches a slope of \(1\) at \(x=0\), and then becomes extremely large and positive as we approach the vertical asymptote at \(x=1\).

In \((1,\infty)\), the slope starts off extremely positive near the vertical asymptote \(x=1\) and then becomes less positive as we move to the right, with our slope approaching \(0\) as \(x\to \infty\)

Lastly \(g'(1)\) does not exist, because \(g(x)\) was not continuous at \(x=1\)

In general, the easiest places to evaluate or estimate \(f'\) are:
(i) where the tangent line is horizontal, so you know the derivative must equal zero
(ii) where the tangent lines are steepest (negative or positive) so you know the max and min values of \(f'\)

Example \(\PageIndex{1}\)

Sketch the graph of the \(y=f'(x)\) for the function \(y=f(x)\) provided below.

Solution

In \((-\infty,-3)\) the slope begins very large and positive, but approaches \(0\) as \(x\to -3^-\)
In \((-3,1)\) the slope starts at zero, becomes negative, and then returns back to zero.
In \((1,\infty)\) the slope begins at zero and becomes very large and positive as \(x\to +\infty\)

Let \(g(x)=f'(x)\) from the previous example. Since we have its graph now, we are be able to sketch the graph of \(y=g'(x)\):

In \((-\infty,-1)\) the slope begins very large and negative, but approaches \(0\) as \(x\to -1^-\)
In \((-1,\infty)\) the slope begins at zero and becomes very large and positive as \(x\to +\infty\)

In case you missed it, we just sketched the graph of the derivative of \(f'(x)\). This is called the second derivative of \(f(x)\). We typically write this as \(f''(x)\)
( If you took a derivative again, it would be the third derivative: \(f'''(x)\) )

Generally we write \(f^{(n)}(x)\) to represent the \(n^{th}\) derivative of a function, especially when \(n\) is large ( so we don't have to count the number of ' we see next to \(f\) )

If you want to find \(f''(x)\) explicitly, you can use the same formula: \( \displaystyle{ f''(a)=\lim_{x\to a} \frac{f'(x)-f'(a)}{x-a}} \)

Example \(\PageIndex{2}\)

Find the second derivative of \(f(x)=\displaystyle{\frac{x^2+1}{x}}\)

Solution

We start by finding \(f'(a)\):

\(\displaystyle{f'(a)=\lim_{x\to a} \frac{f(x)-f(a)}{x-a}=\lim_{x\to a} \frac{\frac{x^2+1}{x}-\frac{a^2+1}{a}}{x-a} \cdot \textcolor{brown}{\frac{ax}{ax}} =\lim_{x\to a} \frac{(x^2+1)a-(a^2+1)x}{(x-a)ax}} \)

\(\displaystyle{=\lim_{x\to a} \frac{a-x}{(x-a)ax}=\lim_{x\to a} \frac{-(x-a)}{(x-a)ax} =\lim_{x\to a} \frac{-1}{ax}=-\frac{1}{a^2}} \)

Now we repeat to find \(f''(a)\):

\(\displaystyle{f''(a)=\lim_{x\to a} \frac{f'(x)-f'(a)}{x-a}=\lim_{x\to a} \frac{\left(-\frac{1}{x^2}\right)-\left(-\frac{1}{a^2}\right)}{x-a} \cdot \textcolor{brown}{\frac{a^2x^2}{a^2x^2}} =\lim_{x\to a} \frac{-a^2+x^2}{(x-a)a^2x^2}} \)

\(\displaystyle{=\lim_{x\to a} \frac{(x-a)(x+a)}{(x-a)a^2x^2}=\lim_{x\to a} \frac{x+a}{a^2x^2}= \frac{2a)}{a^4}=\frac{2}{a^3}} \)

So our second derivative is the function \(f''(x)=\displaystyle{\frac{2}{x^3}}\)

For a function \(f(x)\), we know that \(f'(a)\) represents the slope of the tangent line at the point \((a,f(a))\). This is often referred to as the instantaneous rate of change of \(f(x)\) at \(x=a\).

The second derivative is therefore the rate of change of the slope of the tangent line, so it determines if the slope is increasing or decreasing at \((a,f(a))\).

Later we will see how the values of \(f''(x)\) effect the shape of the graph of \(f(x)\), but for now we will settle for just being able to find it or sketch it!

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