Introduction to Derivatives

When we have a formula for a function, we can determine the slope of the tangent line at a point $(x, f(x))$ by calculating the slope of the secant line through the points $(x, f(x))$ and $(x+h, f(x+h))$, $\mathrm{m}_{\mathrm{sec}}=\frac{\mathrm{f}(\mathrm{x}+\mathrm{h})-\mathrm{f}(\mathrm{x})}{(\mathrm{x}+\mathrm{h})-(\mathrm{x})}$, and then taking the limit of $\mathrm{m}_{\mathrm{sec}}$ as $h$ approaches $0$ (Fig. 8) :

$\mathrm{m}_{\mathrm{tan}}=\lim\limits_{h \rightarrow 0} \mathrm{~m}_{\mathrm{Sec}}=\lim\limits_{h \rightarrow 0} \frac{f(x+h)-f(x)}{(x+h)-(x)}$

Example 2: Find the slope of the line tangent to the graph of $y = f(x) = x^2$ at the point $(2,4)$. (Fig. 9). 

Solution: In this example $x=2$, so $x+h=2+h$ and $f(x+h)=f(2+h)=(2+h)^{2}$

The slope of the tangent line at $(2,4)$ is

$\begin{aligned}&\mathrm{m}_{\tan }=\lim\limits_{h \rightarrow 0} \mathrm{~m}_{\mathrm{sec}}=\lim\limits_{h \rightarrow 0} \frac{f(2+h)-f(2)}{(2+h)-(2)} \\&=\lim\limits_{h \rightarrow 0} \frac{(2+h)^{2}-(2)^{2}}{(2+h)-(2)}=\lim _{h \rightarrow 0} \frac{4+4 h+h^{2}-4}{h} \\&=\lim\limits_{h \rightarrow 0} \frac{4 h+h^{2}}{h}=\lim _{h \rightarrow 0} \frac{h(4+h)}{h}=\lim\limits_{h \rightarrow 0}(4+h)=4\end{aligned}$

The tangent line to the graph of $\mathrm{y}=\mathrm{x}^{2}$ at the point $(2,4)$ has slope 4.

We can use the point–slope formula for a line to find the equation of the tangent line: 

Practice 4: Use the method of Example 2 to show that the slope of the line tangent to the graph of $\mathrm{y}=\mathrm{f}(\mathrm{x})=\mathrm{x}^{2}$ at the point $(1,1)$ is $\mathrm{m}_{\tan }=2$. Also find the values of $\mathrm{m}_{\mathrm{tan}}$ at $(0,0)$ and $(-1,1)$.

It is possible to find the slopes of the tangent lines one point at a time, but that is not very efficient. You should have noticed in the Practice 4 that the algebra for each point was very similar, so let's do all the work once for an arbitrary point $(x, f(x))=\left(x, x^{2}\right)$ and then use the general result for our particular problems. The slope of the line tangent to the graph of $y=f(x)=x^{2}$ at the arbitrary point $\left(x, x^{2}\right)$ is

$\begin{aligned}\mathrm{m}_{\tan } &=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{(x+h)-(x)}=\lim _{h \rightarrow 0} \frac{(x+h)^{2}-(x)^{2}}{(x+h)-(x)}=\lim _{h \rightarrow 0} \frac{x^{2}+2 x h+h^{2}-x^{2}}{h} \\&=\lim _{h \rightarrow 0} \frac{2 x h+h^{2}}{h}=\lim _{h \rightarrow 0} \frac{h(2 x+h)}{h}=\lim _{h \rightarrow 0}(2 x+h)=\mathbf{2} \mathbf{x}\end{aligned}$

The slope of the line tangent to the graph of $y=f(x)=x^{2}$ at the point $\left(x, x^{2}\right)$ is $m_{\tan }=2 x$. We can use this general result at any value of $x$ without going through all of the calculations again. The slope of the line tangent to $y=f(x)=x^{2}$ at the point (4,16) is $m_{\tan }=2(4)=8$ and the slope at $\left(\pi, \pi^{2}\right)$ is $m_{\tan }=2(\pi)=2 \pi$. The value of $x$ determines where we are on the curve ( at $\mathrm{y}=\mathrm{x}^{2}$ ) as well as the slope of the tangent line, $\mathrm{m}_{\tan }=2 \mathrm{x}$, at that point. The slope $m_{\tan }=2 \mathrm{x}$ is a function of $\mathrm{x}$ and is called the derivative of $\mathbf{y}=\mathbf{x}^{2}$.

Simply knowing that the slope of the line tangent to the graph of $y=x^{2}$ is $m_{\tan }=2 x$ at a point $(x, y)$ can help us quickly find the equation of the line tangent to the graph of $\mathrm{y}=\mathrm{x}^{2}$ at any point and answer a number of difficult sounding questions.

Example 3: Find the equations of the lines tangent to $\mathrm{y}=\mathrm{x}^{2}$ at (3,9) and $\left(\mathrm{p}, \mathrm{p}^{2}\right)$.

Solution: At $(3,9)$, the slope of the tangent line is $2 x=2(3)=6$, and the equation of the line is $y-y_{0}=m\left(x-x_{0}\right)$ so $y-9=6(x-3)$ and $y=6 x-9$.

At $\left(p, p^{2}\right)$, the slope of the tangent line is $2 x=2(p)=2 p$, and the equation of the line is $y-y_{o}=m\left(x-x_{o}\right) \quad$ so $y-p^{2}=2 p(x-p)$ and $y=2 p x-p^{2}$.

Example 4: A rocket has been programmed to follow the path $y = x^2$  in space (from left to right along the curve), but an emergency has arisen and the crew must return to their base which is located at coordinates $(3,5)$. At what point on the path $y = x^2$ should the captain turn off the engines so the ship will coast along the tangent to the curve to return to the base? (Fig. 10)

Solution: You might spend a few minutes trying to solve this problem without using the relation $\mathrm{m}_{\tan }=2x$, but the problem is much easier if we do use that result. 

Lets assume that the captain turns off the engine at the point $(p,q)$ on the curve $y = x^2$, and then try to determine what values $p$ and q must have so that the resulting tangent line to the curve will go through the point $(3,5)$. The point $(p,q)$ is on the curve $y = x^2$ , so  $q = p^2$, and the equation of the tangent line, found in Example 3, is $y = 2px – p^2$.

To find the value of $p$ so that the tangent line will go through the point $(3,5)$, we can substitute the values $x = 3$ and $y = 5$ into the equation of the tangent line and solve for $p$ : 

$y = 2px – p^2$  so $5 = 2p(3) – p$  and  $p – 6p + 5 = 0$. 

The only solutions of $p^2 – 6p + 5 = (p – 1)(p – 5) = 0$ are $p = 1$ and $p = 5$, so the only possible points are $(1,1)$ and $(5,25)$. You can verify that the tangent lines to $y = x^2$ at $(1,1)$ and $(5,25)$ go through the base at the point $(3,5)$ (Fig. 11). Since the ship is moving from left to right along the curve, the captain should turn off the engines at the point $(1,1)$. Why not at $(5,25)$? 

Practice 5: Verify that if the rocket engines in Example 4 are shut off at $(2,4)$, then the rocket will go through the point $(3,8)$.