In this section, we discuss the geometrical interpretation of the derivative of a function at a point. Consider the graph of a function $$y=f(x)$$ and let $$P(x_{0},y_{0})$$ be a fixed point on it (Figure 1). Because $$P$$ is on the graph of $$f$$, we know $$y_{0}=f(x_{0})$$. Here, we use the subscript 0 to emphasize that $$x_{0}$$ and $$y_{0}$$ are held constant through the discussion. Let $$Q(x_{1},y_{1})$$ be another point on the curve with $$x_{1}=x_{0}+\Delta x$$ and because $$Q$$ is on the graph of $$f$$ $y_{1}=f(x_{1})=f(x_{0}+\Delta x).$ From this, we subtract $$y_{0}=f(x_{0})$$ to obtain. $\Delta y=y_{1}-y_{0}=f(x_{0}+\Delta x_{0})-f(x_{0}).$ Since $$\Delta x=PR$$ and $$\Delta y=RQ$$, \begin{aligned} \frac{\Delta y}{\Delta x} & =\frac{f(x_{0}+\Delta x)-f(x_{0})}{\Delta x}\\ & =\tan\widehat{RPQ}=\tan\phi\\ & =m_{\text{sec }}\\ & =\text{slope of secant line }PQ.\end{aligned}

• $$\Delta x=PR$$ is positive if $$R$$ is to the right of $$P$$ and negative if $R$ is to the left of $$P$$. $$\Delta y=RQ$$ is positive if $$Q$$ is above $$R$$ and is negative if $$Q$$ is below $$R$$.

• Because the line that connects $$P$$ and $$Q$$ cuts the curve, it is called a “secant line.” The use of secant for this situation originates from the Latin secare meaning to cut, and does not refer to the secant function that we have in trigonometry.

The derivative of $$f$$ at $$x=x_{0}$$ is
\begin{align} \lim_{\Delta x\to0}\frac{\Delta y}{\Delta x} & =\lim_{\Delta x\to0}\frac{f(x_{0}+\Delta x)-f(x_{0})}{\Delta x}\\ & =\left.\frac{dy}{dx}\right|_{x_{0}}=f'(x_{0})\\ &=\text{value of the derivative at }P\tag{a} \end{align} But when we let $$\Delta x\to0$$, the point $$Q$$ will move along the curve and approach nearer and nearer to $$P$$, the secant will turn about $$P$$ and approach the tangent as a limiting position (Figure 2), and we have \begin{align} \lim_{\Delta x\to0}\frac{\Delta y}{\Delta x} & =\lim_{Q\to P}\tan\phi=\tan\theta\\ & =\text{slope of the tangent at }P\tag{b}\end{align} Hence from (a) and (b)
$\bbox[#F2F2F2,5px,border:2px solid black]{\left.\frac{dy}{dx}\right|_{x_{0}}=f'(x_{0})=\text{slope of the tangent line }.}$
Therefore:

The value of the derivative at any point of a curve is equal to the slope of the line drawn tangent to the curve at that point.

It was this tangent problem that led Gottfried Wilhelm Leibniz to the discovery of differential calculus.

Example
Find the equation of the tangent to the parabola $$y=x^{2}$$ at the vertex and at the point where $$x=\frac{1}{2}$$.
Solution
Let $$y=f(x)=x^{2}$$. Differentiating $$f$$, we get \begin{aligned} \frac{dy}{dx}=f'(x) & =\lim_{\Delta x\to0}\frac{f(x+\Delta x)-f(x)}{\Delta x}\\ & =\lim_{\Delta x\to0}\frac{(x+\Delta x)^{2}-x^{2}}{\Delta x}\\ & =\lim_{\Delta x\to0}\frac{\cancel{x^{2}}+2x\Delta x+(\Delta x)^{2}\cancel{-x^{2}}}{\Delta x}\\ & =\lim_{\Delta x\to0}\frac{\bcancel{\Delta x}(2x+\Delta x)}{\bcancel{\Delta x}}\\ & =2x.\end{aligned}

\frac{dy}{dx}=f'(x)=2x=\text{slope of tangent line at any point on the curve.}\tag{i}

To find the slope of the tangent at the vertex, substitute $$x=0$$ in (i), giving $\left.\frac{dy}{dx}\right|_{x=0}=f'(0)=0.$ Therefore, the tangent at the vertex has the slope zero; that is, it is parallel to the axis of $$x$$ and in this case, because it has to pass through $$(0,0)$$, coincides with it.

To find the slope of the tangent at the point $$P$$ where $$x=\frac{1}{2}$$ and $$y=\left(\frac{1}{2}\right)^{2}=\frac{1}{4}$$, substitute $$x=1/2$$ in (i), giving $\left.\frac{dy}{dx}\right|_{x=1/2}=f'(1/2)=1;$ that is, the tangent at the point $$P$$ makes an angle of $$45^{\circ}$$ with the axis of $$x$$. Because the tangent lines passes through $$(x,y)=\left(\frac{1}{2},\frac{1}{4}\right)$$, its equation is $y-\frac{1}{4}=1\cdot\left(x-\frac{1}{2}\right),$ or $y=x-\frac{1}{4}.$

See the following figure