Derivatives

# Origin of Differential Calculus

Differential calculus is a theory originated from the solution of two old problems:

In both the problems, continuous curves are involved and the limiting process is used. So the objects of study in the differential calculus are continuos functions. These problems were solved in a certain sense by Sir Isaac Newton and Gottfried Wilhelm Leibniz and in the process, differential calculus was discovered.

## Tangent Line to a Curve

Let $AB$ be a continuos curve given by $y=ƒ(x)$ and let $P$ and $Q$ be any two points in the curve. Let the co-ordinates of $P$ and $Q$ be $(x, y)$ and $(x’, y’)$.

When a point moves along the curve from the point $P$ to the point $Q$, it moves horizontally through the distance $PR$ and vertically through the distance $RQ$. $PR=LM=OM-OL=x’-x$ $RQ=QM-RM=y’-PL=y’-y$

These quantities $x’-x$ and $y’-y$ are called the increments in $x$ and $y$ respectively and are denoted by $\Delta x$ and $\Delta y$ i.e. $\Delta x=x’-x$ and $\Delta y=y’-y$.

Also, $\Delta y=ƒ(x’)-ƒ(x)=ƒ(x+\Delta x)-ƒ(x)$

If we join the points $P$ and $Q$, we get secant $PQ$ which makes an angle $\theta$ with the $\text{X-axis}$, i.e. $\angle QNM=\theta$. So $\angle QPR=\angle QNM=\theta$ and

$\tan\theta=\frac{QR}{PR}=\frac{\Delta y}{\Delta x}$

which is the slope of the secant $PQ$. As $Q$ moves along the curve and approaches $P$, the secant rotates about $P$. The limiting position of the secant, when $Q$ ultimately coincides with $P$, is the tangent at $P$, making the angle $\phi$ with the $\text{X-axis}$. In that situation, $\Delta x$, $\Delta y$ tend to zero. So

$\lim_{\Delta x \to 0} \frac{\Delta x}{\Delta y}=\lim_{\Delta x \to 0} \tan\theta=\tan\phi$

$\text{or, } \lim_{\Delta x \to 0}\frac{\Delta y}{\Delta x}=\lim_{\Delta x \to 0} \frac{ƒ(x+\Delta x)-ƒ(x)}{\Delta x}=\tan\phi$

Therefore, $\lim_{\Delta x\to 0} \frac{\Delta y}{\Delta x}$ or $\lim_{\Delta x\to 0} \frac{ƒ(x+\Delta x)-ƒ(x)}{\Delta x}$ gives the slope of a tangent to the curve given by the function $ƒ$.

## Instantaneous Velocity

Suppose a particle is moving in a straight line. Then the distance covered by the particle increases with time. So the distance $s$ can be considered to be a function $ƒ$ of the time $t$, and $s=ƒ(t)$.

At times $t$ and $t+\Delta t$, suppose the particle is at the points $P$ and $Q$ respectively such that $AP=s$ and $AQ=s+\Delta s$. Then, $PQ=AQ-AP=s+\Delta s-s=\Delta s$

Also, $\Delta s=s+\Delta s-s=ƒ(t+\Delta t)-ƒ(t)$

So the average velocity, $v_{av}$, during the time interval $(t, t+\Delta t)$ is

$v_{av}=\frac{\Delta s}{\Delta t}=\frac{ƒ(t+\Delta t)-ƒ(t)}{\Delta t}$

Now as $\Delta t \to 0$, $Q$ tends to $P$. So the instantaneous velocity $v$ of the particle at $P$ or in time $t$ is the limit to which $v_{av}$ tends as $\Delta t \to 0$, and

$v=\lim_{\Delta t \to 0}\frac{\Delta s}{\Delta t}=\lim_{\Delta t \to 0} \frac{ƒ(t+\Delta t)-ƒ(t)}{\Delta t}$