Ucale

# Derivative as a rate measurer

If a quantity y depends on and varies with a quantity x, then the rate of change of y with respect to x is dy/dx.

Let y=f(x) be a function of x. Let $\displaystyle \Delta y$ be the change in y corresponding to a small change $\displaystyle \Delta x$ . Then $\displaystyle \frac { \Delta y }{ \Delta x }$ represents the change in y due to a unit change in x. In other words, $\displaystyle \frac { \Delta y }{ \Delta x }$ represents the average rate of change of y w.r.t. x as x changes from x to $\displaystyle x+\Delta x$

As $\displaystyle \Delta x\longrightarrow 0$ the limiting value of this average rate of change of  y with respect to x in interval $\displaystyle \left[ x,x+\Delta x \right]$ becomes the instantaneous rate of change of y w.r.t. x

Thus $\displaystyle \lim _{ \Delta x\longrightarrow 0 }{ \frac { \Delta y }{ \Delta x } }$ = instantaneous rate of change of y w.r.t. x $\displaystyle \Rightarrow \frac { dy }{ dx }$ = rate of change of y w.r.t. x $\displaystyle \left[ \because \lim _{ \Delta x\longrightarrow 0 }{ \frac { \Delta y }{ \Delta x } } =\frac { dy }{ dx } \right]$

The word ”instantaneous” is often dropped .

Hence $\displaystyle \frac { dy }{ dx }$ represents the rate of change of y w.r.t. x for a definite value of x

### Example

Find the rate of change of the area of a circle with respect to its radius

Solution

Let A be the area of the circle. Then $\displaystyle A=\pi { r }^{ 2 }\\ \Rightarrow \frac { dA }{ dr } =2\pi r$

Thus, the rate of change  of the area of the circle w.r.t. its radius r is $\displaystyle 2\pi r$

April 18, 2019
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