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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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