0

We say that derivative of function is the instantaneous rate of change. Then what is the instantaneous rate of change for the straight line? For example the straight line x=2 has derivative 0 and y=2x+3 has derivative equal to 2.... I can't understand this sense of rate of change...

M Mushtaq
  • 11
  • 1
    No, it's the line $y=2$ that has derivative $0$ (with respect to $x$). The line $x=2$ does not define $y$ as a function of $x$ and the derivative doesn't exist. – saulspatz Oct 01 '19 at 10:50
  • Is there any function such that you do understand the "rate of change" of that function? If so, it might help if you would explain how you understand rate of change for that function. If not, it might help if you more carefully explain what you have learned about rate of change so that someone can explain it. – David K Oct 01 '19 at 11:05

2 Answers2

1

It's not $x=2$.

It's $y=2$ which has derivative $0$ when viewed as a function of $x$.
This means the rate of change is $0$ i.e. if you change/increase x by $1$ => $y$ does not change at all.

Then on the other hand $y=2x+3$ has derivative $2$ when viewed as a function of $x$.
This means: change/increase $x$ by 1 => $y$ changes by 2 so the rate of change is $2$.

When viewed this way, it seems quite intuitive, right?

peter.petrov
  • 12,568
0

Let the line be $y=mx+b$, the derivative of the line is: $$y^{'}=\lim_{h\rightarrow 0}\dfrac{m\left(x+h\right)+b-\left(mx+b\right)}{h}=\lim_{h\rightarrow 0}\dfrac{mh}{h}=m$$The derivative (rate of change) is the gradient of the line.

MafPrivate
  • 4,033