Why do we take a derivative?

Solution 1:

If $f(x)$ is a (real) function (of a real variable), then its derivative $f'(a)$ at a point $a$ measures the sensitivity of $f$ to small changes in $x$ around $a$. For example, suppose $x$ is "the height of the sun in the sky" and $f(x)$ is "the length of the shadow cast by this tree." Then $f'(a)$ is a measure of how sensitive the length of the shadow is to a small change in the height of the sun if the height is around $a$. Experimentally, $f'(a)$ will be small if the sun is high in the sky ($a$ is large) while $f'(a)$ will get larger as the sun gets lower in the sky ($a$ is small). The derivative gives us a way to quantify this observation.

I decided not to use the term "rate of change" because I think some students have heard this term so many times that they have become immune to it, so perhaps the above will make more sense.

Solution 2:

I don't know what kind of explanation you have gotten before, if any, so I think that I will start with one of the usuals. Let me first introduce to you the difference quotient: $$\frac{f(x+h)-f(x)}{h}$$ It has a very nice geometrical interpretation.

In this picture, the curve is the graph of a function. I hope that you can understand from my picture that the length of the horizontal red line is $h$, while $f(x+h)-f(x)$ is the length of the vertical red line. So $f(t)$ increases by an amount of $f(x+h)-f(x)$ when $t$ increases by an amount of $h$ from $x$. This means that the average increase on the vertical axis per unit of the horizontal axis is given by the difference quotient $$\frac{f(x+h)-f(x)}{h}.$$ So the difference quotient is the average increase of a $f(t)$ on the interval $[x,x+h]$. Another way to put it is that the difference is the slope of the line passing through the points $(x,f(x))$ and $(x+h,f(x+h))$, like this:

So the green line $g(t)$ can be written as $$g(t) = \frac{f(x+h)-f(x)}{h} t + b$$ where $b=g(0)$ (the point where $g$ crosses the vertical axis).

When we take the derivative of $f$ at $x$ we find the limit of the difference quotient as $h$ goes to $0$, or $$f'(x) = \lim_{h\to 0} \frac{f(x+h)-f(x)}{h}.$$ This is the definition of the derivative. Imagine that $h$ gets smaller and smaller. Then the point $(x+h,f(x+h)-f(x))$ will get closer and closer to the point $(x, f(x))$, and the green line will get closer and closer to the tangent line of $f$ at $x$.

So when $h \to 0$, the difference quotient becomes the slope of the tangentline, and now we have an interpretation of the derivative:

The derivative $f'(x)$ is the is the slope of the tangent line of $f$ at $x$.

Let's finish off by trying to find the derivative of the function $f(x)=x^2$ at the point $x=1$ only with the help of the limit stated above. $$f'(x) = \lim_{h\to 0} \frac{f(x+h)-f(x)}{h} = \lim_{h\to 0} \frac{(x+h)^2-x^2}{h}$$ $$= \lim_{h\to 0} \frac{x^2+2xh+h^2-x^2}{h} = \lim_{h\to 0} \frac{2xh+h^2}{h}$$ $$= \lim_{h\to 0} 2x+h = \underline{2x}$$ When we plug in $x=1$ we get that $f'(1) = 2$. Hope this helps!

Solution 3:

I would recommend reading this: https://web.archive.org/web/20200218154616/http://www.ugrad.math.ubc.ca/coursedoc/math100/notes/derivs/deriv5.html

It's a pretty good, non technical explanation of the derivative and why one might care about it.