(DC3) The Derivative#

By the end of the lesson you will be able to:

  • find the derivative of a function, both at a point and as a function, using the limit definition of the derivative.

Lecture Videos#

Limit Definition of the Derivative#

Introduction#

In the previous learning outcome, we introduced two limits in our discussion of tangent lines and velocity:

Slope of Tangent Line at \(x=a\) is given by \(\displaystyle \lim_{h\to 0} \dfrac{f(a+h)-f(a)}{h}\)

Instantaneous Velocity at \(t=a\) is given by \(\displaystyle \lim_{h\to 0} \dfrac{s(a+h)-s(a)}{h}\)

Notice that these are fundamentally the same limit calculation - the only distinction is the application. In turns out this same limit shows up in other applied fields as well, and so it is important that we give it a special name: the derivative.

Definitions#

Derivative at a Point

\[ f'(a)= \lim_{h\to 0} \dfrac{f(a+h)-f(a)}{h} \]

Here, \(a\) is a specific number.

Derivative as a Function

\[ f'(x)= \lim_{h\to 0} \dfrac{f(x+h)-f(x)}{h} \]

For this definition, we’ve replaced \(a\) with variable \(x\).

Looking at these two definitions, you might be wondering: what’s the difference between \(f'(a)\) and \(f'(x)\)?

  • \(f'(a)\) is a number (e.g. represents the slope of the tagent at a specific point)

  • \(f'(x)\) is a function with variable \(x\) (e.g. its a function that we can evaluate as needed to quickly find the slope of the tangent at different points along the curve.)

We call the process of calculating the derivative, differentiation.

Example 1#

Differentiate the following polynomial using the limit definition of the derivative.

\[ f(x)=x^2-3x+5 \]

How to calculate the derivative using the limit definition

  1. Calculate \(f(x+h)\).

  2. Calculate the difference quotient.

  3. Simplify the difference quotient (until you cancel the \(h\) in the denominator).

  4. Calculate the limit as \(h\to 0\).

Click through the tabs to see the steps of our solution.

In order to start finding the derivative, it is sometimes helpful to first determine and simplify \(f(x+h)\).

\[\begin{split} f(x+h) & = (x+h)^2-3(x+h)+5 \\ &\\ & = x^2+2xh +h^2-3x-3h+5 \\ \end{split}\]

Next, we can use our simplified version of \(f(x+h)\) in our calculation of the difference quotient:

\[\begin{split} \dfrac{f(x+h)-f(x)}{h} & = \dfrac{\big( x^2+2xh+h^2-3x-3h+5\big)-\big(x^2-3x+5\big)}{h}\\ &\\ & = \dfrac{2xh+h^2-3h}{h} \\ & \\ & = \dfrac{h\cdot (2x+h-3)}{h} \\ & \\ & = 2x+h-3 \\ \end{split}\]

With the difference quotient simplified as much as possible, we are finally ready to actually calculate the limit:

\[\begin{split} f'(x) & = \lim_{h\to 0} \dfrac{f(x+h)-f(x)}{h} \\ & \\ & = \lim_{h\to 0} (2x+h-3) \\ & \\ & = 2x +0-3 \\ & \\ &= 2x-3\\ \end{split}\]

The reason that we can use direction evaluation at \(h=0\) when finally calculating this limit, is because the expression \((2x+h+3)\) can be viewed as a polynomial in terms of variable \(h\).

We have successfully differentiated function \(f\) using the limit definition of the derivative. Nice!

\[ f'(x) = 2x-3 \]

Higher Order Derivatives#

Second Derivative

\[ f''(x) = \dfrac{d}{dx}\big[f'(x)\big] \]

The second derivative is defined to be the derivative of the first derivative.

Third Derivative

\[ f'''(x) = \dfrac{d}{dx}\big[f''(x)\big] \]

The third derivative is defined to be the derivative of the second derivative.

In this way we can keep defining higher and higher order derivatives:

\[ f(x) \longrightarrow f'(x) \longrightarrow f''(x) \longrightarrow f'''(x)\longrightarrow f^{(4)}(x) \longrightarrow \cdots \]

Derivative Notation#

What are some of the different ways we indicate derivatives and differentiation?

Operators
\[ \dfrac{d}{dx}\bigg[\cdot \bigg] \qquad \dfrac{d^2}{dx^2}\bigg[\cdot \bigg] \]
First Derivative of \(y=f(x)\)

Prime Notation \(\quad y'\qquad f'(x)\)

Leibniz Notation \(\quad \dfrac{dy}{dx} \qquad \dfrac{d}{dx}f(x)\)

Second Derivative of \(y=f(x)\)

Prime Notation \(\quad y''\qquad f''(x)\)

Leibniz Notation \(\quad \dfrac{d^2y}{dx^2} \qquad \dfrac{d^2}{dx^2}f(x)\)

Evaluation at \(x=a\)

We can also use a vertical line to denote evaluation:

\[ y\biggr\rvert_{x=a} = f(a) \quad\qquad \dfrac{dy}{dx} \biggr\rvert_{x=a} = f'(a) \quad\qquad \dfrac{d^2y}{dx^2} \biggr\rvert_{x=a} = f''(a) \]

Example 2#

Find the derivative of the given function using the limit definition of derivative.

\[ f(x)=x^3-4x \]

Click through the tabs to see the steps of our solution.

We start this derivative calculation by first determining \(f(x+h)\).

\[\begin{split} f(x+h) & = (x+h)^3-4(x+h)\\ $ = (x^3+3x^2h+3xh^2+h^3) - 4x-4h \\ \end{split}\]

Next step is to calculate the difference quotient:

\[\begin{split} f'(x) & = \lim_{h\to 0} \dfrac{f(x+h)-f(x)}{h} \\ &\\ & = \lim_{h\to 0} \dfrac{\big( x^3+3x^2h+3xh^2+h^3 - 4x-4h\big)-\big(x^3-4x\big)}{h}\\ &\\ & = \lim_{h\to 0} \dfrac{3x^2h+3xh^2+h^3 -4h}{h} \\ \end{split}\]

The next goal is to get rid of the \(h\) in the denominator by canceling it. This means we should try to factor out any common terms in the numerator … and we see that all of the terms up there have an \(h\). So we can factor this out and then cancel.

\[\begin{split} f'(x) & = \lim_{h\to 0} \dfrac{3x^2h+3xh^2+h^3 -4h}{h} \\ & \\ & = \lim_{h\to 0}\dfrac{h\cdot (3x^2+3xh+h^2 -4)}{h} \\ & \\ & = \lim_{h\to 0}\big( 3x^2+3xh+h^2 -4 \big) \\ \end{split}\]

And then here, we have a polynomial in terms of \(h\). This means we can use direct evaluation to calculate this limit.

\[\begin{split} & = 3x^2+3x(0)+(0)^2-4 \\ & = 3x^2-4 \end{split}\]

We have successfully differentiated a power function of degree \(3\) and found:

\[ f'(x) = 3x^2-4 \]

Example 3#

Find the derivative of the given function using the limit definition of derivative.

\[ f(x)=\sqrt{x} \]

Click through the tabs to see the steps of our solution.

We start this derivative calculation by first determining \(f(x+h)\).

\[ f(x+h) = \sqrt{x+h} \]

Next step is to calculate the difference quotient:

\[\begin{split} f'(x) & = \lim_{h\to 0} \dfrac{f(x+h)-f(x)}{h} \\ &\\ & = \lim_{h\to 0} \dfrac{\sqrt{x+h}-\sqrt{x}}{h}\\ \end{split}\]

At this point, we might notice that this is an indeterminate form, so we need to use our usual algebraic trick of rationalizing the numerator with the conjugate.

\[\begin{split} & = \lim_{h\to 0} \dfrac{\sqrt{x+h}-\sqrt{x}}{h}\cdot \dfrac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}} \\ & \\ & = \lim_{h\to 0} \dfrac{(x+h)-(x)}{h(\sqrt{x+h}+\sqrt{x})} \\ & \\ & = \lim_{h\to 0} \dfrac{h}{h(\sqrt{x+h}+\sqrt{x})} \\ & \\ & = \lim_{h\to 0}\dfrac{1}{\sqrt{x+h}+\sqrt{x}} \\ \end{split}\]

Once the \(h\)-term in the denominator is canceled, we are most likely ready for the limit.

\[ = \lim_{h\to 0}\dfrac{1}{\sqrt{x+h}+\sqrt{x}} = \dfrac{1}{\sqrt{x+0}+\sqrt{x}} = \dfrac{1}{2\sqrt{x}} \]

Notice that we are able to plug-in \(h=0\) here in order to evaluate this limit. The reason for this, is that \(\tfrac{1}{\sqrt{x+h}+\sqrt{x}}\) is a continuous function. (Square root functions are continuous and no division by \(0\) here.)

We now have our derivative:

\[ f'(x) = \dfrac{1}{2\sqrt{x}} \]

Example 4#

Find the derivative of the given function using the limit definition of derivative.

\[ f(x)=\dfrac{1-x}{2+x} \]

Click through the tabs to see the steps of our solution.

We start this derivative calculation by first determining \(f(x+h)\).

\[ f(x+h) = \dfrac{1-(x+h)}{2+(x+h)} \]

Next step is to calculate the difference quotient:

\[\begin{split} f'(x) & = \lim_{h\to 0} \dfrac{f(x+h)-f(x)}{h} \\ & \\ & = \lim_{h\to 0} \dfrac{\tfrac{1-x-h}{2+x+h}-\tfrac{1-x}{2+x}}{h}\\ \end{split}\]

In order to help us see what’s going on with the numerator, let’s rewrite the division by \(h\) as multiplication by its reciprocal \(1/h\) instead.

\[\begin{split} & = \lim_{h\to 0} \dfrac{1}{h}\cdot\left[\dfrac{1-x-h}{2+x+h}-\dfrac{1-x}{2+x}\right]\\ \end{split}\]

Next we are going to work on combining the two fractions in the brackets by getting a common denominator and doing the multiplication in the numerator.

\[\begin{split} & = \lim_{h\to 0} \dfrac{1}{h}\cdot\left[\dfrac{(1-x-h)(2+x)}{(2+x+h)(2+x)}-\dfrac{(2+x+h)(1-x)}{(2+x+h)(2+x)}\right]\\ & \\ & = \lim_{h\to 0} \dfrac{1}{h}\cdot\left[\dfrac{2-2x-2h+x-x^2-xh}{(2+x+h)(2+x)}-\dfrac{(2+x+h-2x-x^2-xh)}{(2+x+h)(2+x)}\right]\\ & \\ & = \lim_{h\to 0} \dfrac{1}{h}\cdot\left[\dfrac{(2-2x-2h+x-x^2-xh)-(2+x+h-2x-x^2-xh)}{(2+x+h)(2+x)}\right]\\ & \\ \end{split}\]

With the fractions combined, we can work on doing the subtraction (which will greatly simplify things).

\[\begin{split} & = \lim_{h\to 0} \dfrac{1}{h}\cdot\left[\dfrac{-2h-h}{(2+x+h)(2+x)}\right]\\ & \\ & = \lim_{h\to 0} \dfrac{1}{h}\cdot\left[\dfrac{-3h}{(2+x+h)(2+x)}\right]\\ & \\ & = \lim_{h\to 0} \dfrac{-3}{(2+x+h)(2+x)}\\ \end{split}\]

Once the \(h\)-term in the denominator is canceled, we are most likely ready for the limit.

\[\begin{split} = \lim_{h\to 0} \dfrac{-3}{(2+x+h)(2+x)} &= \dfrac{-3}{(2+x+0)(2+x)} \\ &\\ &= \dfrac{-3}{(2+x)(2+x)} = \dfrac{-3}{(2+x)^2} \end{split}\]

Notice that we are able to use direct evaluation here and plug in \(h=0\) (this is rational function in terms of \(h\)).

We have successfully differentiated a rational function and found:

\[ f'(x) = \dfrac{-3}{(2+x)^2} \]