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Deriving Polynomials Using the Power Rule and Simple Derivative Properties

Business Calculus Basic Derivative Rules Deriving Polynomials Using the Power Rule and Simple Derivative Properties

The Power Rule for Monomials

Let’s start out by seeing if there is a pattern to the derivatives of powers of $x$; that is, maybe we can come up with a rule for deriving functions of the form $f(x)=x^n$ (where $n$ is any positive whole number, simply by looking at the derivatives of $x^1$, $x^2$, $x^3$, $x^4$, etc. See the table below for the final computations of the derivatives of these sorts of functions. Note that the video above goes through how one gets the results for a few of them.

$f(x)=x^n$	$f'(x)=\lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h}$
$x^1$	$f'(x)=1$
$x^2$	$f'(x)=2x$
$x^3$	$f'(x)=3x^2$
$x^4$	$f'(x)=4x^3$
$x^5$	$f'(x)=5x^4$

As you can probably tell, the pattern for computing $f'(x)$ from $f(x)=x^n$ is given by the following rule. It turns out that the rule that can be determined from the pattern you see above also holds when the exponent is not a positive whole number as well. All this is baked into the following rule.

The Power Rule

Let $f(x)=x^n$ where $n$ is any real number (could be positive, negative, fraction, decimal, etc.). Then the derivative of $f$ is given by:
$$f'(x)=n\cdot x^{n-1}$$
or, using Leibniz notation (aka $\frac{d}{dx}$ notation),
$$\frac{d}{dx}f(x)=n\cdot x^{n-1}$$

So, what this rule is saying is “to find the derivative of a function of the form $f(x)=x^n$, move the exponent down to the front of the $x$ (multiplying it by what’s already there, if anything) and subtract $1$ from the exponent (no matter what the exponent $n$ originally was).

As said before, the power rule basically tells us to “move the exponent to the front (multiplying by what is already there, if anything) and mindlessly subtract $1$ from the exponent. This gives us:

$$f'(x)=3.52x^{3.52-1}=3.52x^{2.52}

Deriving Functions with Numbers Out in Front

This was hinted at when describing how to derive powers of $x$ above. The following rule helps us derive constant multiples of functions, or “functions with numbers out in front.”

Deriving a Constant Multiple of a Function

Let $f$ be a function whose derivative exists at all $x$-values in some open interval $I$. Furthermore, let $c$ be a real number. Then, the derivative of $c\cdot f(x)$ is given by
$$(c\cdot f(x))’=c\cdot f'(x)$$
or, using Leibniz notation:
$$\frac{d}{dx} c\cdot f(x)=c\cdot \frac{d}{dx} f(x)$$

What this rule is saying is essentially: “derive the ‘function part,’ and bring the constant coefficient along for the ride, multiplying it by the end-result of your differentiation of the ‘main function part.” This is illustrated in the example below.

Using the idea of the rule above for deriving constant multiples of a function, we will ignore the $2.5$ coefficient out in front while we derive the $x^6$ and multiply that $2.5$ to the result. This gives us

$$\frac{d}{dx}\ 2.5x^6=2.5\cdot(6x^5)=15\cdot x^5$$

Yep. That simple. just bring the coefficient along for the ride and multiply it at the end.

Deriving Polynomials and Linear Combinations of Functions

The following property of derivatives vastly expands the number of functions that we can find the derivative of using the power rule.

Deriving Linear Combinations of Functions

Let $f$ and $g$ be functions whose derivative exists at all $x$-values in some open interval $I$. Furthermore, let $c$ and $d$ be real numbers. Then the derivative of $c\cdot f(x)\pm d\cdot g(x)$ (called a linear combination of $f(x)$ and $g(x)$) is
$$(c\cdot f(x)\pm d\cdot g(x))’=c\cdot f'(x)\pm d\cdot g'(x)$$
or, using Leibniz notation:
$$\frac{d}{dx} (c\cdot f(x)\pm d\cdot g(x)) = c\cdot \frac{d}{dx}f(x)+\pm d\cdot \frac{d}{dx}g(x)$$

Basically, what this rule is saying is that when you have functions that are being added or subtracted, who may also have a number out in front of the functions, you can derive just the “main functions” and bring the constant coefficients along for the ride.

Furthermore, the above rule can be generalized to where several functions are being added or subtracted, with constant coefficients. All this good stuff is exhibited in the example below.

To derive this, we simply derive the exponential expressions $x^2$, $-4x^5$, and $3.5x^3$ using the ideas we’ve developed, and then add all the results at the end. Thus:

$$\begin{align}\frac{d}{dx}\ (x^2-4x^5+3.5x^3) &= 2x^1-4(5x^4)+3.5(3x^2)\\&= 2x-20x^4+10.5x^2\end{align}$$

Finding Derivatives at Specific $x$-Values Using Rules

Deriving Linear Combinations of Functions

Let $f$ be a function whose derivative exists for all $x$ on some interval $I$, and let $x=c$ be an $x$-value inside $I$. Then, to find the derivative of $f$ at $x=c$, simply compute the derivative of $f$ using whatever methods or rules you like, and then plug $x=c$ into the result.

According to the above method, we first find the derivative of $f$ and then plug $2$ in at the end… super easy.

$$\begin{align}f'(x)&=3(4x^3)\\&=12x^3\end{align}$$

Thus, $f'(2)=12(2)^3=96$. Done.

Finding Tangent Lines (is now easier)

The glorious thing about having new, simpler, rules for finding derivatives is that it makes short(er) work of finding tangent lines and doing other derivative-related things. Below is an example of finding the equation of a tangent line using the power rule.

Remember: to find a tangent line, we need two ingredients: a point on the line (specifically, the point where the line touches the graph of $f$) and the slope of the line (which comes from the derivative of $f$).

To find the slope of the tangent line, compute the derivative of $f$ at $x=1$. Using the power rule, we compute the derivative of $f$ to be

$$\begin{align}\frac{d}{dx}\ f(x)&=\frac{d}{dx}\ 3x^2+3x^3\\&=3\cdot(2x)+3\cdot(3x^2)\\&=6x+9x^2\end{align}$$

Then, to find $f'(1)$, just plug $1$ into the derivative above! Super simple.

$$\begin{align} f'(1)&=6(1)+9(1)^2\\&=15\end{align}$$

This is your $m$ (i.e. your slope of the tangent line). Now, to find a point on the tangent line, we are already given the $x$ value of that point, namely $x=1$. Now to find the $y$-value of the point, plug $x=1$ into the original function (not the derivative). Thus,

$$\begin{align}f(1)&=3(1)^2+3(1)^3\\&=6\end{align}$$

So, in summary, our ingredients for the tangent line are: $m=15$, $x_0=1$ and $y_0=6$. Plug this all into point-slope form $(y-y_0)=m(x-x_0)$ as shown below, and then solve for $y$. The algebra is skipped.

$$(y-6)=15(x-1)$$

which gives us equation of the tangent line $y=15x-9$.