MA005: Calculus I

A Useful Formula: $D( x^n )$

Functions which include powers of $x$ are very common (every polynomial is a sum of terms which include powers of $x$), and, fortunately, it is easy to calculate the derivatives of such powers. The "pattern" emerging from the first few examples in this section is, in fact, true for all powers of $x$. We will only state and prove the "pattern" here for positive integer powers of $x$, but it is also true for other powers as we will prove later. 

Theorem: If $n$ is a positive integer, then $D( x^n ) = n.x^{n–1}$. 

This theorem is an example of the power of generality and proof in mathematics. Rather than resorting to the definition when we encounter a new power of $x$ (imagine using the definition to calculate the derivative of $x^{307}$ ), we can justify the pattern for all positive integer exponents $n$, and then simply apply the result for whatever exponent we have. We know, from the first examples in this section, that the theorem is true for $n= 1, 2$ and $3$, but no number of examples would guarantee that the pattern is true for all exponents. We need a proof that what we think is true really is true. 

Proof of the theorem: Since $f(x) = x^n$, then $f(x+h) = (x+h)^n$, and in order to simplify $f(x+h) – f(x) = (x+h)^n – x^n$, we will need to expand $(x+h)^n$. However, we really only need to know the first two terms of the expansion and to know that all of the other terms of the expansion contain a power of $h$ of at least 2. The Binomial Theorem from algebra says (for $n > 3$) that $(x+h)^n = x^n + n.^{xn–1}h + a.x^{n–2}h2 + b.x^{n–3}h3 +... + h^n$ where $a$ and $b$ represent numerical coefficients. 

(Expand $(x+h)^n$ for at least a few different values of n to convince yourself of this result.) 

Then $\mathbf{D}(\mathrm{f}(\mathrm{x})) \equiv \lim\limits_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}=\lim\limits_{h \rightarrow 0} \frac{(x+h)^{n}-x^{n}}{h} \quad$ then expand to get 

$=\lim\limits_{h \rightarrow 0} \frac{\left\{x^{n}+n \cdot x^{n-1} h+a \cdot x^{n-2} h^{2}+b \cdot x^{n-3} h^{3}+\ldots+h\right.}{h}$ eliminate $x^n – x^n$

$=\lim\limits_{h \rightarrow 0} \frac{\left\{n \cdot x^{n-1} h+a \cdot x^{n-2} h^{2}+b \cdot x^{n-3} h^{3}+\ldots+h^{n}\right\}}{h}$ factor $h$ out of the numerator

$=\lim\limits_{h \rightarrow 0} \frac{h \cdot\left\{n \cdot x^{n-1}+a \cdot x^{n-2} h+b \cdot x^{n-3} h^{2}+\ldots+h^{n-1}\right\}}{h}$ divide by the factor $h$

$=\lim\limits_{h \rightarrow 0}\left\{n \cdot x^{n-1}+a \cdot x^{n-2} h+b \cdot x^{n-3} h^{2}+\ldots+h^{n-1}\right\}$ separate the limits

$=\mathrm{n} \cdot \mathrm{x}^{\mathrm{n}-1}+\lim\limits_{h \rightarrow 0}\left\{\mathrm{a} \cdot \mathrm{x}^{\mathrm{n}-2} \mathrm{~h}+\mathrm{b} \cdot \mathrm{x}^{\mathrm{n}-3} \mathbf{h}^{2}+\ldots+\mathbf{h}^{\mathbf{n}-1}\right\} \quad$ each term has a factor of $\mathrm{h}$, and $\mathrm{h} \rightarrow 0$

$=\mathrm{n} \cdot \mathrm{x}^{\mathrm{n}-1}+\mathbf{0}=\mathbf{n} \cdot \mathbf{x}^{\mathbf{n}-\mathbf{1}}$ so $\mathbf{D}\left(\mathrm{x}^{\mathrm{n}}\right)=\mathbf{n} \cdot \mathbf{x}^{\mathbf{n}-\mathbf{1}}$

Practice 7: Use the theorem to calculate $D( x5 )$, $d ( x2 )$, $D( x100 )$, $d ( t31 )$, and $D( x0 )$. 

We will occasionally use the result of the theorem for the derivatives of all constant powers of $x$ even though it has only been proven for positive integer powers, so far. The result for all constant powers of $x$ is proved in Section 2.9 

Example 5: Find $\mathbf{D}(1 / \mathrm{x})$ and $\frac{\mathrm{d}}{\mathrm{dx}}(\sqrt{\mathrm{x}})$.

Solution: $\quad \mathbf{D}\left(\frac{1}{\mathrm{x}}\right)=\mathbf{D}\left(\mathrm{x}^{-1}\right)=-1 \mathrm{x}^{(-1)-1}=-1 \mathrm{x}^{-2}=\frac{-1}{\mathrm{x}^{2}} \cdot \frac{\mathbf{d}}{\mathbf{d x}}(\sqrt{\mathrm{x}})=\mathbf{D}\left(\mathrm{x}^{1 / 2}\right)=(1 / 2) \mathrm{x}^{-1 / 2}=\frac{1}{2 \sqrt{\mathrm{x}}}$

These results can be obtained by using the definition of the derivative,  but the algebra is slightly awkward.

Practice 8: Use the pattern of the theorem to find $\mathbf{D}\left(\mathrm{x}^{3 / 2}\right), \frac{\mathbf{d}}{\mathbf{d x}}\left(\mathrm{x}^{1 / 3}\right), \mathrm{D}\left(\frac{1}{\sqrt{\mathrm{x}}}\right)$ and $\frac{\mathbf{d}}{\mathrm{dt}}\left(\mathrm{t}^{\pi}\right)$.

Example 6: It costs $\sqrt{\mathrm{x}}$ hundred dollars to run a training program for $\mathrm{x}$ employees.

(a) How much does it cost to train 100 employees? 101 employees? If you already need to train 100 employees, how much additional will it cost to add 1 more employee to those being trained?

(b) For $f(x)=\sqrt{x}$, calculate $f^{\prime}(x)$ and evaluate $f^{\prime}$ at $x=100$. How does $f^{\prime}(100)$ compare with the last answer in part (a)?

Solution: (a) Put $f(x)=\sqrt{x}=x^{1 / 2}$ hundred dollars, the cost to train $x$ employees. Then $f(100)$=$1000 and $f(101)$=$1004.99, so it costs $4.99 additional to train the 101^st employee.

(b) $\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{2} \mathrm{x}^{-1 / 2}=\frac{1}{2 \sqrt{\mathrm{x}}} \quad$ so $\mathrm{f}^{\prime}(100)=\frac{1}{2 \sqrt{100}}=\frac{1}{20} \quad$ hundred dollars =$5.00.

Clearly $f '(100)$ is very close to the actual additional cost of training the 101^st employee.