The Definition of a Derivative

Read this section to understand the definition of a derivative. Work through practice problems 1-8.

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 101st 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 101st employee.