Subject: Calculus

Derivatives

Derivatives are an important and an interesting subject to be studied. Aside from getting the value of the rate of change of a function, derivatives are also responsible for necessary applications like slope of a function, related rates and finding the mimimum and maximum of a function. As mentioned earlier, derivatives are constructed on the concepts of limit, so let's start with the formal definition of the derivative.

Definition

If f is a function of an independent variable (ie., x), then the derivative of f at x_o, denoted by f'(x_o), is given by

f'(x_0)=\lim_{h \to 0}\frac{f(x_o+h)-f(x_o)}{h},

if this limit exists. If f'(x_o) exists, then f is said to be differentiable at x_o. The function f is said to be differentiable if it is differentiable at each point in the domain of f. Given the above definition, we can quickly form a systematic method in obtaining the derivative of f at x and we call it the three-step rule.

The Three-step Rule

First Step:

We need to simplify the expression:

f(x+h)-f(x)

Second Step:

Divide by h the results we obtained from the first step. That is evaluating the expression:

\frac{f(x+h)-f(x)}{h} such that h\neq0

Third Step:

We will then evaluate

\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}

So, we can now get the derivative of any function by following the three-step rule mentioned above.

Example #1

Find the derivative of f(x)=2x by following the thee-step rule.

Solution: By following the steps above, we have

First Step:

The term f(x+h)-f(x) will then become 2(x+h)-(2x)=2x+2h-2x=2h.

Second Step:

We will divide our result from the first step by h. Thus, \frac{2h}{h}=2.

Third Step:

We then evaluate the limit of the result from step two which is \lim_{h \to 0}2=2.

Therefore, the derivative of f(x)=2x is 2.

Example #2

Find the derivative of the function f(x)=x^{2}.

Solution: By doing the same steps above, we have

First Step

f(x+h)-f(x) = (x+h)^{2}-(x)^{2}=x^{2}+2xh+h^{2}-x^{2}=2xh+h^{2}.

Second Step

Dividing the above result by h, \frac{2xh+h^{2}}{h}=\frac{h(2x+h)}{h}=2x+h.

Third Step

Evaluating the above result, \lim_{h \to 0}2x+h=2x.

Therefore, the derivative of f(x)=x^{2} is 2x.

Example #3

Find the derivative of the function f(x)=x^{3}.

Solution: By doing the same steps above, we have

First Step

f(x+h)-f(x) = (x+h)^{3}-(x)^{3}=x^{3}+3x^2h+3xh^{2}+h^3-x^{3}=3x^2h+3xh^{2}+h^3.

Second Step

Dividing the above result by h, \frac{3x^2h+3xh^{2}+h^3}{h}=3x^2+3xh+h^2.

Third Step

Evaluating the above result, \lim_{h \to 0}3x^2+3xh+h^2=3x^2.

Therefore, the derivative of f(x)=x^{3} is 3x^2.

Example #4

Find the derivative of the function f(x)=\cos{x}.

Solution: By doing the same steps above, we have

First Step

f(x+h)-f(x) = \cos(x+h)-\cos{x}

Using the trigonometric identity of cosines, we have

\cos(x+h) = \cos{x}\cos{h}-\sin{x}\sin{h}

Thus,

f(x+h)-f(x) = \cos(x+h)-\cos{x} = \cos{x}\cos{h}-\sin{x}\sin{h}

= \cos{x}(1-\cos{h})-\sin{x}\sin{h}

Second Step

Dividing the above result by h, \frac{ \cos{x}(1-\cos{h})-\sin{x}\sin{h}}{h}=\frac{\cos{x}(1-\cos{h})}{h}-\frac{\sin{x}\sin{h}}{h}

Third Step

Evaluating the above result, \lim_{h \to 0}\frac{\cos{x}(1-\cos{h})}{h}-\frac{\sin{x}\sin{h}}{h}.

Accordingly, our limit theorems tells us that the above limit can be written as

\lim_{h \to 0}\frac{\cos{x}(1-\cos{h})}{h}-\lim_{h \to 0}\frac{\sin{x}\sin{h}}{h}

then

\cos{x}\cdot \lim_{h \to 0}\frac{1-\cos{h}}{h}-\sin{x}\cdot \lim_{h \to 0}\frac{\sin{h}}{h}

We know that \lim_{h \to 0}\frac{1-\cos{h}}{h}=0 and \lim_{h \to 0}\frac{\sin{h}}{h}=1, thus

\lim_{h \to 0}\frac{\cos{x}(1-\cos{h})}{h}-\lim_{h \to 0}\frac{\sin{x}\sin{h}}{h} = -\sin{x}

Therefore, the derivative of f(x)=\cos{x} is -\sin{x}.

Example #5

Find the derivative of the function f(x)=e^{x}.

Solution: By doing the same steps above, we have

First Step

f(x+h)-f(x) = e^{x+h}-e^x=e^{x}\cdot e^{h} - e^x = e^x(e^h -1).

Second Step

Dividing the above result by h, \frac{e^x(e^h -1)}{h}.

Third Step

Evaluating the limit of the above result, \lim_{h \to 0}\frac{e^x(e^h -1)}{h}.

\lim_{h \to 0}\frac{e^x(e^h -1)}{h}=e^x\cdot \lim_{h \to 0}\frac{e^h -1}{h}

but accordingly, \lim_{h \to 0}\frac{e^h -1}{h}=1

Thus,

\lim_{h \to 0}\frac{e^x(e^h -1)}{h} = e^x

Therefore, the derivative of f(x)=e^x is e^x.

The process of finding the derivative of a given function is called differentiation. So when we are asked to differentiate a given function, then it automatically means to find the derivative of that function.

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