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Shortcuts for Finding Derivatives

Using the derivative formula to find the derivative of a function is a hassle. There are many shortcuts to finding the derivative. The first one is the Power Rule. It can be used with functions which are equal to a polynomial like $x^3 + 5x^2 - 7$ . Each term’s coefficient is multiplied by the exponent and the exponent decreases by one. Let’s look at $f(x)=3x^2$ . The derivative’s coefficient will be 6 because the coefficient of the term, 3, is multiplied by the exponent, 2. The exponent will be 1 because the exponent is decreased by one. Therefore, the derivative is $f^|(x)=6x$ . If the function has more than one term, use the same procedure for each term. Now let’s use the power rule to find the derivative of $g(x)=2x^4-5x^2+3x-2$ . The derivative is $g^|(x)=8x^3-10x+3$ . It is important to notice that the constant, -2, is eliminated since x has an exponent of zero. Also, 3x becomes 3 since anything raised to the power of 0 is equal to 1.

power_rule

Another useful shortcut is the Product Rule. It is useful when the definition is two polynomials multiplied by each other but cannot easily be simplified such as $f(x) = (x^2 - 2)(3x^2 + x - 4)$ . If u is assigned to the first polynomial and v is assigned to the second polynomial, the derivative is $u\frac{dv}{dx}+v\frac{du}{dx}$ . So in $f(x)$ $u = x^2 - 2$ and $v = 3x^2 + x - 4$ . Using the power rule, $\frac{du}{dx} = 2x$ and $\frac{dv}{dx} = 6x + 1$ . Therefore, by using the product rule $f^|(x) = (x^2 - 2)(6x + 1) + (3x^2 + x - 4)(2x) = 12x^3 + 3x^2 - 20x - 2$ .

product_rule

The last rule I am going to discuss for now is the Quotient Rule. The Quotient Rule is used when a polynomial is being divided by another polynomial. Where $y=\frac{u}{v}$ , $\frac{dy}{dx} = \frac{ v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$ . For example, let $f(x)=\frac{x^2+2}{x-1}$ . The u is $x^2 + 2$ and v is $x - 1$ . Once again utilizing the power rule, we find that $\frac{du}{dx} = 2x$ and $\frac{dv}{dx} = 1$ . Plugging these values into the quotient rule yeilds $f^|(x) = \frac{(x-1)(2x)-(x^2+2)(1)}{(x-1)^2} = \frac{x^2-2x-2}{x^2-2x+1}$ .

quotient_rule

Proper use of these rules save time. On some occasions, multiple rules will need to be applied. There is another rule I have not yet mentioned called the chain rule. It makes taking the derivative of a function within a function possible. That post will be saved for after the explanation of the usefulness of derivatives.

Written by todizzle91

August 13, 2009 at 10:35 am

Posted in Calculus

Intro to Derivatives

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The derivative of f(x) at point a is the slope of the tangent line at point a or $\displaystyle f^|(x)=\lim_{h \to 0}\frac{f(x+h) - f(x)}{h}$ . Allow me to explain.
circle_line_types
As seen in the picture above, a secant line is a line that passes through two points of a function and a tangent line touches only one. The function to find the slope of a secant line that passes through is $m=\frac{f(x+h)-f(x)}{h}$ where h is the change in x or $\Delta x$ . This is simply the slope formula, $m=\frac{y_2-y_1}{x_2-x_1}$ , modified. Taking the limit as h approaches 0 means that the $\Delta x$ would “become zero”, so the line would only pass through that one point. That is a tangent line.

Derivative Formula GIF

The picture above should help make it clearer. It illustrates using the derivative formula to find the slope of the tangent line at x=1 on $f(x) = -x^2 + 5$ . The red line is the secant line as h approaches zero from the left (values of h are -3, -2, and -1). The blue line is the secant line as h approaches zero from the right (values of h are 3 and 2). The purple line is the tangent line; it touches the function at only one point. Now I will show you how it is done algebraically. I have written it out since I believe that is easier to understand.

deriv_alg_1

The first thing I do is replace f(x) with its value $-x^2 + 5$ replacing x with x + h in the first term. I then multiply everything out so I am left with a polynomial. The $x^2$ ’s and 5’s cancel out. Now it is possible to cancel out an h in each term. After doing so, I plugged in zero for h to leave me with the derivative -2x.

deriv_alg_2

The derivative only tells you the slope of the tangent line, not the equation. Since you know the slope of the line and a point on the line, it is not difficult to find the equation of the tangent line, g(x) in this case.

There are many tricks and shortcuts to finding the derivative of a line. The derivative formula is usually the most difficult. Things to remember: The derivative of $f(x)$ is $f^|(x)$ . A derivative of the derivative yields the second derivative or $f^{||}(x)$ . The derivative of y is written as $\frac{dy}{dx}$ . In the next two posts I will delve into some of the shortcuts to finding the derivative of a function and why derivatives are useful.

Written by todizzle91

February 16, 2009 at 12:20 pm

Posted in Calculus

Algebraically Finding a Limit

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Finding a limit may be as simple as plugging a number into a function, or it may require manipulating the function if it is undefined at that point. $f(x) = 3x +1$ is a linear function continuous at all points. The limit as x approaches any point a is f(a), $\displaystyle \lim_{x \to a}f(x) = f(a)$ . To find the limit as x approaches 2 in f(x), just plug x into f(x), $\displaystyle\lim_{x \to 2}f(x)=f(2)=3(2)+1=7$ .

Now look at the rational function $g(x)=\frac{x+1}{x-1}$ . g(x) is continuous at every point except when x=1. In order to solve the limit as x approaches 1, g(x) must be manipulated to defined at 1. Multiplying g(x) by $\frac{x-1}{x-1}$ will return the exact same function with the exception of g(1) being defined, (x+1)(x-1). The limit can be taken using this function so $\displaystyle \lim_{x \to 1}g(x)=\lim_{x \to 1}(x+1)(x-1)=0$ .

Another good example is $\displaystyle\lim_{x \to \pm 2}h(x)$ if $h(x)=\frac{x+2}{x^2-4}$ . h(x) is undefined at -2 and 2 so the function needs to be manipulated to be defined at those two points. $x^2-4$ is the difference of two squares so it can be factored to $(x-2)(x+2)$ . The numerator and denominator both share the factor (x+2) so it can be cancelled leaving the function $\frac{1}{x-2}$ . Now it is clear that $\displaystyle\lim_{x \to -2}h(x)=\lim_{x \to -2}\frac{1}{x-2}=-\frac{1}{4}$ ; however the new function is still undefined at 2. Because the function cannot be simplified any more, the limit as x approaches 2 does not exits, $\displaystyle\lim_{x \to 2}h(x) = DNE$ .

Written by todizzle91

December 5, 2008 at 10:36 am

Posted in Calculus

Continuity and Discontinuity

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A function is said to be continuous at a if $\displaystyle \lim_{x \to a} f(x) = f(a)$ . A continuous function’s points are all connected. Below are examples of functions that are not continuous.

The graph above is an example of jump discontinuity. There is jump continuity at each integer. The limit from the positive and negative direction of each is a finite number, but they are not equal. Let’s look at the point where $x=2$ . $\displaystyle \lim_{x \to 2^-}y=1$ and $\displaystyle \lim_{x \to 2^+}y=2$ . Because both numbers are not infinite and are not equal, there is jump discontinuity where $x=2$ .

The graph above is an example of infinite discontinuity. This is when the limit of a point from the negative or positive side is infinite or does not exist. There is infinite discontinuity at the point where $x=0$ . $\displaystyle \lim_{x \to 0^-}y=\infty$ and $\displaystyle \lim_{x \to 0^+}y=\infty$ . Even though the limit from both sides equal infinity, the two lines never actually meet.

Above is an example of removable discontinuity. Removable discontinuity is when the limit from both sides of a point are equal and finite, but the function is undefinded at this point. There is a hole in the graph where $x=-1$ . $\displaystyle \lim_{x \to -1^-}y=-2$ and $\displaystyle \lim_{x \to -1^+}y=-2$ . Even though the limits are equal where $x=-1$ , there is no point there.

Written by todizzle91

October 2, 2008 at 9:49 pm

Posted in Calculus

Intro to Limits

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A limit is what f(x) approaches as x approaches a value from both directions. The limit of a function f(x) as x approaches 0 is written as $\displaystyle\lim_{x \to 0}f(x)$ .

f(x) is pictured above. It is evident that $\displaystyle\lim_{x \to 3}f(x)=5$ . As x gets closer to the value 3, f(x) gets closer to 5.

f(x) never needs to equal the number x approaches. $\displaystyle\lim_{x \to 4}g(x)=3$ even though $g(4)=5$ .

If there is a vertical asymptote asymptote at x=0, $\displaystyle\lim_{x \to 0}f(x)=DNE$ . DNE stands for “does not exist”. The line approaching x from the negative side goes to $-\infty$ . From the positive side, the line appproaches $\infty$ . Because they are not approaching the same point, the limit is said not to exist.

Limits can pertain to only the line coming from only one side. It is written as $\displaystyle\lim_{x \to 0^+}f(x)$ if it only pertains to parts of the line where $x>0$ or as $\displaystyle\lim_{x \to 0^-}f(x)$ if it only pertains to parts of the line where $x<0$ . If $f(x)=\frac{1}{x}$ , then $\displaystyle\lim_{x \to 0^+}f(x) = +\infty$ and $\displaystyle\lim_{x \to 0^-}f(x) = -\infty$ .