# The Science That Draws Necessary Conclusions

To Learn Math

## 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.

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 $f'(x) = (x^2 - 2)(6x + 1) + (3x^2 + x - 4)(2x) = 12x^3 + 3x^2 - 20x - 2$.

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 $f'(x) = \frac{(x-1)(2x)-(x^2+2)(1)}{(x-1)^2} = \frac{x^2-2x-2}{x^2-2x+1}$.

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 13th, 2009 at 10:35 am

Posted in Calculus

### 3 Responses to 'Shortcuts for Finding Derivatives'

1. plz update me with limit and derivatives

prem

21 Mar 10 at 2:10 am

2. You are a cool young man. Thank you for explaining this so clearly.

Juan Wiltocruz

20 May 11 at 9:58 am

3. This was really helpful, especially when my math teacher did properly explain each rule. Thank you. And now I know where to come, if he does it again.

Kaitlin

28 Nov 12 at 3:21 am