The following material should be read prior to attending lab. You are responsible for preparing for lab so that you don't slow down your group.
This section focuses on how to simplify algebraic expressions. In calculus, you will often need to simplify algebraic expressions when differentiating so that you can find a second derivative, find where the derivative equals 0, or to solve an application problem. This might be a good time to go back and review Lab 1 on factoring expressions. Two really important things to remember when simplifying expressions are:
In general, when you learn the product rule, quotient rule, and chain rule, your calculus instructor may not make you simplify your answers. However, later on in the course, you will NEED to simplify your answers so you can think of simplifying your answers early in the course as practice for when you need it. The following examples will illustrate types of problems that you may need to simplify in your calculus course.
| Take out the common factors of $e^{x}$ and $x^{1/2}$. Remember that you always factor out the smallest power of x and 1/2 is smaller than 3/2. | $x^{3/2}e^{x} + \frac{3}{2}x^{1/2}e^{x}$ | $=$ | $e^{x}x^{1/2}\left(x + \frac{3}{2}\right)$ | When you factor out $x^{1/2}$, what are you left with in the first term? You need to find the power for x that you need so that $x^{1/2}x^{\Box{}} = x^{3/2}$. Using the laws of exponents and keeping the base and adding exponents will help you see that the number needed in the box is 1. | |
| $ = $ | $ \sqrt{x}e^{x}\left(\frac{2x \,+\, 3}{2}\right) $ |
| Often when we look at this denominator, we might be tempted to expand it by multiplying it out—DO NOT DO THIS! Often with problems such as these, you want the denominator in factored form because something may reduce later on. In this example, there is no reduction, but we still should leave it in factored form. | $\frac{(1 \,-\, t^2)(3t^2) \,-\, t^3(-2t)}{(1 \,-\, t^2)^2}$ | $=$ | $\frac{t^2[3(1 \,-\, t^2) \,+\, 2t^2]}{(1 \,-\, t^2)^2}$ | There is a common factor of $t^2$ in both terms in the numerator so we factor that out. Then we simplify the portion in brackets by using the distributive property and combining like terms. | |
| $ = $ | $ \frac{t^2[3 \,-\, 3t^2 \,+\, 2t^2]}{(1 \,-\, t^2)^2} $ | $=$ | $\frac{t^2[3 \,-\, t^2]}{(1 \,-\, t^2)^2}$ |
Eventually, you will recognize the previous examples as derivatives that were found using the product and quotient rules. Another important derivative rule is the chain rule and the next two examples focus on it, combined with the product and quotient rules.
| $\small{(2t - 5)^4[-3(8t^2 - 5)^{-4}(16t)] + (8t^2 - 5)^{-3}[4(2t - 5)^3(2)]}$ | ||
| $ = $ | $\small{8(2t - 5)^3(8t^2 - 5)^{-4}[-6t(2t - 5) + (8t^2 - 5)]}$ | We first factor out the common factors of 8, $(2t - 5)^3$, and $(8t^2 - 5)^{-4}$. This leaves us with $-6t(2t - 5)$ in the first term and $(8t^2 - 5)$ in the second term. |
| $ = $ | $\small{8(2t - 5)^3(8t^2 - 5)^{-4}[-12t^2 + 30t + 8t^2 - 5]}$ | |
| $ = $ | $\small{8(2t - 5)^3(8t^2 - 5)^{-4}[-4t^2 + 30t - 5]}$ | |
| $ = $ | $\frac{8(2t \,-\, 5)^3(-4t^2 \,+\, 30t \,-\, 5)}{(8t^2 \,-\, 5)^4}$ | Then we use algebra to simplify the expression in brackets and write the final answer with positive exponents only. |
| $\frac{\sin^2(x^2 \,+\, 1)(2x) \,-\, (x^2 \,+\, 1)[2 \, \sin(x^2 \,+\, 1)\cos(x^2 \,+\, 1)(2x)]}{\sin^4(x^2 \,+\, 1)}$ | ||
| $ = $ | $\frac{2x \, \sin(x^2 \,+\, 1)[\sin(x^2 \,+\, 1) \,-\, 2(x^2 \,+\, 1)\cos(x^2 \,+\, 1)]}{\sin^4(x^2 \,+\, 1)}$ | Here we factor out the common factors from each term in the numerator. What are they? |
| $ = $ | $\frac{2x[\sin(x^2 \,+\, 1) \,-\, 2(x^2 \,+\, 1)\cos(x^2 \,+\, 1)]}{\sin^3(x^2 \,+\, 1)}$ | There isn't much simplification in this problem, but we can reduce the power in the denominator by dividing out one of our common factors. |