This week, we learned the product rule and quotient rule for taking a derivative. Each rule follows a formula and all you have to do is plug in the equations into the formulas when necessary and work it out.
The product rule formula: f(x)g’(x) + g(x)f’(x)
When using the product rule, the problem must state that you take the derivative of two separate equations multiplied by each other.
**note: g’(x) or f’(x) means take the derivative of g(x) or f(x)
Ex. 1) (x+3)(x^2+6x)
(x+3)(2x+6) + (x^2+6x)(1)
2x^2 + 6x + 6x + 18 + x^2 + 6x
3x^2 + 18x + 18
= 3(x^2 + 6x +6)
**note: make sure the answer is as fully simplified as possible
The quotient rule formula: (g(x)f’(x) – f(x)g’(x)) / (g(x))^2
When using the quotient rule, the problem must state that you take the derivative of two equations divided by each other.
**note: if the denominator is just a single term, it is not considered an equation and you do not use the quotient rule; you simply rewrite the problem and take the derivative normally
Ex. 2) (x^3+4) / (x-2)
((x-2)(3x^2) – (x^3+4)(1)) / ((x-2)^2)
((3x^3-6x^2) – (x^3+4)) / ((x-2)^2)
**note: to keep things in simplest terms, do not factor out the bottom
(3x^3 – 6x^2 – x^3 – 4) / ((x-2)^2)
**note: be sure to distribute the negative throughout the second equation
(2x^3 – 6x^2 – 4) / ((x – 2)^2)
(2(x^3 – 3x^2 – 2)) / ((x – 2)^2)
We also learned that in word problems, if it says to find the velocity or the instantaneous velocity, it means to take the derivative of the position function. If it says to find the average velocity, it means to find the slope by plugging in the two x-values into the position function to get the two y-values and using the slope formula (y2 – y1) / (x2 – x1).
Anyway, I pretty much understood everything this week. Now, I wasn't at school Thursday, so I missed that lesson, but I'm sure I'll catch up.
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