On the 21st, we learned about what it means for a point or to be differentiable. A point x is differentiable if f'(x) exists. Some examples of non-differentiable points are when there is a "corner" in the graph of the function, when the graph is discontinuous at a point, or when the tangent line for a point is vertical. Later we learned about the converse and contrapositive of a statement Statement: If A, then B Converse: If B, then A Contrapositive If not B, then not A The converse of a statement is not always true...
In class on Friday, October 24, 2008, we reviewed the topics we learned during the week, including concavity. When f(x) is concave up, f’(x) is increasing and its tangent line is below the curve. Also the sign of f’’(x) is positive. When f(x) is concave down, f’(x) is decreasing and its tangent is above the curve. Also the sign of f’’(x) is negative. Then after we reviewed, we started working with partners on a worksheet about concavity.
On October 28th, 2008, the class did two things: reviewed homework from the previous night and learned about the Product Rule. We reviewed the questions #37 (from pg287). The answer was that f would be increasing on the interval of x>3. After reviewing the homework, we did a lab on the Product Rule. After calculating that (gh)' does not equal g' * h', we started finding out how the two would correctly multiply together. We found out that the product rule is g(x)*h'(x) + h(x)*g'(x). It will always be used with at least two different functions. -chris
On Wednesday, October 29, our class reviewed the Product Rule by doing a series of problems in a review packet. Besides finding the derivative of simple equatons like f(x)=7x(cosx) we also found the derivative of piece-wise functions with two or more derivatives. In order to do these piecewise functions we had to review how to find the derivative of an absolute value.
On Nov. 6, our class learned to find f'(x) for a composite function by using the chain rule. The chain rule is when you take the derivatives of each of the functions and then multiply them together. For instance, you would solve for the derivative of sin(x^2) by first solving for the derivative of the inner function, which is 2x, and then solving for the derivative of the outer function, which is cos(x^2), and then multiplying the two derivatives to get 2xcos(x^2). Another way of stating the chain rule is that if h' is derivative of f(g(x)) then h'=f'(g(x)*g'(x).
On Nov. 10th, we learned to combine the chain rule and the product/quotient rules. We learned to break function a(x)=f(x)*g(d(x)) or a(x)=g(d(x))/f(x) into its different parts, and find the derivative for the whole function. In each, the chain rule is used to find the derivative for the composite function, and that composite function is used as the second function in the product or quotient rule.
On November 11, we started class by reviewing a problem we did for homework on last night’s worksheet. And we found that it was complicated because there was no explicit function like f(x). Instead it was something with the x and y together on one side not solving for one in particular. So our goal for today was to study implicit differentiation, where we have an equation that does not have any explicitly defined equations but it does have implicit ones that you can solve for. So before we started that we had to review finding the derivative of something first so we had four examples that we solved for. 1. d/dx (tan (f(x))) = sec^2 (f(x)) * f’(x) 2. d/dx (sec (g(x))) = tan (g(x)) * sec (g(x)) * g’(x) 3. d/dx (e^h(x)) = e^h(x) * h’(x) 4. d/dx (f(x))^N = N (f(x))^N-1 * f’(x) Next we began to start to find out what implicit differentiation was. So first you need an equation that has implicitly defined functions such as x^2 + y^2 = 25. So next there are two ways that you can follow. The first was the slow method: First change y to f(x): x^2 + (f(x))^2 = 25 Then derive and solve for f’(x): 2x +2f(x) * f’(x) = 0 2f(x) * f’(x) =-2x f’(x) = -x/ f(x) Or you can do it the fast way: Instead keep the y as a y (but don’t forget that this y has to parts to it and you still need to find its derivative as dy/dx): x^2 + y^2 = 25 Next derive and solve for dy/dx: 2x +2y * dy/dx = 0 2y * dy/dx = -2x dy/dx = -x/y So now it is much easier to plug in numbers to these equations rather than the first one and we got it into one clear function of the derivatives of y and of x. So before we moved on to practice problems we did one more example which you can do for practice. So now use implicit differentiation to solve for the derivative of x^2 * y + y^3 = 7 – x. (Answer is dy/dx = (-1 – 2xy) / (x^2 + 3y^2))
Today we reviewed the homework from last night. Next we reviewed the derivatives of trig functions. d/dx (tanx) = sec^2(x) d/dx (cotx) = -csc^2(x) d/dx (secx) = sec(x)tan(x) d/dx (cscx)= -csc(x)cot(x)
Nest, we took out the worksheet from yesterday (November 11) and learned how to find the maximum and minimums of an ellipsis We took the equation we found for dy/dx and set it equal to zero 0 = (y-5x) / (y-x) We then set it equal to y: y=5x The line y=5x intersects with the ellipse at its maximum and its minimum, so we plugged in 5x for y in the original equation: 5x^2 – 2x(5x) + (5x)^2=13 and solved for x: x= + or – the square root of 13/20 In order to find the y coordinate, it’s a better choice to plug this value into the x space of the equation y=5x instead of the equation of the ellipse, and this will give you the points of the maximums and minimums on the ellipse Mr. Hansen handed out a worksheet on implicit differentiation Here we had to find the dy/dx formula, find the maximum and minimum y-values, and find the maximum and minimum x-values To find the max and min x-values, the tangent line of the ellipse is vertical; therefore, the slope is undefined A slope is undefined when the denominator equals zero You set the denominator of the dy/dx equation equal to zero: 0=y-x so y=x Then, you plug in x for all the y values in the original equation of the ellipse, so x= + or – the square root of 8/3 To find the y-coordinate, you plug this back in for x, so y also equals + or – the square root of 8/3
On November 13th, We reviewed implicit differentiability. We also found out that on the graph (x^2/25)+(y^2/9)=1 if you shoot a beam from (-4,0) and the beam hits the wall it will always then hit(4,0) and vice versa. We also used implicit differentiability found out the derivative of an oval shaped graph using the quotient rule.
11 comments:
On the 21st, we learned about what it means for a point or to be differentiable. A point x is differentiable if f'(x) exists. Some examples of non-differentiable points are when there is a "corner" in the graph of the function, when the graph is discontinuous at a point, or when the tangent line for a point is vertical. Later we learned about the converse and contrapositive of a statement
Statement:
If A, then B
Converse:
If B, then A
Contrapositive
If not B, then not A
The converse of a statement is not always true...
In class on Friday, October 24, 2008, we reviewed the topics we learned during the week, including concavity. When f(x) is concave up, f’(x) is increasing and its tangent line is below the curve. Also the sign of f’’(x) is positive. When f(x) is concave down, f’(x) is decreasing and its tangent is above the curve. Also the sign of f’’(x) is negative. Then after we reviewed, we started working with partners on a worksheet about concavity.
On October 28th, 2008, the class did two things: reviewed homework from the previous night and learned about the Product Rule. We reviewed the questions #37 (from pg287). The answer was that f would be increasing on the interval of x>3. After reviewing the homework, we did a lab on the Product Rule. After calculating that (gh)' does not equal g' * h', we started finding out how the two would correctly multiply together. We found out that the product rule is g(x)*h'(x) + h(x)*g'(x). It will always be used with at least two different functions.
-chris
On Wednesday, October 29, our class reviewed the Product Rule by doing a series of problems in a review packet. Besides finding the derivative of simple equatons like f(x)=7x(cosx) we also found the derivative of piece-wise functions with two or more derivatives. In order to do these piecewise functions we had to review how to find the derivative of an absolute value.
On Nov. 6, our class learned to find f'(x) for a composite function by using the chain rule. The chain rule is when you take the derivatives of each of the functions and then multiply them together. For instance, you would solve for the derivative of sin(x^2) by first solving for the derivative of the inner function, which is 2x, and then solving for the derivative of the outer function, which is cos(x^2), and then multiplying the two derivatives to get 2xcos(x^2). Another way of stating the chain rule is that if h' is derivative of f(g(x)) then h'=f'(g(x)*g'(x).
On Nov. 10th, we learned to combine the chain rule and the product/quotient rules. We learned to break function a(x)=f(x)*g(d(x)) or a(x)=g(d(x))/f(x) into its different parts, and find the derivative for the whole function. In each, the chain rule is used to find the derivative for the composite function, and that composite function is used as the second function in the product or quotient rule.
On November 11, we started class by reviewing a problem we did for homework on last night’s worksheet. And we found that it was complicated because there was no explicit function like f(x). Instead it was something with the x and y together on one side not solving for one in particular. So our goal for today was to study implicit differentiation, where we have an equation that does not have any explicitly defined equations but it does have implicit ones that you can solve for. So before we started that we had to review finding the derivative of something first so we had four examples that we solved for.
1. d/dx (tan (f(x))) =
sec^2 (f(x)) * f’(x)
2. d/dx (sec (g(x))) =
tan (g(x)) * sec (g(x)) * g’(x)
3. d/dx (e^h(x)) =
e^h(x) * h’(x)
4. d/dx (f(x))^N =
N (f(x))^N-1 * f’(x)
Next we began to start to find out what implicit differentiation was. So first you need an equation that has implicitly defined functions such as x^2 + y^2 = 25. So next there are two ways that you can follow. The first was the slow method:
First change y to f(x): x^2 + (f(x))^2 = 25
Then derive and solve for f’(x): 2x +2f(x) * f’(x) = 0
2f(x) * f’(x) =-2x
f’(x) = -x/ f(x)
Or you can do it the fast way:
Instead keep the y as a y (but don’t forget that this y has to parts to it and you still need to find its derivative as dy/dx): x^2 + y^2 = 25
Next derive and solve for dy/dx: 2x +2y * dy/dx = 0
2y * dy/dx = -2x
dy/dx = -x/y
So now it is much easier to plug in numbers to these equations rather than the first one and we got it into one clear function of the derivatives of y and of x. So before we moved on to practice problems we did one more example which you can do for practice. So now use implicit differentiation to solve for the derivative of
x^2 * y + y^3 = 7 – x.
(Answer is dy/dx =
(-1 – 2xy) / (x^2 + 3y^2))
Today we reviewed the homework from last night.
Next we reviewed the derivatives of trig functions.
d/dx (tanx) = sec^2(x)
d/dx (cotx) = -csc^2(x)
d/dx (secx) = sec(x)tan(x)
d/dx (cscx)= -csc(x)cot(x)
Nest, we took out the worksheet from yesterday (November 11) and learned how to find the maximum and minimums of an ellipsis
We took the equation we found for dy/dx and set it equal to zero 0 = (y-5x) / (y-x)
We then set it equal to y: y=5x
The line y=5x intersects with the ellipse at its maximum and its minimum, so we plugged in 5x for y in the original equation: 5x^2 – 2x(5x) + (5x)^2=13 and solved for x: x= + or – the square root of 13/20
In order to find the y coordinate, it’s a better choice to plug this value into the x space of the equation y=5x instead of the equation of the ellipse, and this will give you the points of the maximums and minimums on the ellipse
Mr. Hansen handed out a worksheet on implicit differentiation
Here we had to find the dy/dx formula, find the maximum and minimum y-values, and find the maximum and minimum x-values
To find the max and min x-values, the tangent line of the ellipse is vertical; therefore, the slope is undefined
A slope is undefined when the denominator equals zero
You set the denominator of the dy/dx equation equal to zero: 0=y-x so y=x
Then, you plug in x for all the y values in the original equation of the ellipse, so x= + or – the square root of 8/3
To find the y-coordinate, you plug this back in for x, so y also equals + or – the square root of 8/3
On November 13th, We reviewed implicit differentiability. We also found out that on the graph (x^2/25)+(y^2/9)=1 if you shoot a beam from (-4,0) and the beam hits the wall it will always then hit(4,0) and vice versa. We also used implicit differentiability found out the derivative of an oval shaped graph using the quotient rule.
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