This past week I was stressing out completly for the exam. I'm so thankful that we got to push that exam back, i probably would have failed other wise.
Thanks to john I finally learned how to optimize:
1. identify primary and secondary
(primary the one your maximizing or minimizing)(secodary the other one)
2. solve secondary for 1 variable and plug into primary
3.take derivative. plug into secondary equation.
- The problems were some of the last problems on chapter three study guide
I missed friday when we learned all of the new stuff,
anyone feel like taking the time to explain to me what
the heck we're doing and why we're doing it??
On Friday we learned how to do implicit derivatives, which you probably know by now since we had a test on them today.
ReplyDeleteAnyway, implicit derivatives involve both xs and ys. The steps to doing an implicit derivative are:
1. Take the derivative of both sides, just like you would a normal derivative.
2. Everytime you take the derivative of y, you note it with either dy/dx or f'.
3. Solve for dy/dx or f'.
An example is:
x^2+y^2= 9
derivative: 2x + 2y dy/dx= 0
Subtract 2x over and divide by 2y.
dy/dx= -2x/2y
The twos will cancel and your implicit derivative is:
dy/dx= -x/y
Implicit derivatives involve both x's and y's, unlike normal derivatives.
ReplyDelete1: So, first you have to take the derivative of whatever they give you as you normally would.
2: Whenever you take the derivative of y, you have to note it with dy/dx or y' (I prefer y').
3: Solve for dx/dy or y'
(if you want to find slope plug in an x and y value)
example: y^3+y^2-5y-x^2=-4
First you just take the derivative, but don't forget to not the derivatives of the y's! So you get: 3y^2(y')+2y(y')-5(y')-2x=0
Then you have to solve for y', so you get:
y'(3y^2+2y-5)=2x which then is further solved for to get y'=2x/(3y^2+2y-5)
And that's all!!! this problem is finished.
we did implicit derivatives on friday. basically you take the first derivative then you substitute dy/dx for your y-values, then solve for dy/dx.
ReplyDeleteFriday we did implicit derivatives. Implicit derivatives except you have a x and a y to take the derivative of.
ReplyDeleteSteps:
1. Take the derivative like normal
2. Every time you take the derivative of y note it with dy/dx.
3. Solve for dy/dx
Friday we learned how to do implicit derivatives which hopefully you have learned by now. They involve both x's and y's. You just take the derivative like you normally would, but when taking the derivative of a y value you always note it with dy/dx. After taking the derivative you solve for dy/dx, and that's it :) If you want to find slope and are only given the x value just plug in the x value into the original function in order to find the y.
ReplyDeleteok, so aimee i taught this to you throughout the week. it's just implicit derivatives.
ReplyDelete1. figure out if it's an implicit derivative (has x & y values)
2. take the derivative of both sides like normal
3. behind all the derivatives taken of y-values, note it with dy/dx or y'
4. solve for dy/dx or y'