Blog 6

Were stilllll on chapter two, and this week we learned about implicit derivatives. An implicit derivative is derivatives with two variables. The steps for a derivative and implicit derivative do not change. It simply means “with respect to”. For example d/dx- the derivative would be with respect to x. An example to solve a simple one: d/dx (x^3)= 3x^2. When you take a derivative of anything but an x, you write dy/dx, ds/dx, dr/dx **the top letter being your variable. You would gather all dy/dx and solve for dy/dx. So basically you take the derivative like usual, instead when you take the derivative of y you would have to put dy/dx behind it. After that you would combine like terms (if any), then solve for dy/dx. **Product, Quotient, and Chain rule may be used in implicit derivatives.

Example:
X^2+y^2=9

*First thing you would do is take the derivative. (Remember the derivative of all constants=0)
2x+2y(dy/dx)=0
*Next solve for y, so bring 2x to the other side
2y(dy/dx)=-2x
*Then divide by 2y
Leaving you with: dy/dx=x/y

**Now if you would have a point, you would just plug it in to the final answer. Say the point was (3,5)
The answer would then become: 3/5

**Nowww, if you would have to find the second derivative you would have to take the final answer which was x/y and solve for it. In this case you would use quotient rule. Keep in mind that when solving in again you still have to put dy/dx when taking the derivative of y.

Y(-1)-(x)(1dy/dx)/y^2
-y-x(dy/dx)/y^2
**From here you can plug in the original dy/dx
-y-x(x/y)/y^2 = -y-x^2/y/y^2
*Sandwich it and the answer would be –y^2-x^2/y^3
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Now for something I don’t understand. I don’t understand the last thing we learned with all the word problems. When we do them in class I understand them. But if I would try one on my own I would be lost.