Taylor blog #6

This week we learned about implicit derivatives with two variables.
The key thing to remember when taking implicit derivatives is that the steps are the same as taking a regular derivative.
“with respect to” is a phrase you will see often when dealing with implicit derivatives.
The phrase “with respect to” gives you a clue as to what you will be solving for
When you see “d/dx” you also must be able to recognize that this is a symbol for “with respect to x”
When you take a derivative of anything besides x you will write it in the asterisk spot of the following example: D*/Dx
In other words, the top letter next to the d is the variable you are taking the derivative of
For example:
With respect to y: dy/dx
With respect to r: dr/dx
With respect to s: ds/dx
To solve a problem you must follow three general steps:
First take the derivative
Second gather all d*/dx
Third solve for d*/dx
Example:::
Sqrtxy= x^2y+1
(xy)1/2=x^2y+1
X^ 1/2 y^ ½ = x^2 y +1
First take the derivative
X^1/2 (1/2y^-1/2 dy/dx) + y^1/2 (1/2x^-1/2)= x^2 (1dy/dx+y(2x))
1/2x^1/2y^-1/2 dy/dx + 1/2x ^-1/2 y^1/2 = x^2 dy/dx + 2xy
Second gather all d*/dx
1/2x^1/2y^-1/2 dy/dx- X^2 dy/dx= 2xy - 1/2x^-1/2y^1y^1/2
Third solve for d*/dx

Dy/dx = 2xy -1/2x^-1/2y^1/2 / ½ x^1/2 y^-1/2 – x^2

Example with trig:
4cosx siny=1
4[cosx(cosy dy/dx)+ siny (-sinx)] = 0
4cosxx cosy dy/dx – 4 siny sin x =0
4 cosx cosy dy/dx= 4siny sinx
Dy/dx= 4 siny sinx / 4 cosx cos y
Dy/dx= tan y tan x
We also learned how to take the second derivative when dealing with implicit derivatives
The steps to solving the second derivative of an implicit derivative are:
Take the first derivative
Solve for dy/dx
Take the derivative of dy/dx (this will be d^2y/dx^2)
Plug in the result of when you solved for dy/dx into the d^2y/dx^2 equation for dy/dx
Solve again
**** sometimes when you solve the numerator will turn out to be your original equation, if so, plug in what the original equation equaled and simplify.
If not then d^2y/dx^2 will be what you found when you “solved again”
Example:
x^2 + y^2 = 25
Take the first derivative
2x + 2y dy/dx = 0
Solve for dy/dx
Dy/dx= -2x/2y therefore -x/y
Take the derivative of dy/dx
- Y(1)- [ x(1dy/dx)]/y^2
- Y-x dy/dx / y^2
Plug in the result of when you solved for dy/dx into the d^2y/dx^2 equation for dy/dx

- Y-x ( -x/y)/ y^2

Solve again

- Y^2 + x^2/y/y^2/1
-y^2 +x^2/ y^3
**** remember that sometimes when you solve the numerator will turn out to be your original equation, if so, plug in what the original equation equaled and simplify.

Therefore:
-25/y^3