This week in AP Calc, we learned a new process. The process by which we find dy/dx is called implicit differentiation. Here is an example...
Find dy/dx of y^2=x
To find dy/dx, first you would differentiate both sides with respect of x.
y^2 = x
becomes
2y(dy/dx) = 1
Then you want your dy/dx on one side of the equation alone.
Final Answer -> = dy/dx = 1/(2y)
I also learned that the derivative of a circle can be found quite simply.
for example.. d/dx(x^2+y^2) = d/dx (any constant number)
would always go to dy/dx = -x/y
From here to find the slope of any point, you can simply just plug it in to the x and y.
From all this we can say that the Implicit Differentiation Process is as follows..
d^2y/dx^2 = y" for homework purposes as well
Find dy/dx of y^2=x
To find dy/dx, first you would differentiate both sides with respect of x.
y^2 = x
becomes
2y(dy/dx) = 1
Then you want your dy/dx on one side of the equation alone.
Final Answer -> = dy/dx = 1/(2y)
I also learned that the derivative of a circle can be found quite simply.
for example.. d/dx(x^2+y^2) = d/dx (any constant number)
would always go to dy/dx = -x/y
From here to find the slope of any point, you can simply just plug it in to the x and y.
From all this we can say that the Implicit Differentiation Process is as follows..
- Differentiate both sides of the equation with respect to x
- Collect the terms with dy/dx on one side of the equation
- Factor out dy/dx
- Solve for dy/dx
d^2y/dx^2 = y" for homework purposes as well