Find dy/dx
9. x^2 + y^2=2x
10. y^2= 3x+2y
Last 2 questions for the night lol
review implicit derivatives and the chain rule.
x^2 + y^2 = 2x
2x + 2y y' = 2
y' = (1-x)/y
y^2 = 3x+2y
2y y' = 3 + 2y'
y' = 3/(2y-2)
Just remember that you haven't learned very many rules yet. So, any problem you are assigned will have to use one of those few rules.
Double pi, nope lol I suck at it. First 3 weeks I was doing great and getting A's on my quizzes. Week 4 is starting to get to me lol.
No worries! Let's find the derivative of each equation with respect to x.
1. For the equation x^2 + y^2 = 2x:
To find dy/dx, we'll use implicit differentiation, which entails differentiating both sides of the equation with respect to x.
Differentiating x^2 + y^2 = 2x with respect to x:
d/dx(x^2) + d/dx(y^2) = d/dx(2x)
Using the power rule, the derivative of x^2 is 2x.
Next, we'll differentiate y^2 with respect to x. Since y^2 is a function of x, we need to apply the chain rule: d/dx(y^2) = d/dy(y^2) * dy/dx.
Derivative of y^2 with respect to y: d/dy(y^2) = 2y.
Finally, the derivative of 2x with respect to x is simply 2.
So our equation becomes: 2x + 2y * dy/dx = 2
To find dy/dx, we isolate it on one side of the equation:
2y * dy/dx = 2 - 2x
dy/dx = (2 - 2x) / (2y)
2. For the equation y^2 = 3x + 2y:
First, let's rearrange the equation to isolate y:
y^2 - 2y = 3x
Now, let's differentiate both sides with respect to x:
d/dx(y^2 - 2y) = d/dx(3x)
Using the power rule, the derivative of y^2 is 2y, and the derivative of -2y is -2.
The derivative of 3x with respect to x is simply 3.
So our equation becomes: 2y - 2 * dy/dx = 3
To find dy/dx, let's isolate it on one side of the equation:
-2 * dy/dx = 3 - 2y
dy/dx = (3 - 2y) / (-2)
That's it! Now you have the derivatives of both equations.