Find dy/dx for 5x^2-2xy+7y^2=0
I'm getting confused with signs.
Most likely it is for the signs in -2xy
perhaps if you look at it as +(-2x)(y)
5x^2 + (-2x)(y) + 7y^2 = 0
10x + (-2x)(dy/dx) + y(-2) + 14y dy/dx = 0
dy/dx(14y - 2x) = 2y - 10x
dy/dx = (2y - 10x)/(14y - 2x)
= (y - 5x)/(7y - x)
or, you can think of it as
5x^2-(2xy)+7y^2=0
and since
(2xy)' = 2y+2xy', that gives
10x-(2y+2xy')+14yy'
and go from there.
To find dy/dx for the equation 5x^2 - 2xy + 7y^2 = 0, we will use the implicit differentiation method.
Step 1: Differentiate both sides of the equation with respect to x:
d/dx (5x^2 - 2xy + 7y^2) = d/dx (0)
Step 2: Differentiate each term separately:
d/dx (5x^2) - d/dx (2xy) + d/dx (7y^2) = 0
Step 3: Apply the power rule for differentiation:
10x - 2y - 2xdy/dx + 14yy' = 0
Step 4: Rearrange the terms:
-2xdy/dx + 10x + 14yy' - 2y = 0
Step 5: Solve for dy/dx:
-2xdy/dx + 14yy' = 2y - 10x
dy/dx = (2y - 10x) / (-2x + 14y)
Therefore, the derivative dy/dx for the equation 5x^2 - 2xy + 7y^2 = 0 is (2y - 10x) / (-2x + 14y).
To find dy/dx for the equation 5x^2 - 2xy + 7y^2 = 0, we can use implicit differentiation. This technique allows us to differentiate both sides of the equation with respect to x, treating y as a function of x.
Step 1: Differentiate both sides of the equation with respect to x.
The derivative of the left side of the equation will follow the product rule and chain rule, while the derivative of the right side (which is zero) will be zero.
Applying the product rule and chain rule, we have:
d/dx (5x^2) - d/dx(2xy) + d/dx(7y^2) = 0
Step 2: Simplify each term separately.
d/dx (5x^2) = 10x
d/dx (2xy) = 2y + 2xdy/dx (using product rule)
d/dx (7y^2) = 14y * dy/dx (using chain rule)
Step 3: Combine the simplified terms.
10x - 2y - 2xdy/dx + 14y * dy/dx = 0
Step 4: Move the terms related to dy/dx to one side.
-2xdy/dx + 14y * dy/dx = 2y - 10x
Step 5: Factor out dy/dx.
dy/dx(-2x + 14y) = 2y - 10x
Step 6: Solve for dy/dx.
dy/dx = (2y - 10x) / (-2x + 14y)
And that's the final answer for dy/dx.