Find dy/dx for implicit differentiation
y^2-xy+x^2=7
y^2-xy+x^2=7
2y dy/dx - x(dy/dx) - y(dx/dx) + 2x (dx/dx) = 0 , remember dx/dx = 1
simplify and solve for dy/dx
show me your steps so I can check your work.
2y-x (dy/dx)-y(dy/dx)+2x=0
subtract -2y +x and move over to the right side replacing zero
It becomes: -y+2x(dy/dx) = -2y dy/dx + x
Divide both sides by by -2y+x
Answer= 2x-y/ x-2y
I agree IF you place brackets like this ....
(2x-y)/ (x-2y)
To find dy/dx for the given equation using implicit differentiation, we'll differentiate both sides of the equation with respect to x.
Let's start by differentiating the left-hand side of the equation term by term.
For y^2, we can use the chain rule because y is a function of x. Therefore, the derivative of y^2 with respect to x will be 2y * dy/dx.
For -xy, we have to apply the product rule. The derivative of -xy with respect to x will be -y * dx/dx (which is just 1) - x * dy/dx.
Similarly, for x^2, we can apply the power rule. The derivative of x^2 with respect to x will be 2x * dx/dx (again, dx/dx is just 1).
The derivative of the constant term "7" with respect to x will be zero since it is a constant.
So, differentiating the left-hand side term by term, we have:
2y * dy/dx - y - x * dy/dx + 2x = 0
Now, let's rearrange the equation to solve for dy/dx:
Grouping the terms with dy/dx together, we get:
2y * dy/dx - x * dy/dx = y - 2x
Factoring out dy/dx, we have:
(2y - x) * dy/dx = y - 2x
Now, isolate dy/dx by dividing both sides of the equation by (2y - x):
dy/dx = (y - 2x) / (2y - x)
So, the derivative dy/dx for the given equation is (y - 2x) / (2y - x).