the equation is x^2 + y^2 =sqrt(7)
i knw 2x+2y dy/dx= don't knw if sqrt(7)= 0 for this problem or 1/2(7)^-1/2
so wat do i do for sqrt(7) in this problem?
Denke! sqrt7 is a constant. The differential of a constant is zero.
2x+2y dy/dx=0
solve for dy/dx
ok thank you...
so it would be dy/dx=-2x/2y
giving me dy/dx=-x/y
To determine the derivative dy/dx of the given equation x^2 + y^2 = sqrt(7), we need to express y implicitly as a function of x.
First, we differentiate both sides of the equation with respect to x using the chain rule.
On the left side, the derivative of x^2 with respect to x is 2x.
On the right side, the derivative of sqrt(7) with respect to x is zero, because sqrt(7) is a constant.
Then, differentiating y^2 with respect to x requires the chain rule. We get 2y * (dy/dx).
Putting it all together, we have:
2x + 2y * (dy/dx) = 0
Now to determine the value of sqrt(7), you need to substitute the correct value into the equation.
If sqrt(7) = 0, then the equation becomes:
2x + 2y * (dy/dx) = 0
If sqrt(7) = 1/2 * (7)^(-1/2), then the equation becomes:
2x + 2y * (dy/dx) = 1/2 * (7)^(-1/2)
To proceed, you need to check the given information or the context of the problem to determine which value of sqrt(7) is appropriate in this case.