What is the derivative of x2 + y2 = 2xy?

The answer, dy dx = 1 might make us think about the question a bit. For x2 + y2 = 2xy, we get (by differentiating implicitly), dy dx = 1. That’s the same as the derivative of a linear function with slope, 1. Hmmmmm.

How do you solve a x2 + bx + c = 0?

All equations of the form a x 2 + b x + c = 0 can be solved using the quadratic formula: 2 a − b ± b 2 − 4 a c ​ ​. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction. This equation is in standard form: ax^ {2}+bx+c=0.

What is the value of sin 2 θ + cos2 θ?

\\sin^2 heta + \\cos^2 heta = 1. sin2 θ+cos2 θ = 1. In order to prove trigonometric identities, we generally use other known identities such as Pythagorean identities.

What is the equation for the tangent line of y2?

Similarly y2 is really (some f of x)2 so we’ll need the power and chain rules. (Implicit differentiation is using the chain rule .) We could solve for dy dx = −(2x +2y + 1) 2x − 2y, but if we don’t have to, it’s usually easier to substitute numbers now: The tangent line contains the point (1,2) and has slope m = 7 2 so its equation is:

How do you find the slope of the curve?

We start by differentiating, using the product rule, the power rule and implicit differentiation. The slope of the curve is given by evaluating the point within the derivative. Hopefully this helps!

How do you find the derivative of X with respect to R2?

Differentiate with respect to x: d dx (x 2) + d dx (y 2) = d dx (r 2) Let’s solve each term: Use the Power Rule: d dx (x2) = 2x. Use the Chain Rule (explained below): d dx (y2) = 2y dy dx. r 2 is a constant, so its derivative is 0: d dx (r2) = 0. Which gives us: 2x + 2y dy dx = 0. Collect all the dy dx on one side.