We're going to find the derivative of the original function, so the derivative of in terms of X of X ^2 plus the derivative in terms of X of the quantity XY. Hence we have to use the product rule minus the derivative in terms of X of y ^2 equaling the derivative in terms of X of the constant 4. So when we evaluate this, we get 2X. The DXDX is one plus the derivative of the X times the Y plus the derivative of the Y times the X product rule -2 YDYDX equaling the derivative of a constant which is 0. So 2X plus the derivative in terms of X of X is 1D XDX times the Y plus the derivative of Y in terms of X is DYD X * X -2 YDYD X = 0. So we're going to factor everything that has a DYDX set in it. So we're going to get DYDX times the quantity X -, 2 Y. We're going to take everything that didn't have a DYDX in it to the other side. So negative two X -, y. And then we're going to divide. So we get the derivative equals negative two X -, y / X - 2 Y. Now the question says we want to figure out the tangent line and the normal line at the given point of four, six. So we're going to take that derivative and we're going to stick in the .46. Every time we see X put in four. Every time we see Y, we put in six. Simplify it up. We get a slope of 7 fourths. So now we use point slope y - 6 = 7 fourths X - 4. Take everything to one side to solve for Y. So y = 7 fourths X - 1. The normal line is exactly the same as the tangent line with the exception of. We need the perpendicular, so we need the negative reciprocal for its slope y - 6 = -4 sevenths times the quantity X - 4. We simplify that up. We get Y equal -4 sevenths X + 58 sevenths.