d(x^2)/dx= 2x and d(y^2)/dx= 2y(dy/dx) but d(-xy)/dx is NOT -dy/dx+ dy/dx!hauk-gwai said:Homework Statement
Find the equation of the tangent line to the graph of x^2 - xy + y^2 = 19 (where y=y(x)) at (3,-2).
Homework Equations
The Attempt at a Solution
So, this is what I did:
d/dx(x^2-xy+y^2) = (19)d/dx
2x*dx/dx-dy/dx+dy/dx+2y(dy/dx)=0
2x+2y(dy/dx)=0
And I am stuck, the answer to this question is:
y+2=8/7(x-3)
and I don't know how to get the slope.
The equation of the tangent line to a given point on a graph is y = mx + b, where m is the slope of the tangent line and b is the y-intercept.
To find the slope of the tangent line to a graph, you can use the derivative of the function at the given point. Alternatively, you can use the slope formula: m = (y2 - y1) / (x2 - x1), where (x1, y1) and (x2, y2) are points on the tangent line.
Yes, the equation of the tangent line can change for different points on the same graph. This is because the slope of the tangent line and the y-intercept can vary depending on the point chosen.
No, the equation of the tangent line and the equation of the graph at a given point are not the same. The equation of the graph represents the entire curve, while the equation of the tangent line only represents the slope and y-intercept at a specific point on the curve.
Yes, you can find the equation of the tangent line to a graph at a point using its coordinates. You will need to use the slope formula or the derivative to find the slope of the tangent line, and then use the point-slope formula to find the equation of the line.