# Ploting points, tangent lines

• bonzy87
In summary, to verify that (1,1) is a point on the graph of y + ln xy = 1, you would need to substitute x=1 and y=1 into the equation and show that it satisfies it. To find the equation of the tangent line at (1,1), you would need to find the gradient of the graph at x=1 and substitute it into the equation of a straight line along with the coordinates of (1,1) to solve for c.

#### bonzy87

Verify that (1,1) is a point on the graph of y + ln xy = 1 and find the equation of the tangent line at (1,1) to this graph

For the 2nd part, consider a graph of y=f(x). The gradient of the graph at the point x=a would be $y'(a)$ The notation y' just means $$\frac{dy}{dx}$$. Now the equation of a straight line is y=mx + c, where m and c is the gradient and the y-intercept to be determined. You can find the gradient of the graph f(x) at x=a by evaluating y'(a). To find c, just substitute in the coordinates of the point where you want to find the tangent line, as well as m, into the equation of the straight line and solve for c. That's how it's done.