Find the gradient at the point (1,1) of the tangent line to the curve x^2+y^2-3x+4y-3=0.

To find the gradient at the point (1,1) of the tangent line to the curve, you need to find the derivative of the curve equation with respect to x and evaluate it at this point.

Step 1: Start by rearranging the given equation into the standard form of a circle equation:

x^2 + y^2 - 3x + 4y - 3 = 0
x^2 - 3x + y^2 + 4y = 3
x^2 - 3x + 9/4 + y^2 + 4y + 4 = 3 + 9/4 + 4
(x^2 - 3x + 9/4) + (y^2 + 4y + 4) = 19/4
(x - 3/2)^2 + (y + 2)^2 = 19/4

So, the given equation represents a circle centered at (3/2, -2) with a radius of sqrt(19)/2.

Step 2: Differentiate both sides of the equation implicitly with respect to x:

d/dx [(x - 3/2)^2 + (y + 2)^2] = d/dx (19/4)
2(x - 3/2) + 2(y + 2) * dy/dx = 0
(x - 3/2) + (y + 2) * dy/dx = 0

Step 3: Now, we need to find the slope (dy/dx) at the point (1,1). Plug in the values of x = 1 and y = 1 into the equation:

(1 - 3/2) + (1 + 2) * dy/dx = 0
-1/2 + 3 * dy/dx = 0

Step 4: Solve for dy/dx:

3 * dy/dx = 1/2
dy/dx = 1/6

Therefore, the gradient at the point (1,1) of the tangent line to the curve is 1/6.