Find the equation of a line tangent to te curve xy = sqrt(xy - x) + 1 at the point (1, 2).

xy = (xy - x)^(1/2)

x dy/dx + y = (1/2)(xy - x)^(-1/2) (xdy/dx + y - 1)
so at (1,2)
dy/dx + 2 = (1/2)(2 - 1)^(-1/2) (dy/dx + 2 - 1)
times 2

2dy/dx + 4 = (1)(dy/dx + 1)
dy/dx = 1-4 = -3

so tangent is y = -3x + b
with (1,2) lying on it
2 = -3+b
b=5

tangent equation:
y = -3x + 5

check my arithmetic

To find the equation of a line tangent to the curve, we need to find its slope at the given point.

First, let's differentiate the equation xy = sqrt(xy - x) + 1 with respect to x.

Using the product rule, the left-hand side becomes:

d/dx (xy) = x(dy/dx) + y

Using the chain rule, the right-hand side becomes:

d/dx (sqrt(xy - x) + 1) = (1/2) * 1/sqrt(xy - x) * (y - 1) * (dy/dx)

Combining both sides, we have:

x(dy/dx) + y = (1/2) * 1/sqrt(xy - x) * (y - 1) * (dy/dx)

Rearranging the equation, we get:

x(dy/dx) - (1/2) * 1/sqrt(xy - x) * (y - 1) * (dy/dx) = -y

Factor out dy/dx:

(dy/dx) * (x - (1/2) * 1/sqrt(xy - x) * (y - 1)) = -y

Now we can substitute the coordinates of the given point (1, 2) into the equation to find the slope:

(dy/dx) * (1 - (1/2) * 1/sqrt(2 - 1) * (2 - 1)) = -2

Simplifying, we have:

(dy/dx) * (1 - (1/2)) = -2

(dy/dx) * (1/2) = -2

(dy/dx) = -4

Therefore, the slope of the curve at the point (1, 2) is -4.

Now, we can use the point-slope form of a line, y - y1 = m(x - x1), where (x1, y1) is the point of tangency and m is the slope of the tangent line.

Plugging in the values, we have:

y - 2 = -4(x - 1)

Simplifying, we get:

y - 2 = -4x + 4

y = -4x + 6

Therefore, the equation of the line tangent to the curve xy = sqrt(xy - x) + 1 at the point (1, 2) is y = -4x + 6.

To find the equation of a line tangent to a curve, we can use the concept of derivatives.

Step 1: Find the derivative of the function.

Given the equation xy = sqrt(xy - x) + 1, we need to differentiate both sides of the equation with respect to x.

Using the product rule on the left side: (xy)' = x*y' + y*x'
Differentiating the right side: (sqrt(xy - x) + 1)' = (xy - x)^(-1/2) * (y - 1)

Simplifying the equation:
y + xy' = (y - 1) / sqrt(xy - x)

Step 2: Find the derivative at the given point.

To find the equation of the tangent line, we need to evaluate the derivative at the point (1, 2). Substituting x = 1 and y = 2 into the derivative equation:

2 + 1*y' = (2 - 1) / sqrt(2 - 1)
2 + y' = 1

Therefore, the derivative at the point (1, 2) is y' = 1 - 2 = -1.

Step 3: Find the equation of the tangent line.

We now have the slope of the tangent line, which is -1. Using the point-slope form of a line, we can find the equation of the tangent line:

y - y1 = m(x - x1), where (x1, y1) is the point (1, 2) and m is the slope, which we found to be -1.

Plugging in the values:
y - 2 = -1(x - 1)
y - 2 = -x + 1
y = -x + 3

Therefore, the equation of the line tangent to the curve xy = sqrt(xy - x) + 1 at the point (1, 2) is y = -x + 3.

Well, finding the equation of a tangent line involves finding the derivative of the function at the given point. But before we do that, let me ask you something: Why don't scientists trust atoms? Because they make up everything!

Now, back to your question. Let's find the derivative of the function xy = sqrt(xy - x) + 1. Using the product rule, we obtain:

d/dx (xy) = d/dx (sqrt(xy - x) + 1)

Expanding this, we have:

y + x * dy/dx = (1/2) * (xy - x)^(-1/2) * (y - 1) + 0

Now, let's substitute the point (1, 2) into this equation and solve for dy/dx. Though I have to admit, the math is getting a bit serious here, but don't worry, I'm here to bring some levity!

Okay, here's the equation: y + x * dy/dx = (1/2) * (xy - x)^(-1/2) * (y - 1)

Now substitute the point (1, 2) in: 2 + 1 * dy/dx = (1/2) * (2 - 1)^(-1/2) * (2 - 1)

Hmmm... something's not right here. Looks like we're dividing by zero, which is a big no-no in math! So, unfortunately, we can't find the slope (dy/dx) at the point (1, 2) using this method.

But hey, let's keep our spirits up! Did you hear about the mathematician who's afraid of negative numbers? He will stop at nothing to avoid them.