How do you find the equation of the tangent line to the curve defined by y = √x at the point (36, 6)?

To find the equation of the tangent line to the curve defined by y = √x at the given point (36, 6), follow these steps:

Step 1: Differentiate the Function

The first step is to find the derivative of the function, which gives us the slope of the tangent line. The function is:

y = √x

This can be rewritten using exponent notation:

y = x^1/2

Now, we can differentiate it using the power rule:

dy/dx = (1/2)x^-1/2 = 1/(2√x)

Next, we will evaluate the derivative at x = 36 to find the slope of the tangent line at that point:

dy/dx|_(x=36) = 1/(2√36) = 1/(2*6) = 1/12

Thus, the slope of the tangent line at the point (36, 6) is m = 1/12.

Now, we can use the point-slope form of a line, which is given by:

y - y₁ = m(x - x₁)

Here, (x₁, y₁) = (36, 6) and m = 1/12. Substituting these values into the formula gives us:

y - 6 = (1/12)(x - 36)

Now, let’s simplify the equation:

y - 6 = (1/12)x - (1/12) * 36
y - 6 = (1/12)x - 3

Adding 6 to both sides to get y by itself:

y = (1/12)x + 3

The equation of the tangent line to the curve y = √x at the point (36, 6) is:

y = (1/12)x + 3