Find an equation of the tangent line to the curve at the given point y = √x at point (36, 6)

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

1. Calculate the derivative: The derivative of the function gives us the slope of the tangent line. The function is y = √x, which can also be written as y = x^(1/2).
2. Using the power rule of differentiation, the derivative is:
• y’ = (1/2)x^(-1/2) = 1/(2√x)
3. Evaluate the derivative at x = 36:
• y’|_(x=36) = 1/(2√36) = 1/(2*6) = 1/12
4. So, the slope (m) of the tangent line at the point (36, 6) is 1/12.
5. Use the point-slope form to find the equation of the tangent line: The point-slope form of a line is given by:
• y – y1 = m(x – x1)
6. Here, (x1, y1) = (36, 6) and m = 1/12. Plugging these values into the point-slope form:
• y – 6 = (1/12)(x – 36)
7. Simplifying:
• y – 6 = (1/12)x – 3
• y = (1/12)x + 3
8. Final result: The equation of the tangent line to the curve y = √x at the point (36, 6) is:
• y = (1/12)x + 3