Question

1)y=3x^3, x0=1 2)y=cosx, x0=П/6 3)Y=2sinx, x0=П/4 4) y=2+(x под корнем), x0=4 уравнение касательной

Answer 1

Y=3×1^3=3
y=cosP/6=корень3/2
y=2sinp/4=корень 2
y=2+корень4=4

Answer 2

To find the equation of the tangent line at a point (x0, y0) on a curve represented by a function y=f(x), we need to find the slope of the tangent line (which is the derivative of the function at that point) and then use the point-slope form of a line to write the equation of the tangent line.

1. y=3x^3, x0=1: We take the derivative of y with respect to x: y' = 9x^2 At x0=1, the slope of the tangent line is y'(1) = 9(1)^2 = 9. The point-slope form of a line is y - y0 = m(x - x0), where m is the slope and (x0, y0) is the point. So the equation of the tangent line at (1, 3) is: y - 3 = 9(x - 1) y = 9x - 6

2. y=cosx, x0=П/6: We take the derivative of y with respect to x: y' = -sinx At x0=π/6, the slope of the tangent line is y'(π/6) = -sin(π/6) = -1/2. The point-slope form of a line is y - y0 = m(x - x0), where m is the slope and (x0, y0) is the point. So the equation of the tangent line at (π/6, √3/2) is: y - √3/2 = (-1/2)(x - π/6) y = (-1/2)x + (√3/2 + π/12)

3. y=2sinx, x0=П/4: We take the derivative of y with respect to x: y' = 2cosx At x0=π/4, the slope of the tangent line is y'(π/4) = 2cos(π/4) = 2/√2 = √2. The point-slope form of a line is y - y0 = m(x - x0), where m is the slope and (x0, y0) is the point. So the equation of the tangent line at (π/4, √2) is: y - √2 = √2(x - π/4) y = √2x - (π√2)/4

4. y=2+√x, x0=4: We take the derivative of y with respect to x: y' = 1/(2√x) At x0=4, the slope of the tangent line is y'(4) = 1/(2√4) = 1/4. The point-slope form of a line is y - y0 = m(x - x0), where m is the slope and (x0, y0) is the point. So the equation of the tangent line at (4, 4) is: y - 4 = (1/4)(x - 4) y = (1/4)x + 3