How do you find the equation of the tangent line when x = 1, include the derivative of #Y= arctan(sqrtx)#?

Question

William Abdalla · Accepted Answer

#x-4y-1+pi=0#

As the curve is #y=tan^(-1)(sqrtx)# or #tany=sqrtx#

and when #x=1#, #y=pi/4#, so we are seeking a tangent at #(1,pi/4)#

slope of tangent is given by #(dy)/(dx)#

as #tany=sqrtx#, we have #sec^2y(dy)/(dx)=1/(2sqrtx)#

and #(dy)/(dx)=cos^2y/(2sqrtx)# and slope of tangent is

#1/2xx1/2=1/4#

Hence equation of tangent is

#y-pi/4=1/4(x-1)# or #x-4y-1+pi=0#

graph{(tany-sqrtx)(x-4y-1+pi)=0 [-2.813, 7.187, -1.68, 3.32]}

Isabella Ackerman · Answer

undefined To find the equation of the tangent line when x = 1 for the function y = arctan(sqrt(x)), we need to find the derivative of the function first. The derivative of y with respect to x can be found using the chain rule.

The derivative of y = arctan(sqrt(x)) is given by:

dy/dx = (1 / (1 + x)^(3/2)) * (1 / (2 * sqrt(x)))

Now, we can substitute x = 1 into the derivative to find the slope of the tangent line at x = 1:

dy/dx = (1 / (1 + 1)^(3/2)) * (1 / (2 * sqrt(1)))
       = (1 / 2) * (1 / 2)
       = 1 / 4

Therefore, the slope of the tangent line at x = 1 is 1/4.

To find the equation of the tangent line, we need a point on the line. Since x = 1, we can substitute this value into the original function to find the corresponding y-coordinate:

y = arctan(sqrt(1))
  = arctan(1)
  = π/4

So, the point on the tangent line is (1, π/4).

Using the point-slope form of a line, we can write the equation of the tangent line:

y - y1 = m(x - x1)

Substituting the values we found:

y - π/4 = (1/4)(x - 1)

This is the equation of the tangent line when x = 1 for the function y = arctan(sqrt(x)).