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

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Answer 1 · 2017-08-08T14:03:30

#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]}