What is the equation of the line tangent to # f(x)=x/secx # at # x=pi/3#?

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Answer 1 · 2018-04-11T12:48:52

Since we are talking about a tangent this means we'll want to find the gradient at that point.

#f(x) = x/(sec(x))#

#f(x) = xcos(x)#

#f'(x) = cos(x) - xsin(x)#

Gradient at #x = pi/3# :

#f'(pi/3) = cos(pi/3) - pi/3sin(pi/3)#

#f'(pi/3) = 1/2 - pi/3 xxsqrt3/2#

#f'(pi/3) = 1/2 - pi/(2sqrt3)#

Now the equation of the tangent can be found using this gradient and the point at which the tangent is #(pi/3 , f(pi/3))#:

Using point-gradient formula:

#y-y_1 = m(x - x_1)#

#y - f(pi/3) = (1/2 - pi/(2sqrt3))(x - pi/3)#

#y = (sqrt3 - pi)/(2sqrt3)x - pi/6 + pi^2/(6sqrt3) + pi/3cos(pi/3)#

#y= (sqrt3 - pi)/(2sqrt3)x - pi/6 +pi^2/(6sqrt3) + pi/3 xx 1/2#

This gives you:

#y = (sqrt3 - pi)/(2sqrt3)x + pi^2/(6sqrt3)#