What is the equation of the line tangent to # f(x)=2x-secx # at # x=pi/4#?

Question

What is the equation of the line tangent to #  f(x)=2x-secx  # at #  x=pi/4#?

Scarlett Allison · Accepted Answer

#y-pi/2+sqrt2=(2-sqrt2)(x-pi/4)#

First of all, find the point the tangent line will intersect:

#f(pi/4)=(2pi)/4-sec(pi/4)=pi/2-sqrt2#

The tangent line passes through #P(pi/4,pi/2-sqrt2)#.

The part that requires calculus is finding the slope of the tangent line at #x=pi/4#, which will be equal to #f'(pi/4)#.

Differentiating the function:

#f'(x)=d/dx(2x)-d/dx(secx)#

The derivative of #2x# is #2#.

To find the derivative of #secx#, if you don't have it memorized, I'd use #secx=(cosx)^-1# then differentiate using the power and chain rule:

#d/dx(cosx)^-1=-(cosx)^-2d/dx(cosx)=(-1)/cos^2x(-sinx)#

#=1/cosx*sinx/cosx=secxtanx#

Putting this together,

#f'(x)=2-secxtanx#

And the slope of the tangent line is

#f'(pi/4)=2-sec(pi/4)tan(pi/4)=2-sqrt2(1)=2-sqrt2#.

Putting the point #P(pi/4,pi/2-sqrt2)# and slope #m=2-sqrt2# into the equation of a line:

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

#y-pi/2+sqrt2=(2-sqrt2)(x-pi/4)#

Elijah Abram · Answer

undefined The equation of the line tangent to f(x)=2x-secx at x=pi/4 is y = 2x - 2√2.