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

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Answer 1 · 2017-08-31T19:33:43

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

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:

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

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

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