Find an equation of the tangent line

Question

Find an equation of the tangent line to the graph of the function f(x) = e^sec(x) at the point (π/4, e^sqrt(2))

Answer 1

Please see video.

Answer 2

The derivative is the slope of the line tangent to the curve. Use the Chain Rule to find the derivative:

f'(x) = sec(x) tan(x) e^sec(x) = m

To find the value of the slope at the point (π/4, e^√2), plug in x = π/4:

m = sec(x) tan(x) e^sec(x) = sec(π/4) tan(π/4) e^sec(π/4) = √2 (1) e^√2 = √2 e^√2

Now find the equation of the line (y = mx + b) where m = √2 e^√2 and goes through the given point (π/4, e^√2).

y - e^√2 = 2 e^√2 (x - π/4)

y = 2 e^√2 (x - π/4) + e^√2

y = 2 e^√2 x - πe^√2 /2 + e^√2

Answer 3

y = exp(sec(x))

ln y = sec x

y'/y = sec x tanx

y' = sec x tanx * exp(sec(x))

= sec pi/4 tan pi/4 exp ( sec(pi/4))

= 2/ sqrt(2) * 1 exp ( 2/ sqrt(2))

= sqrt(2) exp( sqrt(2))

B = y - mx

= exp(sqrt(2)) - sqrt(2) exp( sqrt(2)) * pi/4

= exp(sqrt(2) [1 + sqrt(2)*pi/4]

Y = MX + B = sqrt(2)exp(sqrt(2)) X + exp(sqrt(2) [1 + sqrt(2)*pi/4]

= exp(sqrt(2) [ sqrt(2) X + 1 + sqrt(2)*pi/4]