How do you find the equation of the tangent line to the curve #y=2secx# at x =pi/4?

Question

William Abdalla · Accepted Answer

#y=sqrt(2)x+2sqrt(2)-sqrt(2)/2*pi#

The searched Tangent line has the equation

#y=mx+n# to get we compute #f'(x)#

#f'(x)=2sec(x)tan(x)#

so we get

#f'(pi/4)=2sec(pi/4)tan(pi/4)=2sqrt(2)# thus #m=2sqrt(2)# To get the Point we Substitute #x=pi/4# in our function:

#f(pi/4)=2sec(pi/4)=2sqrt(2)#

so we have

#2sqrt(2)=2sqrt(2)+n#

#n=2sqrt(2)-sqrt(2)/2pi#

so the equation of our line is given by

#y=2sqrt(2)x+2sqrt(2)-sqrt(2)/2pi#

Christopher Allers · Answer

undefined To find the equation of the tangent line to the curve y=2secx at x = pi/4, we need to find the derivative of the function y=2secx and evaluate it at x = pi/4. The derivative of y=2secx is dy/dx = 2secx * tanx. Evaluating this at x = pi/4, we get dy/dx = 2sec(pi/4) * tan(pi/4). Simplifying, sec(pi/4) = sqrt(2) and tan(pi/4) = 1. Therefore, dy/dx = 2 * sqrt(2) * 1 = 2sqrt(2).

Now, we have the slope of the tangent line, which is 2sqrt(2). To find the equation of the tangent line, we use the point-slope form of a line, y - y1 = m(x - x1), where (x1, y1) is the point on the curve. Plugging in the values x1 = pi/4, y1 = 2sec(pi/4) = 2sqrt(2), and m = 2sqrt(2), we get y - 2sqrt(2) = 2sqrt(2)(x - pi/4). Simplifying further, the equation of the tangent line to the curve y=2secx at x = pi/4 is y = 2sqrt(2)x - sqrt(2)pi + 2sqrt(2).