What is the equation of the line tangent to #f(x)=4secx–8cosx# at #x=3#?

1 Answer
Dec 10, 2017

Equation of line tangent can be formed by #y_1-y=m(x_1-x)# where #m# is the slope and #x_1 and y_1# represent the x-coordinate and y-coordinate of the point of intersection of #f(x)# and tangent.

We know that slope of line tangent to #y=f(x)# is #f'(x)=dy/dx# .
So, to do this question, we have to find #f'(x)# first.

#f(x)=4secx-8cosx#
#f'(x)=4secxtanx-8(-sinx)#
#=4secxtanx +8sinx#

When #x=3#, the slope of line tangent #(m)=f'(3)#
#=4sec(3)tan(3)+8sin(3)#
and the y-coordinate of this point is #f(3)#
#=4sec(3)-8cos(3)#

We get all the information we need and we can plug them into the equation.
Equation of line tangent at #x=3#:
#4sec(3)-8cos(3)-y=[4sec(3)tan(3)+8sin(3)](3-x)#