How do you find the equation of tangent line to the curve # f(x) = 9tanx# at the point #(pi/4, f(pi/4)) #?

1 Answer
May 11, 2017

#36x-2y+18-9pi=0#

Explanation:

We have to find equation of tangent to curve #f(x)=9tanx# at point #(pi/4,f(pi/4))# i.e. #(pi/4,9tan(pi/4))# and as #tan(pi/4)=1#, this point is #(pi/4,9)#.

The slope of tangent is given by value of first derivative of #f(x)#, at that point I.e. at #(pi/4,9)#.

As #(df)/(dx)=9sec^2x# we have #f'(pi/4)=9sec^2(pi/4)=9xx(sqrt2)^2=18#

As slope is #18# and tangent passes through #(pi/4,9)#, its equation is

#(y-9)=18(x-pi/4)# i.e. #36x-2y+18-9pi=0#.

graph{(36x-2y+18-9pi)(y-9tanx)=0 [-4, 4, -20, 20]}