How do you differentiate #f(x)=(tanx-1)/secx# at #x=pi/3#?

Question

1618 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-17T00:08:42

#(1 + sqrt(3))/2#

Rewrite in sine and cosine.

#f(x) = (sinx/cosx- 1)/(1/cosx)#

#f(x) = ((sinx - cosx)/cosx)/(1/cosx)#

#f(x) = sinx - cosx#

We differentiate this using #d/dx(sinx) = cosx# and #d/dx(cosx) = -sinx#.

#f'(x) = cosx - (-sinx)#

#f'(x) = cosx + sinx#

We now evaluate #f'(pi/3)#:

#f'(pi/3) = cos(pi/3) + sin(pi/3)#

#f'(pi/3) = 1/2 + sqrt(3)/2#

#f'(pi/3) = (1 + sqrt(3))/2#

Hopefully this helps!