Derivatives of Trigonometric Functions

Question

f(x) = -5cosx + 7tanx

f '(x) =

f ′(π/4)

Answer 1

Solution:

f(x) = -5cosx + 7tanx

f'(x) = ?

let take the first term -5cosx, it is of the form uv where u = -5 and v=cosx The derivative of uv is

d(uv) = du(v)+dv(u) , du =0, and dv= -sinx then d(-5cosx) = (0)(cosx) + (-sinx)(-5) = 5sinx

Lets take the 2nd term 7tanx, using the same principles u = 7, du = 0, v=tan x, dv= d(tanx)

to find d(tanx) recall that tan x = sinx/cosx thenu using the d(u/v) rule = du(v)-dv(u)/v² ,

here again u = sin x, du = cosx, v = cosx, dv = -sinx substitute

d(tanx) = cosx cosx - (-sinx)(sinx) / cos²x = cos²x + sin²x / cos²x = 1 / cos²x = sec²x

and d( 7 tanx) = (0)(tanx)+(sec²x)(7) = 7 sec²x

then f'(x) = 5 sinx + 7 sec²x

and f'(π/4) = 5 sin(π/4) + 7 sec²(π/4), but sec²(π/4)= 1/ cos²(π/4)

and f'(π/4) = 5 sin(π/4) + 7 (1/cos²(π/4) = 5(√2/2) + 7[ 1/ (√2/2)²] = 5√2/2 + 7 / (2/4) = 5√2/2 + 14