F(x)=secx tanx/sinx ?

Question

show that f'(x)= 1/xsinx (sec^3 x + (1-1/x) secx tan x )

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Answer 1 · 2018-05-01T14:54:56

It is not true, so it cannot be shown.

#f(x) = sec^2x#, so #f'(x) = 2sec^2x\ tanx#

#f'(pi/4) = 4#

But

#1/(pi/4)sin(pi/4)(sec^3(pi/4)+(1-1/(pi/4))sec(pi/4)tan(pi/4)) != 4#

and

#1/((pi/4)sin(pi/4))(sec^3(pi/4)+(1-1/(pi/4))sec(pi/4)tan(pi/4)) != 4#