How do you differentiate $f (x) = {[\frac{sin x + tan x}{sin x cos x}]}^{3}$ $f(x) =[(sin x + tan x)/(sin xcos x)]^3$ ?

Question

1518 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-05-07T06:17:28

$color(red)(f' (x)=3*sec^3 x*tan x*(1+sec x)^2(1+2*sec x))$

Start from the given $f (x)=[(sin x+ tan x)/(sin x cos x)]^3$ and simplify it first

$f(x)=[sin x/(sin x cos x)+tan x/(sin x cos x)]^3$

$f(x)=[cancelsin x/(cancelsin x cos x)+(cancelsin x/cos x)*1/(cancelsin x cos x)]^3$

$f(x)=[1/( cos x)+(1/cos x)*1/(cos x)]^3$

$f(x)=[1/( cos x)+1/cos^2 x]^3$

$f(x)=[sec x+sec^2 x]^3$

At this point we differentiate using the power formula
$d/dx(u^n)=n*u^(n-1)*d/dx(u)$

$f(x)=[sec x+sec^2 x]^3$

$f' (x)=d/dx[sec x+sec^2 x]^3$

$f' (x)=3*[sec x+sec^2 x]^(3-1)*d/dx(sec x+sec^2 x)$

$f' (x)=3*[sec x+sec^2 x]^2*(sec x*tan x+2*(sec x)^(2-1)d/dx(sec x))$

$f' (x)=3*[sec x+sec^2 x]^2*(sec x*tan x+2*sec x*(sec x*tan x))$

$f' (x)=3*[sec x+sec^2 x]^2*(sec x*tan x+2*sec^2 x*tan x)$

$color(red)(f' (x)=3*sec^3 x*tan x*(1+sec x)^2(1+2*sec x))$

God bless...I hope the explanation is useful.

How do you differentiate f(x) =[(sin x + tan x)/(sin xcos x)]^3 f(x)=[sinx+tanxsinxcosx]3 f(x) =[(sin x + tan x)/(sin xcos x)]^3 ?