Prove that (tanx)' = sec^2 x
W need to find the derivative of tanx.
We know that tanx = sinx/cosx
Since tan x is a quotient of two functions, then we will use the quotient rule to find the derivative.
We know that if f(x)= u/v , then f'(x)= (u'v - uv')/v^2
==> Let tanx = u/v such that:
u= sinx ==> u' = cosx
v= cosx ==> v' = -sinx
==> (tanx)' = ( cosx*cosx - sinx*-sinx)/cos^2 x
==> (tanx)' = (cos^2x + sin^2 x)/cos^2 x
But we know that cos^2 x + sin^2 x = 1
==> (tanx)' = 1/cos^2 x = (1/cosx)^2
Also we know that sec x = 1/cos x
==> (tanx)' = sec^2 x........