The derivative of #y = tan(x)sec(x)# with respect to #x# is #(dy)/(dx) = sec^3(x) + sec(x)tan^2(x)#
To perform this differentiation, we will need to use the Product rule, which states that given the product of two functions, #u(x)v(x)#, here represented as #f(x) = u(x)v(x)#...
#(df)/(dx) = (du)/(dx)v(x) + u(x) (dv)/(dx)#
Or, more simply:
#f' = u'v + uv'#
By setting #u(x) = tan(x)# and #v(x) = sec(x)#, we obtain:
#(df)/(dx) = (d/dx(tan x))(sec x) + (tan x)(d/dx(sec x))#
The derivative of the function #tan(x)#, for all #x# where the function is continuous and differentiable, is #sec^2(x)#. The derivative of the function #sec(x)# is #sec(x)tan(x)#. (If uncertain how we arrived at this, proofs are provided here: http://www.math.com/tables/derivatives/more/trig.htm). Thus, we obtain...
#f'(x) = sec^3(x) + sec(x)tan^2(x)#
This equation can be further simplified if desired...
#f'(x) = sec(x)(sec^2(x) + tan^2(x))#
At this point, if desired, one can manipulate trigonometric identities, specifically #sec^2(x) = tan^2(x) +1#, to obtain...
#f'(x) = sec(x)(2tan^2(x) +1)#
or
#f'(x) = sec(x)(2sec^2(x) -1)#
Source for Trigonometric derivative proofs:
"Proofs: Derivative Trig Functions." Math .com. Math .com, 2000-2005. Web. 28 August 2014.
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