Question

What is the derivative of `y=tan(x) sec(x)`?

Answer 1

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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