Find the derivative ... h(x) = e^2x-3(tan(x^2 + 1)

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You need to differentiate the given function with respect to  x, using the chain rule, such that:

`(u(v(x)))' = u'(v(x))*v'(x)*x'`

Reasoning by analogy yields:

`h'(x) = (e^(2x))'*(2x)' - 3(tan(x^2+1))'(x^2+1)'`

`h'(x) = 2(e^(2x)) - 3(1/(cos^2(x^2+1)))(2x)`

Factoring out 2 yields:

`h'(x) = 2(e^(2x) - (3x)/(cos^2(x^2+1)))`

You should remember that `sec x = 1/(cos x) => sec^2 x = 1/(cos^2 x),`  hence, you may use the alternate result substituting `sec^2(x^2+1)`  for `1/(cos^2(x^2+1))`  such that:

`h'(x) = 2(e^(2x) - (3x)(sec^2(x^2+1)))`

Hence, evaluating the derivative of the given function yields `h'(x) = 2(e^(2x) - (3x)/(cos^2(x^2+1))) `

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