Prove that the d(tan(x))/dx is equal to sec^2(x).

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Prove that the d(tan(x))/dx is equal to sec^2(x).

You can express tan(x) as sin(x)/cos(x).

Therefore, tan(x)= sin(x)/ cos(x)

The quotient rule can be applied here as there is a function of x in the numerator and denominator.

Quotient Rule: (v*(du/dx) - u*(dv/dx))/v²

Let u =sin(x) and v=cos(x) and hence (du/dx)= cos(x) and (dv/dx)= -sin(x).

Therefore:

d(tan(x))/dx= (cos(x)*cos(x))-(sin(x)*(-sin(x))/(cos²(x))

=(cos²(x)+sin²(x))/(cos²(x))

Using the trig identity, cos²(x)+sin²(x)=1, the numerator of the fraction can be tidied and heavily simplified.

d(tan(x))/dx= 1/(cos²(x))

As 1/(cos(x)) is equal to sec(x), 1/(cos²(x)) is equal to sec²(x).

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