Prove that the derivative of tan(x) is sec^2(x).

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Let y = tan(x)

Recall the definition of tan(x) as sin(x)/cos(x)

Therefore y = sin(x)/cos(x)

Use the quotient rule, which states that for y = f(x)/g(x), dy/dx = (f'(x)g(x) - f(x)g'(x))/g2(x) with f(x) = sin(x) and g(x) = cos(x).

Recall the derivatives of sin(x) as cos(x) and cos(x) as -sin(x)

This gives:

dy/dx = (cos(x)*cos(x) + sin(x)*sin(x)) / cos2(x)

Recall the trigonometric identity sin2(x) + cos2(x) = 1

Therefore dy/dx = 1/cos2(x) = sec2(x)

QED

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