deriv of tanx
sec squared x
The derivative of tan(x) can be found by applying the quotient rule of differentiation. Recall that the quotient rule states:
If y = f(x)/g(x), then y’ = [f'(x)g(x) – f(x)g'(x)]/[g(x)]^2.
Now, let f(x) = sin(x) and g(x) = cos(x), such that tan(x) = sin(x)/cos(x). Applying the quotient rule, we get:
tan'(x) = [cos(x)cos(x) – sin(x)(-sin(x))]/[cos(x)]^2
tan'(x) = [cos^2(x) + sin^2(x)]/cos^2(x)
tan'(x) = 1/cos^2(x)
We can further simplify this expression using the trigonometric identity: cos^2(x) = 1 + tan^2(x). Substituting this identity into the expression for tan'(x), we get:
tan'(x) = 1/[1 + tan^2(x)]
Therefore, the derivative of tan(x) is 1/[1 + tan^2(x)], or equivalently, sec^2(x).
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