use the derivatives of y=sinx and y=cosx to develop the derivative of y=tanx
tan = sin/cos
tan' = (cos*cos + sin*sin)/cos^2 = 1/cos^2 = sec^2
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To derive the derivative of y = tan(x), we can use the derivatives of y = sin(x) and y = cos(x) together.
Let's start by recalling the basic trigonometric identity:
tan(x) = sin(x) / cos(x)
Now, let's differentiate both sides of this equation with respect to x.
On the left side, since tan(x) is a single term, its derivative is straightforward:
d/dx(tan(x)) = sec^2(x)
On the right side, we have a quotient of two functions: sin(x) and cos(x). To differentiate this quotient, we need to apply the quotient rule.
The quotient rule states that if we have a function f(x) divided by g(x), the derivative of this quotient is given by:
d/dx[f(x) / g(x)] = [g(x) * f'(x) - f(x) * g'(x)] / [g(x)]^2
Applying this rule to sin(x) / cos(x), we have:
d/dx(sin(x) / cos(x)) = [(cos(x) * cos(x)) - (sin(x) * (-sin(x)))] / [cos(x)]^2
= [(cos^2(x) + sin^2(x)) / cos^2(x)]
= 1 / cos^2(x)
Now, we know that sec^2(x) is equal to 1 / cos^2(x).
Therefore, we can conclude that:
d/dx(tan(x)) = sec^2(x)
So, the derivative of y = tan(x) is given by dy/dx = sec^2(x).