d/dx tan(x)
To find the derivative of tan(x), use the quotient rule:
The quotient rule states that if you have a function u(x) and v(x), the derivative of the quotient of u(x) and v(x) is given by:
(d/dx)(u(x)/v(x)) = (v(x)(d/dx)(u(x)) – u(x)(d/dx)(v(x))) / (v(x))^2
In this case, u(x) = sin(x) and v(x) = cos(x)
To find the derivative of tan(x), use the quotient rule:
The quotient rule states that if you have a function u(x) and v(x), the derivative of the quotient of u(x) and v(x) is given by:
(d/dx)(u(x)/v(x)) = (v(x)(d/dx)(u(x)) – u(x)(d/dx)(v(x))) / (v(x))^2
In this case, u(x) = sin(x) and v(x) = cos(x). Now we can proceed with the derivative:
(d/dx) tan(x) = (d/dx)(sin(x)/cos(x))
= (cos(x)(d/dx)(sin(x)) – sin(x)(d/dx)(cos(x))) / (cos(x))^2
Next, we need to find the derivatives of sin(x) and cos(x):
(d/dx)(sin(x)) = cos(x) (use the derivative of sin(x) = cos(x))
(d/dx)(cos(x)) = -sin(x) (use the derivative of cos(x) = -sin(x))
Substituting these values back into the derivative equation:
(d/dx) tan(x) = (cos(x)cos(x) – sin(x)(-sin(x))) / (cos(x))^2
= (cos(x)^2 + sin(x)^2) / (cos(x))^2
Recall that sin(x)^2 + cos(x)^2 = 1 (from the Pythagorean Identity)
Therefore, the derivative simplifies to:
(d/dx) tan(x) = 1 / (cos(x))^2
Alternatively, we can also express this result as:
(d/dx) tan(x) = sec(x)^2
So, the derivative of tan(x) with respect to x is equal to 1 divided by the square of the cosine of x, or equivalently, the square of the secant of x.
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