derivative of tan(x)
The derivative of the tangent function, \(\tan(x)\), can be found using the quotient rule
The derivative of the tangent function, \(\tan(x)\), can be found using the quotient rule. The quotient rule states that for functions of the form \(\frac{f(x)}{g(x)}\), the derivative is given by:
\[
\left(\frac{f'(x) \cdot g(x) – f(x) \cdot g'(x)}{(g(x))^2}\right)
\]
Let’s apply this rule to find the derivative of \(\tan(x)\).
First, we need to express \(\tan(x)\) as a quotient. Using the identity \(\tan(x) = \frac{\sin(x)}{\cos(x)}\), we can write:
\[
\tan(x) = \frac{\sin(x)}{\cos(x)}
\]
Now, let’s differentiate the numerator and denominator separately. The derivative of \(\sin(x)\) is \(\cos(x)\), and the derivative of \(\cos(x)\) is \(-\sin(x)\).
Using the quotient rule, the derivative of \(\tan(x)\) can be written as:
\[
\left(\frac{\cos(x) \cdot \cos(x) – \sin(x) \cdot (-\sin(x))}{(\cos(x))^2}\right)
\]
Simplifying this expression, we get:
\[
\frac{\cos^2(x) + \sin^2(x)}{\cos^2(x)}
\]
Using the trigonometric identity \(\cos^2(x) + \sin^2(x) = 1\), we can further simplify the expression:
\[
\frac{1}{\cos^2(x)}
\]
Recall that \(\sec^2(x) = \frac{1}{\cos^2(x)}\), so we can rewrite the derivative as:
\[
\sec^2(x)
\]
Therefore, the derivative of \(\tan(x)\) is \(\sec^2(x)\).
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