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# How do you take the derivative of $\tan \left( \dfrac{\pi x}{2} \right)?$

Last updated date: 20th Jun 2024
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Hint: The quotient rule is method of determining the differentiation of a function in which the ratio of two function is differentiable by applying variant rule,
$\tan \left( x \right)=\dfrac{\sin x}{\cos x}\to \tan \left( x \right)$
$=\dfrac{{{\cos }^{2}}\left( x \right)+{{\sin }^{2}}\left( x \right)}{{{\cos }^{2}}\left( x \right)}$
The quotient rule is applicable when in the given problem one function is divided to another function.

Complete step by step solution:We know that,
For every function $h(x)$ that can be written as $f(g(x)),h'(x)=f'(g(x)).g'(x)$
In this case, we have
$f(x)=\tan (x),g(x)=\dfrac{\pi .x}{2}$
The derivative for $g(x)$ can be easily compute,
$g'(x)=\dfrac{\pi }{2}$
Since, the leading coefficient is the derivative for a first degree polynomial.
The derivative for $f(x)$ you can easily check it from the reference table or try to remember it, or we can use quotient rules.
$\tan \left( x \right)=\dfrac{\sin \left( x \right)}{\cos \left( x \right)}\to \tan \left( x \right)$
$=\dfrac{{{\cos }^{2}}\left( x \right)+{{\sin }^{2}}\left( x \right)}{{{\cos }^{2}}\left( x \right)}$
Using identity of ${{\cos }^{2}}\left( x \right)+{{\sin }^{2}}\left( x \right)=1$we get that,
$\tan '\left( x \right)=\dfrac{1}{{{\cos }^{2}}\left( x \right)}$
$\therefore \tan '\left( x \right)={{\sec }^{2}}\left( x \right)$
So, the derivative of
$\tan \left( \dfrac{\pi .x}{2} \right)=\dfrac{\pi }{2}.{{\sec }^{2}}\left( \dfrac{\pi .x}{2} \right)$

For example, the derivative $\tan '\left( x \right)=\dfrac{1}{{{\cos }^{2}}\left( x \right)}$
${{\sec }^{2}}\left( x \right)$ another in differentiation is rate of change of function or it is a process for finding derivatives. There is another basic rule, also known as chain rule. They provide a way to composite function.