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Question: How do you take the derivative of\[\tan (\sqrt x )\]?...

How do you take the derivative oftan⁡(x)\tan (\sqrt x )?

Explanation

Solution

In such trigonometric equation you have to simplify the given bracket and then can move to direct solution, because we know the direct differentiation of the trigonometric identity, and since brackets have an algebraic expression so it also needs to be differentiated along with the identity.

Formulae Used:
ddx(tanx) = cotx\dfrac{{\text{d}}}{{{\text{dx}}}}\left( {{\text{tanx}}} \right){\text{ = cotx}}

Complete step by step solution:
For the given equation tan⁡(x)\tan (\sqrt x )

We have to differentiate both the terms trigonometric identity as well as the algebraic expression written under it, on solving we get:

⇒tan⁡(x)=tan⁡x12 \Rightarrow \tan (\sqrt x ) = \tan {x^{\dfrac{1}{2}}}

On putting derivative side with respect to “x” and differentiating first the trigonometric identity then along with the algebraic term, both the identity result should be in product form because in question also we are having both the terms in product, and similarly we have to solve this. We get:

{\dfrac{1}{2}{x^{\dfrac{1}{2} - 1}}} \right) = \cot {x^{\dfrac{1}{2}}}\left( {\dfrac{1}{2}{x^{\dfrac{{1 - 2}}{2}}}} \right) = \cot {x^{\dfrac{1}{2}}}\left( {\dfrac{1}{2}{x^{\dfrac{{ - 1}}{2}}}} \right) = \cot {x^{\dfrac{1}{2}}}\left( {\dfrac{1}{{2\sqrt x }}} \right) = \dfrac{1}{{2\sqrt x }}\cot {x^{\dfrac{1}{2}}}$$ **This is our required differentiation for the given term in question.** **Additional Information:** Here you can convert the identity also but that process will become more complicated as compared to this, but the result obtained will be the same, that is if you check for any angle you would get the same result from that differentiation, and what we have done here. **Note:** In order to deal with differentiation you need to be very clear with the basics of differentiation and especially when you are dealing with trigonometric identities then you must have to learn the basic differentiation formula for all the identities we are having in trigonometry, then only you can proceed with the question.