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Question: Differentiate the following function w.r.t.x: \(\tan ({\sin ^{ - 1}}x)\)...

Differentiate the following function w.r.t.x:
tan⁡(sin⁡−1x)\tan ({\sin ^{ - 1}}x)

Explanation

Solution

Hint : To solve the given complex function, we will apply the chain rule of differentiation. As, there are two parts of the given function, we will start with assuming the parts of function, such as y=tan⁡(sin⁡−1x).y = \tan ({\sin ^{ - 1}}x). Similarly, u=(sin⁡−1x).u = ({\sin ^{ - 1}}x).Then on differentiating all the parts of the function with respect to the variable, and then finally we will connect all the values, to get our required answer.

Complete step-by-step answer :
We have been given a complex function tan⁡(sin⁡−1x).\tan ({\sin ^{ - 1}}x). We will solve the given function using the chain rule of differentiation, because, as we can see one function is inside of another function.
We know that, the general formula of chain rule of differentiation is,
⇒dydx=dydu×dudx.....eq.(1)\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{du}} \times \dfrac{{du}}{{dx}}.....eq.(1)
Let the given function be
⇒y=tan⁡(sin⁡−1x)....eq.(2)\Rightarrow y = \tan ({\sin ^{ - 1}}x)....eq.(2)
We can see in the function, that (sin⁡−1x)({\sin ^{ - 1}}x) is a function of tan function.
Now, let u=(sin⁡−1x)....eq.(3)u = ({\sin ^{ - 1}}x)....eq.(3)
So, we get eq.(2)eq.(2) as y=tan⁡uy = \tan u
So, on differentiating y=tan⁡uy = \tan u, with respect to u, we get
⇒dydu=sec⁡2u.................(∵d(tan⁡x)dx=sec⁡2x)\Rightarrow \dfrac{{dy}}{{du}} = {\sec ^2}u.................(\because \dfrac{{d(\tan x)}}{{dx}} = {\sec ^2}x)
Also differentiating u=(sin⁡−1x),u = ({\sin ^{ - 1}}x), with respect to x, we get
⇒dudx=11−x2.....................(∵d(sin⁡−1x)dx=11−x2)\Rightarrow \dfrac{{du}}{{dx}} = \dfrac{1}{{\sqrt {1 - {x^2}} }}.....................(\because \dfrac{{d({{\sin }^{ - 1}}x)}}{{dx}} = \dfrac{1}{{\sqrt {1 - {x^2}} }})
Now, on applying these values in eq. (1),eq.{\text{ (}}1), we get
⇒dydx=dydu×dudx ⇒dydx=sec⁡2u×11−x2  \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{du}} \times \dfrac{{du}}{{dx}} \\\ \Rightarrow \dfrac{{dy}}{{dx}} = {\sec ^2}u \times \dfrac{1}{{\sqrt {1 - {x^2}} }} \\\
On putting the value of u in above equation, we get
⇒dydx=sec⁡2(sin⁡−1x)×11−x2\Rightarrow \dfrac{{dy}}{{dx}} = {\sec ^2}({\sin ^{ - 1}}x) \times \dfrac{1}{{\sqrt {1 - {x^2}} }}
⇒dydx=sec⁡2(sin⁡−1x)1−x2\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{{{\sec }^2}({{\sin }^{ - 1}}x)}}{{\sqrt {1 - {x^2}} }}
Thus, dydx=sec⁡2(sin⁡−1x)1−x2\dfrac{{dy}}{{dx}} = \dfrac{{{{\sec }^2}({{\sin }^{ - 1}}x)}}{{\sqrt {1 - {x^2}} }} is our required answer.

Note : Differentiation is the process of finding the derivative of a function. We were given a complex function, for that chain rule of differentiation is used, since the function is in the form of f(g(x))f\left( {g\left( x \right)} \right), i.e., one function is inside of one another function.
One tip for students is to learn basic formulas of differentiation by daily practising, so that while solving these types of questions you don’t get stuck in between.