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

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

Explanation

Solution

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

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

Thus,dydx=−cos⁡(tan⁡−1e−x)1+(e−x)2(e−x)\dfrac{{dy}}{{dx}} = \dfrac{{ - \cos ({{\tan }^{ - 1}}{e^{ - x}})}}{{1 + {{({e^{ - x}})}^2}}}({e^{ - x}}) is our required answer.

Note:
Here in the question, we are asked about the differentiation, let us understand about it in detail. So, differentiation is the process of finding the derivative or rate of change of a function. In complex functions, like what we are given here, we need to use the chain rule of differentiation, just like we did here.
But there is a confusion which mainly occurs when to use the chain rule, so, just check simply whether the function is in the form of f(g(x))f\left( {g\left( x \right)} \right) or not. If it is, then apply the rule undoubtedly.