Question

Differentiate ln(tan $x$ ) with respect to ${\sin ^{ - 1}}({e^x})$ .
(A) $\dfrac{{{e^{ - x}}\sqrt {1 - {e^{2x}}} }}{{\sin x\,\,.\,\,\cos x}}\,$
(B) $\dfrac{{{e^{ - x}}\sqrt {1 - {e^{2x}}} }}{{\sin x\,\,.\,\,\cot x}}\,$
(C) $\dfrac{{{e^{ - x}}\sqrt {1 + {e^{2x}}} }}{{\sin x\,\,.\,\,\cos x}}\,$
(D) $\dfrac{{{e^{ - x}}\sqrt {1 + {e^{2x}}} }}{{\sin x\,\,.\,\,\sec x}}\,$

Answer 1

Hint : Whenever we need to differentiate one trigonometric function with respect to another. For this type of problem we first let one function as ‘u’ and the other function as ‘v’. Then dividing their derivative to obtain required differentiation of the functions with respect to each other.
Formulas used: $\dfrac{d}{{dx}}\ln (A) = \dfrac{1}{A}\dfrac{d}{{dx}}(A),\,\,\,\dfrac{d}{{dx}}{\sin ^{ - 1}}A = \dfrac{1}{{\sqrt {1 - {A^2}} }}\dfrac{d}{{dx}}(A),\,\,\dfrac{d}{{dx}}\tan x = {\sec ^2}x$ and $\dfrac{d}{{dx}}({e^x}) = {e^x}$

Complete step-by-step answer :
Consider u = $\ln (\tan x)$ and v = ${\sin ^{ - 1}}({e^x})$
Now, differentiating ‘u’ with respect to x. We have,
Let $u{\text{ }} = {\text{ ln}}\left( {tanx} \right)$

\dfrac{{du}}{{dx}} = \dfrac{d}{{dx}}\left\\{ {\ln \left( {\tan x} \right)} \right\\} \\\ \Rightarrow \dfrac{{du}}{{dx}} = \dfrac{1}{{\tan x}}.{\sec ^2}x \\\ \Rightarrow \dfrac{{du}}{{dx}} = \dfrac{{\cos x}}{{\sin x}}.\dfrac{1}{{{{\cos }^2}x}} \\\ \Rightarrow \dfrac{{du}}{{dx}} = \dfrac{1}{{\sin x.\cos x}}..................(i) \\\

Now, differentiating ‘v’ with respect of x. We have
\dfrac{{dv}}{{dx}} = \dfrac{d}{{dx}}\left\\{ {{{\sin }^{ - 1}}({e^x})} \right\\} \\\ \Rightarrow \dfrac{{dv}}{{dx}} = \dfrac{1}{{\sqrt {1 - {{({e^x})}^2}} }}.\dfrac{d}{{dx}}({e^x})\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\\{ {\because \dfrac{d}{{dx}}{{\sin }^{ - 1}}A = \dfrac{1}{{\sqrt {1 - {A^2}} }}.\dfrac{d}{{dx}}(A)} \right\\} \\\
$\Rightarrow \dfrac{{dv}}{{dx}} = \dfrac{1}{{\sqrt {1 - {e^{2x}}} }}({e^x}) \\\ or \\\ \dfrac{{dv}}{{dx}} = \dfrac{{{e^x}}}{{\sqrt {1 - {e^{2x}}} }}................................(ii) \\\$
Now, dividing equation (i) by equation (ii) we have
$\dfrac{{\dfrac{{du}}{{dx}}}}{{\dfrac{{dv}}{{dx}}}} = \dfrac{{\dfrac{1}{{\sin x.\cos x}}}}{{\dfrac{{{e^x}}}{{\sqrt {1 - {e^{2x}}} }}}} \\\ \Rightarrow \dfrac{{du}}{{dv}} = \dfrac{1}{{\sin x.\cos x}}.\dfrac{{\sqrt {1 - {e^{2x}}} }}{{{e^x}}} \;$
Or we can write
$\dfrac{{du}}{{dx}} = \dfrac{{{e^{ - x}}\sqrt {1 - {e^{2x}}} }}{{\sin x.\cos x}}$
Therefore, from above we see that derivative of function $\ln \left( {\tan x} \right)$ with respect to the function ${\sin ^{ - 1}}({e^x})$ is $\dfrac{{{e^{ - x}}\sqrt {1 - {e^{2x}}} }}{{\sin x.\cos x}}$ .
Hence, from the given four options we see that the correct option is (A).
So, the correct answer is “Option A”.

Note : For inverse trigonometric functions we can differentiate them in two ways. First we can apply direct differentiating inverse formulas or taking inverse functions as other variables and solving them in terms of other variables and finally simplifying to get its derivative.