Question

Find the differential coefficient of the given function ${\log _e}\sqrt {\dfrac{{1 + \sin x}}{{1 - \sin x}}}$ is

$$\left( 1 \right){\text{ cosecx}} \\\ \left( 2 \right){\text{ secx}} \\\ \left( 3 \right){\text{ tanx}} \\\ \left( 4 \right){\text{ cosx}} \\\$$

Answer 1

First of all we have to define the properties of the logarithmic function we are using to solve the given problem
${\log _e}{a^b} = b\log a$
$\log \left( {\dfrac{a}{b}} \right) = \log a - \log b$
After that use the rule of derivatives
Product Rule:- $\dfrac{d}{{dx}}\left( {f\left( x \right)g\left( x \right)} \right) = \dfrac{{df\left( x \right)}}{{dx}}.g\left( x \right) + f\left( x \right)\dfrac{{dg\left( x \right)}}{{dx}}$
Sum Rule:- $d\left( {f\left( x \right) + g\left( x \right)} \right) = d\left( {f\left( x \right)} \right) + d\left( {g\left( x \right)} \right)$
Constant Rule:- $\dfrac{d}{{dx}}\left( {cf\left( x \right)} \right) = c\dfrac{d}{{dx}}f\left( x \right)$
And lastly use the derivatives of the standard functions

$$d\left( {\sin x} \right) = \cos x \\\ d\left( {\log x} \right) = \dfrac{1}{x} \\\$$

Complete step by step answer:
The function is given as ${\log _e}\sqrt {\dfrac{{1 + \sin x}}{{1 - \sin x}}}$
Let the function be equal to $y$ then
$y = {\log _e}\sqrt {\dfrac{{1 + \sin x}}{{1 - \sin x}}}$
It can also be written as
$y = {\log _e}{\left( {\dfrac{{1 + \sin x}}{{1 - \sin x}}} \right)^{\dfrac{1}{2}}}$
Using the property of logarithmic function that is ${\log _e}{a^b} = b\log a$
Here $a = \dfrac{{1 + \sin x}}{{1 - \sin x}}$ and $b = \dfrac{1}{2}$
$y = \dfrac{1}{2}{\log _e}\left( {\dfrac{{1 + \sin x}}{{1 - \sin x}}} \right)$
Again, using another property of logarithmic function that is $\log \left( {\dfrac{a}{b}} \right) = \log a - \log b$
Here $a = 1 + \sin x$ and $b = 1 - \sin x$
$y = \dfrac{1}{2}\left( {\log \left( {1 + \sin x} \right) - \log \left( {1 - \sin x} \right)} \right)$
Now, on differentiating with respect to $x$we get
$\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {\dfrac{1}{2}\left( {\log \left( {1 + \sin x} \right) - \log \left( {1 - \sin x} \right)} \right)} \right)$
Using the derivative rule
$\dfrac{d}{{dx}}\left( {f\left( x \right) - g\left( x \right)} \right) = \dfrac{d}{{dx}}f\left( x \right) - \dfrac{d}{{dx}}g\left( x \right)$and
$\dfrac{d}{{dx}}\left( {cf\left( x \right)} \right) = c\dfrac{d}{{dx}}f\left( x \right)$ we get

$$\dfrac{{dy}}{{dx}} = \dfrac{1}{2}\dfrac{d}{{dx}}\left( {\log \left( {1 + \sin x} \right) - \log \left( {1 - \sin x} \right)} \right) \\\ \dfrac{{dy}}{{dx}} = \dfrac{1}{2}\left( {\dfrac{d}{{dx}}\log \left( {1 + \sin x} \right) - \dfrac{d}{{dx}}\log \left( {1 - \sin x} \right)} \right) \\\$$

Using the derivative of a standard function that is

$$d\left( {\sin x} \right) = \cos x \\\ d\left( {\log x} \right) = \dfrac{1}{x} \\\$$

We get

$$\dfrac{{dy}}{{dx}} = \dfrac{1}{2}\left( {\dfrac{1}{{1 + \sin x}} \times \dfrac{d}{{dx}}\left( {1 + \sin x} \right) - \dfrac{1}{{1 - \sin x}} \times \dfrac{d}{{dx}}\left( {1 - \sin x} \right)} \right) \\\ \dfrac{{dy}}{{dx}} = \dfrac{1}{2}\left( {\dfrac{1}{{1 + \sin x}}\left( {0 + \cos x} \right) - \dfrac{1}{{1 - \sin x}}\left( {0 - \cos x} \right)} \right) \\\$$

On simplifying we get

$$\dfrac{{dy}}{{dx}} = \dfrac{1}{2}\left( {\dfrac{{\cos x}}{{1 + \sin x}} + \dfrac{{\cos x}}{{1 - \sin x}}} \right) \\\ \\\$$

Taking L.C.M on the right-hand side of the equation

$$\dfrac{{dy}}{{dx}} = \dfrac{1}{2}\left( {\dfrac{{\cos x\left( {1 - \sin x} \right) + \cos x\left( {1 + \sin x} \right)}}{{\left( {1 + \sin x} \right)\left( {1 - \sin x} \right)}}} \right) \\\ \\\$$

Taking common $\cos x$ from the right-hand side of the equation
$\dfrac{{dy}}{{dx}} = \dfrac{1}{2}\left( {\cos x} \right)\left( {\dfrac{{1 - \sin x + 1 + \sin x}}{{1 - {{\sin }^2}x}}} \right)$
$\dfrac{{dy}}{{dx}} = \dfrac{1}{2}\left( {\cos x} \right)\left( {\dfrac{2}{{1 - {{\sin }^2}x}}} \right)$
Using trigonometric identity $1 - {\sin ^2}x = {\cos ^2}x$
$\dfrac{{dy}}{{dx}} = \dfrac{1}{2}\left( {\cos x} \right)\left( {\dfrac{2}{{{{\cos }^2}x}}} \right)$
On simplifying we get

$$\dfrac{{dy}}{{dx}} = \dfrac{{\cos x}}{{{{\cos }^2}x}} \\\ = \dfrac{1}{{\cos x}} \\\$$

We know that $\sec x = \dfrac{1}{{\cos x}}$
Using this in the equation we get

$$\dfrac{{dy}}{{dx}} = \dfrac{1}{{\cos x}} = \sec x \\\ \dfrac{{dy}}{{dx}} = \sec x \\\$$

Put the value of $y$ that is $y = {\log _e}{\left( {\dfrac{{1 + \sin x}}{{1 - \sin x}}} \right)^{\dfrac{1}{2}}}$ in the above equation we get
$\dfrac{{d\left( {lo{g_e}{{\left( {\dfrac{{1 + \sin x}}{{1 - \sin x}}} \right)}^{\dfrac{1}{2}}}} \right)}}{{dx}} = \sec x$
Hence, the differential coefficient of the given function ${\log _e}\sqrt {\dfrac{{1 + \sin x}}{{1 - \sin x}}}$is $\sec x$

So, the correct answer is “Option 2”.

Note:
In the question, we cannot directly differentiate with respect to $x$ otherwise it gets complicated. Before differentiation, we have to use the properties of the logarithmic function. We can directly solve by differentiating the given function but it is a lengthy one.
Trigonometric identities are equalities that involve trigonometric functions and are useful whenever trigonometric functions are involved in an expression or an equation. Six basic trigonometric ratios are sine, cosine, tangent, cosecant, secant, and cotangent and all the fundamental trigonometric identities are derived from the six trigonometric ratios.