
The differential coefficient of $\log (\tan x)$ is
$
{\text{A}}{\text{. 2sec2}}x \\
{\text{B}}{\text{. 2cosec2}}x \\
{\text{C}}{\text{. 2se}}{{\text{c}}^2}x \\
{\text{D}}{\text{. 2cose}}{{\text{c}}^2}x \\
$
Answer
614.7k+ views
Hint:- In this question first we need to let the given function equal to $y$. Then, we have to find a differential coefficient which is also called derivative $\{ \dfrac{{d(f(x))}}{{dx}}\} $. After finding the derivative we have to convert it into form which matches with one of the given options.
Complete step-by-step answer:
Let $y = $$\log (\tan x)$. --- Eq.1
Now the differential coefficient of $y$ is given by
$ \Rightarrow \dfrac{{dy}}{{dx}}$
Now on differentiating eq.1 with respect to $x$
We get
$
\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{d\{ \log (\tan x)\} }}{{dx}} \\
\\
$ ---- eq.2
We know
$ \Rightarrow \dfrac{{d\log x}}{{dx}} = \dfrac{1}{x}$ ---- eq.3
$ \Rightarrow \dfrac{{d\tan x}}{{dx}} = {\sec ^2}x$ ----eq.4
Using eq.3 and eq.4 and chain rule, we get
$ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{\tan x}}.{\sec ^2}x$
We know $\tan x = \dfrac{{\sin x}}{{\cos x}}{\text{ and se}}{{\text{c}}^2}x = \dfrac{1}{{{{\cos }^2}x}}$
The above equation can be written as
$
\Rightarrow \dfrac{{dy}}{{dx}} = \left( {\dfrac{{\cos x}}{{\sin x}}} \right).\dfrac{1}{{{{\cos }^2}x}} \\
\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{\sin x.\cos x}} \\
$
Multiply “2” in both denominator and numerator we get
$ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{2}{{2\sin x.\cos x}}$ ---- eq.5
We know,
$ \Rightarrow 2\sin x.\cos x = \sin 2x$
And$\dfrac{1}{{\sin 2x}} = {\text{cosec2}}x$
Using above equations we can rewrite eq.5 as
$ \Rightarrow \dfrac{{dy}}{{dx}} = 2{\text{cosec2}}x$
Therefore, the differential coefficient of $\log (\tan x)$ is $2{\text{cosec2}}x$.
Hence option B. is correct.
Note:- Whenever you get this type of question the key concept to solve this is to learn concept the differential coefficient (derivative$\{ \dfrac{{d(f(x))}}{{dx}}\} $) and the derivatives of most basic functions . And remember one more thing that a coefficient is usually a constant quantity, but the differential coefficient of function is a constant function only if function is a linear function.
Complete step-by-step answer:
Let $y = $$\log (\tan x)$. --- Eq.1
Now the differential coefficient of $y$ is given by
$ \Rightarrow \dfrac{{dy}}{{dx}}$
Now on differentiating eq.1 with respect to $x$
We get
$
\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{d\{ \log (\tan x)\} }}{{dx}} \\
\\
$ ---- eq.2
We know
$ \Rightarrow \dfrac{{d\log x}}{{dx}} = \dfrac{1}{x}$ ---- eq.3
$ \Rightarrow \dfrac{{d\tan x}}{{dx}} = {\sec ^2}x$ ----eq.4
Using eq.3 and eq.4 and chain rule, we get
$ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{\tan x}}.{\sec ^2}x$
We know $\tan x = \dfrac{{\sin x}}{{\cos x}}{\text{ and se}}{{\text{c}}^2}x = \dfrac{1}{{{{\cos }^2}x}}$
The above equation can be written as
$
\Rightarrow \dfrac{{dy}}{{dx}} = \left( {\dfrac{{\cos x}}{{\sin x}}} \right).\dfrac{1}{{{{\cos }^2}x}} \\
\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{\sin x.\cos x}} \\
$
Multiply “2” in both denominator and numerator we get
$ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{2}{{2\sin x.\cos x}}$ ---- eq.5
We know,
$ \Rightarrow 2\sin x.\cos x = \sin 2x$
And$\dfrac{1}{{\sin 2x}} = {\text{cosec2}}x$
Using above equations we can rewrite eq.5 as
$ \Rightarrow \dfrac{{dy}}{{dx}} = 2{\text{cosec2}}x$
Therefore, the differential coefficient of $\log (\tan x)$ is $2{\text{cosec2}}x$.
Hence option B. is correct.
Note:- Whenever you get this type of question the key concept to solve this is to learn concept the differential coefficient (derivative$\{ \dfrac{{d(f(x))}}{{dx}}\} $) and the derivatives of most basic functions . And remember one more thing that a coefficient is usually a constant quantity, but the differential coefficient of function is a constant function only if function is a linear function.
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