Solveeit Logo

Question

Question: The differential coefficient of \(\log (\tan x)\) is \( {\text{A}}{\text{. 2sec2}}x \\\ {\...

The differential coefficient of log(tanx)\log (\tan x) is
A. 2sec2x B. 2cosec2x C. 2sec2x D. 2cosec2x  {\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 \\\

Explanation

Solution

Hint:- In this question first we need to let the given function equal to yy. Then, we have to find a differential coefficient which is also called derivative d(f(x))dx\\{ \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=y = log(tanx)\log (\tan x). --- Eq.1
Now the differential coefficient of yy is given by
dydx\Rightarrow \dfrac{{dy}}{{dx}}
Now on differentiating eq.1 with respect to xx
We get
dydx=dlog(tanx)dx   \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{d\\{ \log (\tan x)\\} }}{{dx}} \\\ \\\ ---- eq.2
We know
dlogxdx=1x\Rightarrow \dfrac{{d\log x}}{{dx}} = \dfrac{1}{x} ---- eq.3
dtanxdx=sec2x\Rightarrow \dfrac{{d\tan x}}{{dx}} = {\sec ^2}x ----eq.4
Using eq.3 and eq.4 and chain rule, we get
dydx=1tanx.sec2x\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{\tan x}}.{\sec ^2}x
We know tanx=sinxcosx and sec2x=1cos2x\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
dydx=(cosxsinx).1cos2x dydx=1sinx.cosx  \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
dydx=22sinx.cosx\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{2}{{2\sin x.\cos x}} ---- eq.5
We know,
2sinx.cosx=sin2x\Rightarrow 2\sin x.\cos x = \sin 2x
And1sin2x=cosec2x\dfrac{1}{{\sin 2x}} = {\text{cosec2}}x
Using above equations we can rewrite eq.5 as
dydx=2cosec2x\Rightarrow \dfrac{{dy}}{{dx}} = 2{\text{cosec2}}x
Therefore, the differential coefficient of log(tanx)\log (\tan x) is 2cosec2x2{\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 (derivatived(f(x))dx\\{ \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.