Question

If $y = {\log _{\cos x}}\sin x$, then $\dfrac{{dy}}{{dx}}$ is equal to:
(A) $\dfrac{{\left[ {\cot x\log \left( {\cos x} \right) + \tan x\log \left( {\sin x} \right)} \right]}}{{{{\left( {\log \left( {\cos x} \right)} \right)}^2}}}$
(B) $\dfrac{{\left[ {\tan x\log \left( {\cos x} \right) + \cot x\log \left( {\sin x} \right)} \right]}}{{{{\left( {\log \left( {\cos x} \right)} \right)}^2}}}$
(C) $\dfrac{{\left[ {\cot x\log \left( {\cos x} \right) + \tan x\log \left( {\sin x} \right)} \right]}}{{{{\left( {\log \left( {\sin x} \right)} \right)}^2}}}$
(D) None of these

Answer 1

Hint: In the given problem, we are required to differentiate $y = {\log _{\cos x}}\sin x$ with respect to x. Since, $y = {\log _{\cos x}}\sin x$ is a composite function, so we will have to apply chain rule of differentiation in the process of differentiating $y = {\log _{\cos x}}\sin x$ . So, differentiation of $y = {\log _{\cos x}}\sin x$ with respect to x will be done layer by layer using the chain rule of differentiation. The derivative of \[\log \left( x \right)\] with respect to x must be remembered. We will also use the quotient rule of differentiation.

Complete answer: So, we have, $y = {\log _{\cos x}}\sin x$.
Now, we first simplify the logarithmic function using the law of logarithm $\dfrac{{\log x}}{{\log y}} = {\log _y}x$. So, we get,
 $ \Rightarrow y = \dfrac{{\log \sin x}}{{\log \cos x}}$
Now, we have to differentiate both the sides with respect to x, we get,
$ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left[ {\dfrac{{\log \sin x}}{{\log \cos x}}} \right]$
Now, we use the quotient rule of differentiation $\dfrac{d}{{dx}}\left[ {\dfrac{{f\left( x \right)}}{{g\left( x \right)}}} \right] = \dfrac{{g\left( x \right)\dfrac{d}{{dx}}\left[ {f\left( x \right)} \right] + f\left( x \right)\dfrac{d}{{dx}}\left[ {g\left( x \right)} \right]}}{{{{\left[ {g\left( x \right)} \right]}^2}}}$. So, we get,
$ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{\log \cos x\dfrac{d}{{dx}}\left[ {\log \sin x} \right] - \left[ {\log \sin x} \right]\dfrac{d}{{dx}}\left[ {\log \cos x} \right]}}{{{{\left( {\log \cos x} \right)}^2}}}$
Now, we know that derivative of $\log x$ with respect to x is $\left( {\dfrac{1}{x}} \right)$. So, using chain rule of differentiation \[\dfrac{d}{{dx}}\left[ {f\left( {g\left( x \right)} \right)} \right] = f'\left( {g\left( x \right)} \right)g'\left( x \right)\], we get,
$ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{\log \cos x\left( {\dfrac{1}{{\sin x}}} \right)\left( {\cos x} \right) - \left[ {\log \sin x} \right]\left( {\dfrac{1}{{\cos x}}} \right)\left( { - \sin x} \right)}}{{{{\left( {\log \cos x} \right)}^2}}}$
Now, using the trigonometric formulae $\cot x = \dfrac{{\cos x}}{{\sin x}}$ and $\tan x = \dfrac{{\sin x}}{{\cos x}}$ to simplify the expression, we get,
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{\cot x\log \cos x + \tan x\log \sin x}}{{{{\left( {\log \cos x} \right)}^2}}}\]
So, the derivative of $y = {\log _{\cos x}}\sin x$ is \[\dfrac{{\cot x\log \cos x + \tan x\log \sin x}}{{{{\left( {\log \cos x} \right)}^2}}}\].
So, option (A) is the correct answer.

Note:
The given problem may also be solved using the first principle of differentiation. The derivatives of basic functions must be learned by heart in order to find derivatives of complex composite functions using chain rule of differentiation. The chain rule of differentiation involves differentiating a composite by introducing new unknowns to ease the process and examine the behaviour of function layer by layer. One must know the quotient rule of differentiation to tackle similar problems.