Question

How do you find \[\dfrac{{dy}}{{dx}}\] given \[y = \ln \left( {\cos x} \right)\] ?

Answer 1

Hint: We are asked to differentiate \[y\] with respect to \[x\] . The value of \[y\] is given. Observe the terms involved in \[y\] and recall the concepts of differentiation. Also, the cosine term is present in \[y\] so you will need to recall the value of differentiation of cosine.

Complete step-by-step answer:
Given, \[y = \ln \left( {\cos x} \right)\] .
We are asked to differentiate \[y\] with respect to \[x\] .
Let \[\cos x = z\]
Then \[y\] can be written as,
 \[y = \ln z\] (i)
If we have a function \[f(z)\] and we differentiate it with respect to \[x\] we write it as,
 \[\dfrac{d}{{dx}}\left( {f(z)} \right) = \dfrac{{d\left( {f\left( z \right)} \right)}}{{dz}}\dfrac{{dz}}{{dx}}\]
Using this concept differentiating \[y\] with respect to \[x\] we get,
 \[\dfrac{{dy}}{{dx}} = \dfrac{{d\left( {\ln z} \right)}}{{dz}}\dfrac{{dz}}{{dx}}\] (ii)
Differentiating \[\ln x\] with respect to \[x\] we get \[\dfrac{1}{x}\] . So,
 \[\dfrac{{d\left( {\ln z} \right)}}{{dz}} = \dfrac{1}{z}\]
Putting this value in equation (ii) we get,
 \[\dfrac{{dy}}{{dx}} = \dfrac{1}{z}\dfrac{{dz}}{{dx}}\]
Putting the value of \[z\] in the above equation we get,
 \[\dfrac{{dy}}{{dx}} = \dfrac{1}{{\cos x}}\dfrac{{d\left( {\cos x} \right)}}{{dx}}\]
Differentiating \[\cos x\] with respect to \[x\] we get \[ - \sin x\] . So,
 \[\dfrac{{dy}}{{dx}} = \dfrac{1}{{\cos x}}\left( { - \sin x} \right)\]
 \[ \Rightarrow \dfrac{{dy}}{{dx}} = - \dfrac{{\sin x}}{{\cos x}}\]
 \[ \Rightarrow \dfrac{{dy}}{{dx}} = - \tan x\]
Therefore, the value of \[\dfrac{{dy}}{{dx}}\] is \[ - \tan x\] .
So, the correct answer is “ \[ - \tan x\] ”.

Note: Differentiation can be defined as the rate of change of one variable with respect to another variable. If we are differentiating \[y\] with respect to \[x\] , we get the measure of how the value of \[y\] changes with respect to \[x\] . Remember the derivatives of important functions such as logarithm, exponential and trigonometric functions. Also, remember there are three main functions in trigonometry, these are sine, cosine and tangent. There are three other trigonometric functions which can be written in terms of the main functions, these are cosecant which is inverse of sine, secant which is inverse of cosine and cotangent which is inverse of tangent. Also, while solving questions related to trigonometry, you should always remember the basic formulas of trigonometry.