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

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

Explanation

Solution

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

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

Note : Differentiation can be defined as the rate of change of one variable with respect to another variable. If we are differentiating yy with respect to xx , we get the measure of how the value of yy changes with respect to xx . 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.