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Question

Question: How do you use the chain rule to differentiate \[f\left( x \right)=\cos \left( \ln x \right)\]?...

How do you use the chain rule to differentiate f(x)=cos(lnx)f\left( x \right)=\cos \left( \ln x \right)?

Explanation

Solution

We have been asked to differentiate f(x)=cos(lnx)f\left( x \right)=\cos \left( \ln x \right) using the chain rule. For solving this we have to use the chain rule It states that, the derivative will be equal to the derivative of the outside function with respect to the inside function, times the derivative of the inside function which is mathematically given as f(x)=y(u(x)).u(x)f'\left( x \right)={y}'\left( u\left( x \right) \right).{u}'\left( x \right) where f(x)=y(u(x))f\left( x \right)=y\left( u\left( x \right) \right) .

Complete step by step answer:
From the question given, we have been given that f(x)=cos(lnx)f\left( x \right)=\cos \left( \ln x \right)
First of all, to differentiate the given function using chain rule, we have to know about the chain rule. After knowing about the chain rule, we have to differentiate the given function using the chain rule.
Chain rule:
In differential calculus, we use the chain rule when we have a composite function. It states that, the derivative will be equal to the derivative of the outside function with respect to the inside function, times the derivative of the inside function.
Let us see what that looks like mathematically y(u(x)).u(x){y}'\left( u\left( x \right) \right).{u}'\left( x \right)
By using the above chain rule formula, we have to differentiate the given function.
Let us assume that y=cosuy=\cos u and u=lnxu=\ln x
Then the derivatives of above assumptions are y=sinu\Rightarrow {y}'=-\sin u and u=1x\Rightarrow {u}'=\dfrac{1}{x}
We got all the values which we need to substitute in the chain rule formula.
Substitute all the values we got in the chain rule formula, by substituting the all the values we got in the chain rule formula, we get
f(x)=sinu×1x\Rightarrow {f}'\left( x \right)=-\sin u\times \dfrac{1}{x}
f(x)=sin(lnx)×1x\Rightarrow {f}'\left( x \right)=-\sin \left( \ln x \right)\times \dfrac{1}{x}
f(x)=sin(lnx)x\Rightarrow {f}'\left( x \right)=\dfrac{-\sin \left( \ln x \right)}{x}
Hence, we got the derivative for the given function by using the chain rule.

Note:
We should be well known about the chain rule and its formula. Also, we should be well known about the usage of the chain rule formula in differentiating the function. Also, we should be very careful while doing the calculation part after applying the chain rule formula. Similarly we have many formulae like for f(x)=u(x)×v(x)f\left( x \right)=u\left( x \right)\times v\left( x \right) it will be f(x)=u(x)×v(x)+u(x)×v(x)f'\left( x \right)=u'\left( x \right)\times v\left( x \right)+u\left( x \right)\times v'\left( x \right) and many more.