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Question: What is the derivative of \( f\left( x \right) = \sin \left( {\ln x} \right) \) ?...

What is the derivative of f(x)=sin(lnx)f\left( x \right) = \sin \left( {\ln x} \right) ?

Explanation

Solution

Hint : Here, we have to find the derivative of f(x)=sin(lnx)f\left( x \right) = \sin \left( {\ln x} \right) . Here, there are two functions in the given function. So, we will need to use the chain rule to find its derivative. According to chain rule, we need to differentiate both these functions and then multiply them in order to find the derivative of the main function. The chain rule is expressed as
ddx[f(g(x))]=f(g(x))g(x)\dfrac{d}{{dx}}\left[ {f\left( {g\left( x \right)} \right)} \right] = f'\left( {g\left( x \right)} \right)g'\left( x \right)

Complete step by step solution:
In this question, we have to find the derivative of f(x)=sin(lnx)f\left( x \right) = \sin \left( {\ln x} \right) .
Here, we are given two functions in the given function, sinlnx\sin \ln x and lnx\ln x .
That means, we cannot differentiate the given function directly.
We have to use the chain rule for differentiating the above function. According to the chain rule, when two functions are given, we differentiate one function and then multiply it by the derivative of the second function. The chain rule is expressed as
ddx[f(g(x))]=f(g(x))g(x)\Rightarrow \dfrac{d}{{dx}}\left[ {f\left( {g\left( x \right)} \right)} \right] = f'\left( {g\left( x \right)} \right)g'\left( x \right)
According to our question,
f(x)=sin(lnx)f\left( x \right) = \sin \left( {\ln x} \right) and g(x)=lnxg\left( x \right) = \ln x
So, we have to differentiate both sin(lnx)\sin \left( {\ln x} \right) and lnx\ln x and then multiply them to get the derivative of the given function. Therefore,
f(x)=sin(lnx)\Rightarrow f\left( x \right) = \sin \left( {\ln x} \right)
Now, we know that the derivative of sinx\sin x is cosx\cos x and the derivative of lnx\ln x is 1x\dfrac{1}{x} . Therefore,
f(x)=ddxsin(lnx) f(x)=cos(lnx)1x f(x)=cos(lnx)x   \Rightarrow f'\left( x \right) = \dfrac{d}{{dx}}\sin \left( {\ln x} \right) \\\ \Rightarrow f'\left( x \right) = \cos \left( {\ln x} \right) \cdot \dfrac{1}{x} \\\ \Rightarrow f'\left( x \right) = \dfrac{{\cos \left( {\ln x} \right)}}{x} \;
Hence, the derivative of the given function f(x)=sin(lnx)f\left( x \right) = \sin \left( {\ln x} \right) is cos(lnx)x\dfrac{{\cos \left( {\ln x} \right)}}{x} .
So, the correct answer is “ cos(lnx)x\dfrac{{\cos \left( {\ln x} \right)}}{x} ”.

Note : We learned the chain rule in this question, but there are two more important rules that we should learn. These two rules are
Product rule
Quotient rule
Product rule: The product rule tells us how to differentiate the product of two different functions.
\Rightarrow ddxuv=udvdx+vdudx\dfrac{d}{{dx}}uv = u\dfrac{{dv}}{{dx}} + v\dfrac{{du}}{{dx}}
For Example: ddx2xcosx=2xddxcosx+cosxddx2x=2xsinx+2cosx\dfrac{d}{{dx}}2x\cos x = 2x\dfrac{d}{{dx}}\cos x + \cos x\dfrac{d}{{dx}}2x = - 2x\sin x + 2\cos x
Quotient rule: Quotient rule is used for differentiating problems when one function is divided by another function.
ddx(uv)=vddxuuddxvv2\Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{u}{v}} \right) = \dfrac{{v\dfrac{d}{{dx}}u - u\dfrac{d}{{dx}}v}}{{{v^2}}}
For example: ddx(2xcosx)=cosxddx2x2xddxcosxcos2x=2cosx+2xsinxcos2x\dfrac{d}{{dx}}\left( {\dfrac{{2x}}{{\cos x}}} \right) = \dfrac{{\cos x\dfrac{d}{{dx}}2x - 2x\dfrac{d}{{dx}}\cos x}}{{{{\cos }^2}x}} = \dfrac{{2\cos x + 2x\sin x}}{{{{\cos }^2}x}}