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Question

Question: How to differentiate \( \ln \left( {{x^2} + 1} \right) \)...

How to differentiate ln(x2+1)\ln \left( {{x^2} + 1} \right)

Explanation

Solution

Hint : There are two different formulas you need to use to solve this problem. One is the basic differentiation formula and the other one is to differentiate ln. The basic differentiation formula is xn=nxn1{x^n} = n{x^{n - 1}} and then if we differentiate lnx\ln x we get, 1x\dfrac{1}{x} . Keep this in mind and solve the problem.

Complete step-by-step answer :
Here we have more than one function, so we consider this below mentioned rule, which is known as chain rule. Here we take f(g(x))f(g(x)) as ln(x2+1)\ln ({x^2} + 1) . First differentiate ln(x2+1)\ln ({x^2} + 1) and then differentiate x2+1{x^2} + 1 separately. When we differentiate ln(x2+1)\ln ({x^2} + 1) we get, 1x2+1\dfrac{1}{{{x^2} + 1}} and when we differentiate (x2+1)({x^2} + 1) we get (2x+0)(2x + 0) , as we know if we differentiate a constant term, it will be equal to zero. The formula for chain rule is given by,
ddx[f(g(x))]=df(g(x))dxdg(x)dx d(ln(x2+1))dx=1x2+1(2x+0)=2xx2+1   \dfrac{d}{{dx}}\left[ {f\left( {g\left( x \right)} \right)} \right] = \dfrac{{df\left( {g\left( x \right)} \right)}}{{dx}}{\dfrac{{dg\left( x \right)}}{{dx}}^{^{}}} \\\ \dfrac{{d\left( {\ln \left( {{x^2} + 1} \right)} \right)}}{{dx}} = \dfrac{1}{{{x^2} + 1}}\left( {2x + 0} \right) = \dfrac{{2x}}{{{x^2} + 1}} \;
This is our required solution.
So, the correct answer is “ 2xx2+1\dfrac{{2x}}{{{x^2} + 1}} ”.

Note : Here in this problem, it has a natural log which is represented by ln\ln . The difference between ln\ln and log\log is, in natural ln\ln it has the base ee in it, while in the log, it has the base 1010 . log10{\log _{10}} tells you that what power does 1010 has to be raised to get a number xx and lne{\ln _e} , tells us that what power does ee has to be raised to get a number xx .
This chain rule is applied, when there is function of function in the given equation. It should be remembered forever in our lifetime, because it is important to know this formula to solve many complex equations. ln(x2+1)\ln ({x^2} + 1) is mentioned as function of function, because ln\ln is a function and (x2+1)({x^2} + 1) is also another function and hence we represent it as function of function. And for this case we use chain rule to solve this function.