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Question: How do you find the derivative of \(\ln \left( {1 + {x^2}} \right)\)....

How do you find the derivative of ln⁡(1+x2)\ln \left( {1 + {x^2}} \right).

Explanation

Solution

This is a derivative of the composite function f(g(x))f\left( {g\left( x \right)} \right). We can calculate the derivative of this composite function with the help of following

Formula Used:
ddxf(g(x))=f′(g(x))⋅g′(x)\dfrac{d}{{dx}}f\left( {g\left( x \right)} \right) = f'\left( {g\left( x \right)} \right) \cdot g'\left( x \right)

Complete step by step solution:
In the derivative of composite function f(g(x))f\left( {g\left( x \right)} \right) the two functions are as follows:
f(x)=ln⁡(x) g(x)=x2+1  f\left( x \right) = \ln \left( x \right) \\\ g\left( x \right) = {x^2} + 1 \\\
The chain formula of deriving the derivative of the given function is
ddxf(g(x))=f′(g(x))⋅g′(x)⋯⋯(1)\dfrac{d}{{dx}}f\left( {g\left( x \right)} \right) = f'\left( {g\left( x \right)} \right) \cdot g'\left( x \right) \cdots \cdots \left( 1 \right)
Here,
f(x)=ln⁡(x)f\left( x \right) = \ln \left( x \right)
Now the derivative of this function is
f′(x)=1xf'\left( x \right) = \dfrac{1}{x}
And another function
g(x)=1+x2g\left( x \right) = 1 + {x^2}
And the derivative of this function is
g′(x)=2xg'\left( x \right) = 2x
Therefore, we can calculate the derivative of whole function by the following expression
f′(g(x))=1g(x)f'\left( {g\left( x \right)} \right) = \dfrac{1}{{g\left( x \right)}}
Now substitute the value of g(x)g\left( x \right)in the above expression
f′(g(x))=1x2+1f'\left( {g\left( x \right)} \right) = \dfrac{1}{{{x^2} + 1}}
Now substitute all the values in equation (1) we get

ddx[ln⁡(x2+1)]=2x1+x2\dfrac{d}{{dx}}\left[ {\ln \left( {{x^2} + 1} \right)} \right] = \dfrac{{2x}}{{1 + {x^2}}}
This is the final solution of the given problem.

Note:
We can also find out the derivative of logarithmic functions with the help of the following method.
Let’s suppose
y=ln⁡uy = \ln u
Where uu is the function of xx
Now differentiate the above equation with respect to xx we get
dydx=1ududx⋯⋯(2)\dfrac{{dy}}{{dx}} = \dfrac{1}{u}\dfrac{{du}}{{dx}} \cdots \cdots \left( 2 \right)
And if uu is function of xx then
u=f(x)u = f\left( x \right)
Then in that case the equation (2) can be written in the following form
dydx=f′(x)f(x)\dfrac{{dy}}{{dx}} = \dfrac{{f'\left( x \right)}}{{f\left( x \right)}}
While calculating the derivatives of the logarithmic function it is important to know all the formulas of derivatives and we have to first differentiate the given log function then the inside function.