Question

Differentiate $\log (1 + {x^2})$ with respect to ${\tan ^{ - 1}}x$

Answer 1

Hint: First we have to define what the terms we need to solve the problem are.The process of finding derivative of a function is called differentiation. If x and y are two variables, then the rate of change of x with respect to y is the derivative.We know that a \[tan\] of 90 degrees is defined as infinity. Thus for \[{\tan ^{ - 1}}\] the value is 90 degrees.

Complete step by step answer:
A logarithm function or log operator is used when we have to deal with the powers of a number, to understand it better, we will see an example.
Suppose $y = {z^a}$, where $y,z,a$ are the real numbers.
Then if we apply log on both sides with base z, we will get the following results.
$ \Rightarrow {\log _z}y = a$
The low operator has many properties, some of the properties which we will use in this particular problem with differentiation,
The derivative of the logarithmic function is given by: \[f'\left( x \right) = 1/\left( {{\text{ }}x{\text{ }}ln\left( b \right){\text{ }}} \right)x\] is the function argument. b is the logarithm base. \[ln(b)\]is the natural logarithm of b.
From the given problem suppose take $u = \log (1 + {x^2})$ and $v = {\tan ^{ - 1}}x$
To differentiate $u$with respect to $v$, that is $\dfrac{{du}}{{dv}}$
First differentiate $u$with respect to $x$ we get $\dfrac{{du}}{{dx}} = \dfrac{d}{{dx}}[\log (1 + {x^2})]$ since $\dfrac{d}{{dx}}[\log ( x)] = \dfrac{1}{x}$ --- $(1)$
$ \Rightarrow \dfrac{{du}}{{dx}} = \dfrac{1}{{1 + {x^2}}}\dfrac{d}{{dx}}[\log (1 + {x^2})]$ by $(1)$
$ \Rightarrow \dfrac{{du}}{{dx}} = \dfrac{1}{{1 + {x^2}}}\left[ {\dfrac{d}{{dx}}(1) + \dfrac{d}{{dx}}({x^2})} \right]$(giving the values to inside to the derivative parts)
Since the differentiation of $\dfrac{d}{{dx}}({x^n}) = n{x^{n - 1}}$
$ \Rightarrow \dfrac{{du}}{{dx}} = \dfrac{1}{{1 + {x^2}}}\left[ {0 + 2{x^{2 - 1}}} \right]$
$ \Rightarrow \dfrac{{du}}{{dx}} = \dfrac{1}{{1 + {x^2}}}\left[ {2x} \right]$
Therefore, we get
$ \Rightarrow \dfrac{{du}}{{dx}} = \dfrac{{2x}}{{1 + {x^2}}}$
Now we are going to differentiate $v$with respect to $x$
Therefore \[ \Rightarrow \dfrac{{dv}}{{dx}} = \dfrac{1}{{1 + {x^2}}}\]since deriving the equation with respect to x $\left( {\dfrac{{du}}{{dv}} = \dfrac{{\dfrac{{du}}{{dx}}}}{{\dfrac{{dv}}{{dx}}}}} \right)$
\[ \Rightarrow \dfrac{{du}}{{dv}} = \dfrac{{\dfrac{{2x}}{{1 + {x^2}}}}}{{\dfrac{1}{{1 + {x^2}}}}}\]
\[ \Rightarrow \dfrac{{du}}{{dv}} = \dfrac{{2x}}{{1 + {x^2}}} \times (1 + {x^2})\](bringing lower to upper)
Hence after solving every term, we get \[\dfrac{{du}}{{dv}} = 2x\]

Note: Differentiation is used to study the small change of a quantity with respect to unit change of another. On the other hand, integration is used to add small and discrete data, which cannot be added singularly and represented in a single value. The logarithm is the inverse function to exponentiation. That means the logarithm of a given number x is the exponent to which another fixed number, the base b, must be raised, to produce that number x.