Question

Find the value of \[\dfrac{d}{{dx}}\left( {{e^x}\log\left( {1 + {x^2}} \right)} \right)\].
A. \[{e^x}\left( {\log\left( {1 + {x^2}} \right) + \dfrac{{2x}}{{\left( {1 + {x^2}} \right)}}} \right)\]
B. \[{e^x}\left( {\log\left( {1 + {x^2}} \right) + \dfrac{{2x}}{{\left( {1 - {x^2}} \right)}}} \right)\]
C. \[{e^x}\left( {\log\left( {1 + {x^2}} \right) + \dfrac{x}{{\left( {1 + {x^2}} \right)}}} \right)\]
D. \[{e^x}\left( {\log\left( {1 + {x^2}} \right) - \dfrac{x}{{\left( {1 + {x^2}} \right)}}} \right)\]

Answer 1

Hint: First, consider \[y = {e^x}log\left( {1 + {x^2}} \right)\]. Then differentiate the equation with respect to the variable \[x\] and apply the product rule of differentiation to reach the required answer.

Formula Used:
\[\dfrac{d}{{dx}}\left( {{e^x}} \right) = {e^x}\]
\[\dfrac{d}{{dx}}\left( {\log x} \right) = \dfrac{1}{x}\]
\[\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}\]
Product rule of differentiation: \[\dfrac{d}{{dx}}\left( {uv} \right) = u\dfrac{{dv}}{{dx}} + v\dfrac{{du}}{{dx}}\]

Complete step by step solution:
The given derivative is \[\dfrac{d}{{dx}}\left( {{e^x}\log\left( {1 + {x^2}} \right)} \right)\].
Let consider,
\[y = {e^x}\log\left( {1 + {x^2}} \right)\]
Now differentiate the equation with respect to the variable \[x\] and apply the product rule of differentiation.
We get,
\[\dfrac{{dy}}{{dx}} = \log\left( {1 + {x^2}} \right)\dfrac{d}{{dx}}\left( {{e^x}} \right) + {e^x}\dfrac{d}{{dx}}\left( {\log\left( {1 + {x^2}} \right)} \right)\]
Apply the differential formulas \[\dfrac{d}{{dx}}\left( {{e^x}} \right) = {e^x}\], \[\dfrac{d}{{dx}}\left( {\log x} \right) = \dfrac{1}{x}\] and the chain rule.
\[\dfrac{{dy}}{{dx}} = {e^x}\log\left( {1 + {x^2}} \right) + {e^x}\left( {\dfrac{1}{{\left( {1 + {x^2}} \right)}}\dfrac{d}{{dx}}\left( {1 + {x^2}} \right)} \right)\]
Now apply the formula \[\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}\].
\[ \Rightarrow \dfrac{{dy}}{{dx}} = {e^x}\log\left( {1 + {x^2}} \right) + {e^x}\left( {\dfrac{1}{{\left( {1 + {x^2}} \right)}}\left( {2x} \right)} \right)\]
Simplify the above equation.
Factor out the common term.
\[ \Rightarrow \dfrac{{dy}}{{dx}} = {e^x}\left( {\log\left( {1 + {x^2}} \right) + \dfrac{{2x}}{{\left( {1 + {x^2}} \right)}}} \right)\]
Therefore,
\[\dfrac{d}{{dx}}\left( {{e^x}\log\left( {1 + {x^2}} \right)} \right) = {e^x}\left( {\log\left( {1 + {x^2}} \right) + \dfrac{{2x}}{{\left( {1 + {x^2}} \right)}}} \right)\]

Hence the correct option is A.

Note: Students often make mistakes while using the product rule on the composite functions. To calculate the derivative of a composite function always apply the chain rule of differentiation.