Question

If $y = \dfrac{{{e^x}\log x}}{{{x^2}}}$, then $\dfrac{{dy}}{{dx}}$ is equal to:
A. ${e^x}\dfrac{{\left[ {1 + \left( {x + 2} \right)\log x} \right]}}{{{x^4}}}$
B. ${e^x}\dfrac{{\left[ {1 + \left( {x - 2} \right)\log x} \right]}}{{{x^4}}}$
C. ${e^x}\dfrac{{\left[ {1 - \left( {x - 2} \right)\log x} \right]}}{{{x^3}}}$
D. ${e^x}\dfrac{{\left[ {1 + \left( {x - 2} \right)\log x} \right]}}{{{x^3}}}$

Answer 1

In the given problem, we are required to differentiate $y = \dfrac{{{e^x}\log x}}{{{x^2}}}$ with respect to x. Since, $y = \dfrac{{{e^x}\log x}}{{{x^2}}}$ is a rational function in variable x, so we will have to apply quotient rule of differentiation in the process of differentiating $y = \dfrac{{{e^x}\log x}}{{{x^2}}}$ . Also derivatives of basic algebraic and trigonometric functions must be remembered thoroughly. We also must know the product rule and chain rule of differentiation to solve the given problem.

Complete step by step answer:
Now, $\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {\dfrac{{{e^x}\log x}}{{{x^2}}}} \right)$ .
Now, using the quotient rule of differentiation, we know that $\dfrac{d}{{dx}}\left( {\dfrac{{f(x)}}{{g(x)}}} \right) = \dfrac{{g(x) \times \dfrac{d}{{dx}}\left( {f(x)} \right) - f(x) \times \dfrac{d}{{dx}}\left( {g(x)} \right)}}{{{{\left[ {g\left( x \right)} \right]}^2}}}$ .
So, Applying quotient rule to $\dfrac{d}{{dx}}\left( {\dfrac{{{e^x}\log x}}{{{x^2}}}} \right)$, we get,
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{{x^2}\dfrac{d}{{dx}}\left( {{e^x}\log x} \right) - \left( {{e^x}\log x} \right)\dfrac{d}{{dx}}\left( {{x^2}} \right)}}{{{{\left( {{x^2}} \right)}^2}}}$
Substituting the derivative of ${x^2}$ with respect to x as $2x$,
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{{x^2}\dfrac{d}{{dx}}\left( {{e^x}\log x} \right) - \left( {{e^x}\log x} \right)\left( {2x} \right)}}{{{x^4}}}$
Now, applying product rule of differentiation $\dfrac{d}{{dx}}\left( {f(x) \times g(x)} \right) = g(x) \times \dfrac{d}{{dx}}\left( {f(x)} \right) + f(x) \times \dfrac{d}{{dx}}\left( {g(x)} \right)$, we get,
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{{x^2}\left[ {{e^x}\dfrac{d}{{dx}}\left( {\log x} \right) + \log x\dfrac{d}{{dx}}\left( {{e^x}} \right)} \right] - \left( {{e^x}\log x} \right)\left( {2x} \right)}}{{{x^4}}}$

Now, we know that the derivative of $\log x$ is $\left( {\dfrac{1}{x}} \right)$. So, we get,
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{{x^2}\left[ {{e^x}\left( {\dfrac{1}{x}} \right) + \log x\left( {{e^x}} \right)} \right] - \left( {{e^x}\log x} \right)\left( {2x} \right)}}{{{x^4}}}$
Taking $x{e^x}$ common from all the terms,
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{{e^x}{x^2}\left[ {\left( {\dfrac{1}{x}} \right) + \log x} \right] - {e^x}\left( {\log x} \right)\left( {2x} \right)}}{{{x^4}}}$
Cancelling common factors in numerator and denominator, we get,
\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{{e^x}x\left\\{ {\left[ {1 + x\log x} \right] - \left( {\log x} \right)\left( 2 \right)} \right\\}}}{{{x^4}}}
\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{{e^x}\left\\{ {1 + x\log x - 2\log x} \right\\}}}{{{x^3}}}
Now, simplifying the expression to match the options, we get,
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{{e^x}\left[ {1 + \left( {x - 2} \right)\log x} \right]}}{{{x^3}}}$
$\therefore \dfrac{{dy}}{{dx}} = \dfrac{{{e^x}\left[ {1 + \left( {x - 2} \right)\log x} \right]}}{{{x^3}}}$

Hence, option D is the correct answer.

Note: The given problem may also be solved using the first principle of differentiation. The product rule of differentiation involves differentiating a product of two functions and the chain rule of differentiation involves differentiating a composite by introducing new unknowns to ease the process and examine the behaviour of function layer by layer. The quotient rule involves differentiation of a rational function in some variable. One must know derivatives of some basic functions such as logarithmic and exponential function in order to tackle such problems.