Question

Differentiate with respect to $x$:
$\dfrac{{{e}^{x}}\log x}{{{x}^{2}}}$

Answer 1

Hint: The question can be solved by applying the quotient rule and product rule of differentiation to it which is used if function is the ratio of two functions each of which are differentiable.
This method is used if there are two or more than two functions in the form of product such that each are differentiable.
These are some of the standard derivations which are used to solve this question:
$\dfrac{d}{dx}\left( {{e}^{x}} \right)={{e}^{x}}$ and $\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}$
\[\dfrac{d}{dx}\left( \log x \right)=\dfrac{1}{x}\]

Complete step by step solution:
In the given expression we have the numerator in which two terms are present which are differentiable while in denominator the term is differentiable therefore, we solve it as follows:
Applying quotient and product rule $\dfrac{d}{dx}\left( \dfrac{a}{b} \right)=\dfrac{b\left( \dfrac{da}{dx} \right)-a\left( \dfrac{db}{dx} \right)}{{{b}^{2}}}$ to the expression $u=\dfrac{{{e}^{x}}\log x}{{{x}^{2}}}$ we get:
$\dfrac{du}{dx}=\dfrac{{{x}^{2}}\dfrac{d}{dx}\left( {{e}^{x}}\log x \right)-{{e}^{x}}\log x\dfrac{d}{dx}\left( {{x}^{2}} \right)}{{{\left( {{x}^{2}} \right)}^{2}}}$ ……(1)
Now we know that the differentiation of ${{e}^{x}}$ is $\dfrac{d}{dx}\left( {{e}^{x}} \right)={{e}^{x}}$ and $\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}$
The differentiation of $\log x$ is $\dfrac{1}{x}$
Therefore, the equation (1) can be rewritten by using all the standard derivations of derivatives and we get:
$= \dfrac{{{x}^{2}}\left[ \dfrac{d}{dx}\left( {{e}^{x}} \right)\log x+{{e}^{x}}\dfrac{d}{dx}\left( \log x \right) \right]-{{e}^{x}}\log x\times 2x}{{{x}^{4}}}$
Now we differentiate all the terms, we get:
\[= \dfrac{{{x}^{2}}\left[ \log x\times {{e}^{x}}+{{e}^{x}}\times \dfrac{1}{x} \right]-{{e}^{x}}\log x\times 2x}{{{x}^{4}}}\]
Now we can multiply ${{x}^{2}}$ to all the terms enclosed within brackets and simplify them, we get:
$= \dfrac{{{x}^{2}}\times \log x\times {{e}^{x}}+{{x}^{2}}\times {{e}^{x}}\times \dfrac{1}{x}-{{e}^{x}}\times 2x\times \log x}{{{x}^{4}}}$
Now cancel all the terms that are common and then find the differentiation of the expression we get:
$= \dfrac{{{x}^{2}}\times \log x\times {{e}^{x}}+x\times {{e}^{x}}-{{e}^{x}}\times 2x\times \log x}{{{x}^{4}}}$
$\therefore \dfrac{{{x}^{2}}\log x{{e}^{x}}+x{{e}^{x}}-2x{{e}^{x}}\log x}{{{x}^{4}}}$

Note: The above question can also have one more answer if we divide all the terms by $x$ than the answer obtained is $\dfrac{{{e}^{x}}\left( 1+\left( x-2 \right)\log x \right)}{{{x}^{3}}}$.
We should be careful while applying the quotient rule and not forgetting the negative sign present between terms in the numerator as it could get the wrong answer and we should square the term present in the denominator. Therefore, the correct formula for quotient rule is $\dfrac{d}{dx}\left( \dfrac{a}{b} \right)=\dfrac{b\left( \dfrac{da}{dx} \right)-a\left( \dfrac{db}{dx} \right)}{{{b}^{2}}}$ and for product rule is $\dfrac{d}{dx}\left( ab \right)=a\dfrac{db}{dx}+b\dfrac{da}{dx}$
Another common mistake is while applying the product rule as we have to apply it simultaneously with quotient rule which could be confusing hence, we should apply the formula carefully.