Question

Differentiate the following function with respect to $x$:
${{x}^{n}}{{\log }_{a}}x\text{ }{{e}^{x}}$

Answer 1

Hint: For solving this question we will write the term ${{\log }_{a}}x$ as $\dfrac{\ln x}{\ln a}$ in the given function and then, differentiate $y=\dfrac{1}{\ln a}\left( {{x}^{n}}\ln x\text{ }{{e}^{x}} \right)$ with respect to $x$. After that, we will apply the product rule of differential calculus, and formulas like $\dfrac{d\left( {{x}^{n}} \right)}{dx}=n{{x}^{n-1}}$, $\dfrac{d\left( \ln x \right)}{dx}=\dfrac{1}{x}$ and $\dfrac{d\left( {{e}^{x}} \right)}{dx}={{e}^{x}}$. Then, we will arrange the terms in the result to get the final answer.

Complete step-by-step answer:
We to differentiate the following function with respect to $x$:
$y={{x}^{n}}{{\log }_{a}}x\text{ }{{e}^{x}}$
Now, as we will use the formula ${{\log }_{c}}b=\dfrac{{{\log }_{e}}b}{{{\log }_{e}}c}$ to write ${{\log }_{a}}x=\dfrac{{{\log }_{e}}x}{{{\log }_{e}}a}$ in the above equation. Then,
$\begin{aligned} & y={{x}^{n}}{{\log }_{a}}x\text{ }{{e}^{x}} \\\ & \Rightarrow y={{x}^{n}}\times \dfrac{{{\log }_{e}}x}{{{\log }_{e}}a}\times {{e}^{x}} \\\ \end{aligned}$
Now, we can write ${{\log }_{e}}x=\ln x$ and ${{\log }_{e}}a=\ln a$ in the above equation. Then,
$\begin{aligned} & y={{x}^{n}}\times \dfrac{{{\log }_{e}}x}{{{\log }_{e}}a}\times {{e}^{x}} \\\ & \Rightarrow y=\dfrac{1}{\ln a}\left( {{x}^{n}}\ln x\text{ }{{e}^{x}} \right)..................\left( 1 \right) \\\ \end{aligned}$
Now, before we proceed, we should know the following formulas and concepts of differential calculus:
1. If $y=f\left( x \right)\cdot g\left( x \right)$ , then $\dfrac{dy}{dx}=\dfrac{d\left( f\left( x \right)\cdot g\left( x \right) \right)}{dx}={f}'\left( x \right)\cdot g\left( x \right)+f\left( x \right)\cdot {g}'\left( x \right)$ . This is also known as the product rule of differentiation.
2. If $y={{x}^{n}}$ , then $\dfrac{dy}{dx}=\dfrac{d\left( {{x}^{n}} \right)}{dx}=n{{x}^{n-1}}$ .
3. If $y=\ln x$ , then $\dfrac{dy}{dx}=\dfrac{d\left( \ln x \right)}{dx}=\dfrac{1}{x}$ .
4. If $y={{e}^{x}}$ , then $\dfrac{dy}{dx}=\dfrac{d\left( {{e}^{x}} \right)}{dx}={{e}^{x}}$ .
Now, from the equation (1) we have the following equation:
$y=\dfrac{1}{\ln a}\left( {{x}^{n}}\ln x\text{ }{{e}^{x}} \right)$
Now, we will differentiate the above equation with respect to $x$ . Then,
$\begin{aligned} & y=\dfrac{1}{\ln a}\left( {{x}^{n}}\ln x\text{ }{{e}^{x}} \right) \\\ & \Rightarrow \dfrac{dy}{dx}=\dfrac{d\left( \dfrac{1}{\ln a}\left( {{x}^{n}}\ln x\text{ }{{e}^{x}} \right) \right)}{dx} \\\ \end{aligned}$
Now, as $\dfrac{1}{\ln a}$ is a constant term so, we can write $\dfrac{d\left( \dfrac{1}{\ln a}\left( {{x}^{n}}\ln x\text{ }{{e}^{x}} \right) \right)}{dx}=\dfrac{1}{\ln a}\dfrac{d\left( {{x}^{n}}\ln x\text{ }{{e}^{x}} \right)}{dx}$ . Then,
$\begin{aligned} & \dfrac{dy}{dx}=\dfrac{d\left( \dfrac{1}{\ln a}\left( {{x}^{n}}\ln x\text{ }{{e}^{x}} \right) \right)}{dx} \\\ & \Rightarrow \dfrac{dy}{dx}=\dfrac{1}{\ln a}\dfrac{d\left( {{x}^{n}}\ln x\text{ }{{e}^{x}} \right)}{dx} \\\ \end{aligned}$
Now, we will use the product rule of differentiation to write $\dfrac{d\left( {{x}^{n}}\ln x\text{ }{{e}^{x}} \right)}{dx}=\left( \ln x\text{ }{{e}^{x}} \right)\times \dfrac{d\left( {{x}^{n}} \right)}{dx}+\left( {{x}^{n}}{{e}^{x}} \right)\times \dfrac{d\left( \ln x \right)}{dx}+\left( {{x}^{n}}\ln x \right)\times \dfrac{d\left( {{e}^{x}} \right)}{dx}$ in the above equation. Then,
$\begin{aligned} & \dfrac{dy}{dx}=\dfrac{1}{\ln a}\dfrac{d\left( {{x}^{n}}\ln x\text{ }{{e}^{x}} \right)}{dx} \\\ & \Rightarrow \dfrac{dy}{dx}=\dfrac{1}{\ln a}\times \left( \left( \ln x\text{ }{{e}^{x}} \right)\times \dfrac{d\left( {{x}^{n}} \right)}{dx}+\left( {{x}^{n}}{{e}^{x}} \right)\times \dfrac{d\left( \ln x \right)}{dx}+\left( {{x}^{n}}\ln x \right)\times \dfrac{d\left( {{e}^{x}} \right)}{dx} \right) \\\ \end{aligned}$
Now, we will use the formulas of the differential calculus discussed above, to write $\dfrac{d\left( {{x}^{n}} \right)}{dx}=n{{x}^{n-1}}$ , $\dfrac{d\left( \ln x \right)}{dx}=\dfrac{1}{x}$ and $\dfrac{d\left( {{e}^{x}} \right)}{dx}={{e}^{x}}$ in the above equation. Then,
$\begin{aligned} & \dfrac{dy}{dx}=\dfrac{1}{\ln a}\times \left( \left( \ln x\text{ }{{e}^{x}} \right)\times \dfrac{d\left( {{x}^{n}} \right)}{dx}+\left( {{x}^{n}}{{e}^{x}} \right)\times \dfrac{d\left( \ln x \right)}{dx}+\left( {{x}^{n}}\ln x \right)\times \dfrac{d\left( {{e}^{x}} \right)}{dx} \right) \\\ & \Rightarrow \dfrac{dy}{dx}=\dfrac{1}{\ln a}\left( \left( \ln x\text{ }{{e}^{x}} \right)\times \left( n{{x}^{n-1}} \right)+\left( {{x}^{n}}{{e}^{x}} \right)\times \dfrac{1}{x}+\left( {{x}^{n}}\ln x \right)\times {{e}^{x}} \right) \\\ & \Rightarrow \dfrac{dy}{dx}=\dfrac{1}{\ln a}\left( n{{x}^{n-1}}\ln x\text{ }{{e}^{x}}+{{x}^{n-1}}{{e}^{x}}+{{x}^{n}}\ln x\text{ }{{e}^{x}} \right) \\\ & \Rightarrow \dfrac{dy}{dx}=\dfrac{{{x}^{n-1}}{{e}^{x}}}{\ln a}\left( n\ln x+x\ln x+1 \right) \\\ \end{aligned}$
Now, from the above result we conclude that if $y={{x}^{n}}{{\log }_{a}}x\text{ }{{e}^{x}}$ then, differentiation of $y$ with respect to $x$ will be $\dfrac{dy}{dx}=\dfrac{{{x}^{n-1}}{{e}^{x}}}{\ln a}\left( n\ln x+x\ln x+1 \right)$ .
Thus, $\dfrac{dy}{dx}=\dfrac{{{x}^{n-1}}{{e}^{x}}}{\ln a}\left( n\ln x+x\ln x+1 \right)$ will be our final answer.

Note: Here, the student should first understand what is asked in the question and then proceed in the right direction to get the correct answer quickly. And first, we should write the term ${{\log }_{a}}x$as $\dfrac{\ln x}{\ln a}$ and then, differentiate $y=\dfrac{1}{\ln a}\left( {{x}^{n}}\ln x\text{ }{{e}^{x}} \right)$ with respect to $x$ . Moreover, we should apply the product rule of differential calculus, and formulas like $\dfrac{d\left( {{x}^{n}} \right)}{dx}=n{{x}^{n-1}}$ , $\dfrac{d\left( \ln x \right)}{dx}=\dfrac{1}{x}$ and $\dfrac{d\left( {{e}^{x}} \right)}{dx}={{e}^{x}}$ correctly without any mathematical error to get the correct answer easily.