Question

If $x+y=3{{e}^{2}}$ then $\dfrac{d\left( {{x}^{y}} \right)}{dx}=0$ for $x=$ ?
A. $e$
B. ${{e}^{2}}$
C. ${{e}^{e}}$
D. $2{{e}^{2}}$

Answer 1

Hint: We will take a variable and equate it with ${{x}^{y}}$ then we will take log on both sides and then we will differentiate it with respect to $x$ . We will gain the value of $\dfrac{dy}{dx}$ by differentiating the equation given in the question that is $x+y=3{{e}^{2}}$ and then again put it back in the obtained equation earlier and then equate it to $0.$ Finally, we will see from the options and see which option satisfies the equation.

Complete step by step answer:
Let’s assume that ${{x}^{y}}=t$ , now we will take logarithm on both sides that is $\log \left( {{x}^{y}} \right)=\log t$ . Now, we know that according to the property of logarithm: $\log f{{\left( x \right)}^{n}}=n\log f\left( x \right)$ ,
Therefore: \[y\log x=\log t\Rightarrow y\ln x=\ln t\] ,
We know that: $\dfrac{d\left( f\left( x \right).g\left( x \right) \right)}{dx}=f'\left( x \right)g\left( x \right)+g'\left( x \right)f\left( x \right)$ and $\dfrac{d\left( \log f\left( x \right) \right)}{dx}=\dfrac{1}{f\left( x \right)}.f'\left( x \right)$
We have the equation: \[y\ln x=\ln t\], we will now differentiate this equation with respect to $x$:
$\dfrac{y}{x}+\ln x.\dfrac{dy}{dx}=\dfrac{1}{t}\dfrac{dt}{dx}$ ,
Now let’s take $t$ from the right hand side to the left hand side:
$t\left( \dfrac{y}{x}+\ln x\dfrac{dy}{dx} \right)=\dfrac{dt}{dx}\text{ }..........\text{Equation 1}$

Now, it is given that $x+y=3{{e}^{2}}$ , let’s differentiate this equation with respect to $x$:
$1+\dfrac{dy}{dx}=0\Rightarrow \dfrac{dy}{dx}=-1$ , we will put this value of $\dfrac{dy}{dx}$ in equation 1:
$t\left( \dfrac{y}{x}+\ln x\dfrac{dy}{dx} \right)=\dfrac{dt}{dx}\text{ }\Rightarrow t\left( \dfrac{y}{x}+\ln x\left( -1 \right) \right)=\dfrac{dt}{dx}\Rightarrow t\left( \dfrac{y}{x}-\ln x \right)=\dfrac{dt}{dx}\text{ }........\text{ Equation 2}\text{.}$
We know that ${{x}^{y}}=t$ ,
And it is given that $\dfrac{dt}{dx}=\dfrac{d\left( {{x}^{y}} \right)}{dx}=0$ , we will put this value in equation 2, therefore: $t\left( \dfrac{y}{x}-\ln x \right)=\dfrac{dt}{dx}=0\Rightarrow t\left( y-x\ln x \right)=0$
Now, either $t=0\text{ or }y-x\ln x=0$ , since $t$ or ${{x}^{y}}$ is a non-zero term for given values therefore,
$y-x\ln x=0\Rightarrow y=x\ln x$ , we will now put this value in $x+y=3{{e}^{2}}$ , we will get: $x+x\ln x=3{{e}^{2}}\Rightarrow x\left( 1+\ln x \right)=3{{e}^{2}}\text{ }...........\text{Equation 3}\text{.}$
We will now put and check the values of $x$ from the options and we will find out that $x={{e}^{2}}$ will satisfy equation 3.

Hence, the correct option is B.

Note:
Students might make the mistake while differentiating as there are many properties used in it and we should be careful while applying them. Remember that we said $t$ or ${{x}^{y}}$ to be non zero because it is given in the question that $\dfrac{d\left( {{x}^{y}} \right)}{dx}=0$ , which implies that ${{x}^{y}}$ must be some constant.