Question

For x > 1, if ${{(2x)}^{2y}}=4{{e}^{(2x-2y)}}$, then ${{(1+{{\log }_{e}}2x)}^{2}}\dfrac{dy}{dx}$ is equal to
$\begin{align}
  & \text{a) lo}{{\text{g}}_{e}}2x \\
 & \text{b) }\dfrac{x{{\log }_{e}}2x+{{\log }_{e}}2}{x} \\
 & \text{c)}\text{ xlo}{{\text{g}}_{e}}2x \\
 & \text{d) }\dfrac{x{{\log }_{e}}2x-{{\log }_{e}}2}{x} \\
\end{align}$

Answer 1

Hint: Now we are given with the equation ${{(2x)}^{2y}}=4{{e}^{(2x-2y)}}$ we will simplify the equation by taking log on both side and uspe properties of log to further simplify. Then we will write it in terms of y and we will differentiate the equation to find $\dfrac{dy}{dx}$. To find $\dfrac{dy}{dx}$ we will use the formula $\dfrac{f}{g}=\dfrac{gf'-fg'}{{{g}^{2}}}$ and use the fact that differentiation of f(g(x)) = $f'(g(x)).g'(x)$ . Further we will rearrange terms to find the value of ${{(1+{{\log }_{e}}2x)}^{2}}\dfrac{dy}{dx}$

Complete step-by-step answer:
Now we are given with the equation that ${{(2x)}^{2y}}=4{{e}^{(2x-2y)}}$
Taking log on both the sides we get
$\log {{(2x)}^{2y}}=\log 4{{e}^{(2x-2y)}}$
Now we know that $\log ab=\log a+\log b$ using this we get
$\log {{(2x)}^{2y}}=\log 4+\log {{e}^{(2x-2y)}}$
$\log {{(2x)}^{2y}}=\log {{2}^{2}}+\log {{e}^{(2x-2y)}}$
Now we know a property of log according to which $\log {{a}^{n}}=n\log a$
Hence using this we get
$2y\log \left( 2x \right)=\left( 2x-2y \right)\log e+2\log 2$
Now since the base of log given is e ${{\log }_{e}}e=1$ hence we get
$2y\log \left( 2x \right)=\left( 2x-2y \right)+2\log 2$
Now dividing the whole equation by 2 we get
$y\log \left( 2x \right)=x-y+\log 2$
Now taking y on RHS we get
$\begin{align}
  & y\log \left( 2x \right)+y=x+\log 2 \\
 & \Rightarrow y(1+\log 2x)=(x+\log 2) \\
 & \Rightarrow y=\dfrac{x+\log 2}{1+\log 2x} \\
\end{align}$
Now differentiating on both sides we get.
We know that differentiation of $\dfrac{f}{g}=\dfrac{gf'-fg'}{{{g}^{2}}}$. Using this we get
\[\dfrac{dy}{dx}=\dfrac{(1+\log 2x)(x+\log 2)'-(x+\log 2)\left( (1+\log 2x) \right)'}{{{\left( 1+\log 2x \right)}^{2}}}.................(1)\]
Now differentiation of (x + log 2) = 1 since log 2 is a constant
Substituting this in equation (1) we get
\[\dfrac{dy}{dx}=\dfrac{(1+\log 2x)-(x+\log 2)\left( (1+\log 2x) \right)'}{{{\left( 1+\log 2x \right)}^{2}}}.......................(2)\]
Now let us differentiate \[\left( {{(1+\log 2x)}^{2}} \right)\]
Now we know that differentiation of f(g(x)) is given by $f'(g(x)).g'(x)$ using this we get
\[\begin{align}
  & \left( 1+\log 2x \right)'=.\left( \dfrac{1}{2x}.2 \right) \\
 & \left( 1+\log 2x \right)'=\dfrac{1}{x} \\
\end{align}\]
Now substituting the value of \[\left( {{(1+\log 2x)}^{2}} \right)'\] in equation (2) we get
\[\dfrac{dy}{dx}=\dfrac{(1+\log 2x)-\dfrac{(x+\log 2)}{x}}{{{\left( 1+\log 2x \right)}^{2}}}\]
Now multiplying the whole equation by ${{\left( 1+\log 2x \right)}^{2}}$ we get
\[\begin{align}
  & \dfrac{dy}{dx}{{\left( 1+\log 2x \right)}^{2}}=(1+\log 2x)-\dfrac{(x+\log 2)}{x} \\
 & =\dfrac{x+x\log 2x-x+\log 2}{x} \\
 & =\dfrac{x\log 2x+\log 2}{x} \\
\end{align}\]
Hence the value of ${{\left( 1+\log 2x \right)}^{2}}$= \[\dfrac{x\log 2x+\log 2}{x}\]

So, the correct answer is “Option (b)”.

Note: While using the formula for differentiation of $\dfrac{f}{g}$note that the the differentiation is $=\dfrac{gf'-fg'}{{{g}^{2}}}$ not to be confused with $\dfrac{gf'+fg'}{{{g}^{2}}}$ or $\dfrac{fg'-gf'}{{{g}^{2}}}$.