Question

If we have an expression${{x}^{y}}={{e}^{x-y}}$ then prove that $\dfrac{dy}{dx}=\dfrac{{{\log }_{e}}x}{{{(1+{{\log }_{e}}x)}^{2}}}$

Answer 1

To solve the above question, we use the concepts of logarithms, variable separable method and differentiation. Here in the above question we have given that ${{x}^{y}}={{e}^{x-y}}\,\,$ and we have to show that $\dfrac{dy}{dx}=\dfrac{{{\log }_{e}}x}{{{(1+{{\log }_{e}}x)}^{2}}}$ from the given equation. We need to multiply the given equation by the log then after derivation we have to arrange the equation according to the problem so that we will get to the final answer. To solve this problem, we will need this formula $\left[ \because \dfrac{d(uv)}{dx}=u\dfrac{d(v)}{dx}+v\dfrac{d(u)}{dx} \right]$.

Complete step-by-step solution:
So here we have Given that;
${{x}^{y}}={{e}^{x-y}}\,\,$
On taking logarithm to the base “e” on both sides, we will get;
${{\log }_{e}}({{x}^{y}})={{\log }_{e}}({{e}^{x-y}})$
After applying the log on both sides, we have to use the identity of log which is $\,\left[ \begin{matrix} \because & {{\log }_{e}}{{a}^{m}}=m{{\log }_{e}}a \\\ \end{matrix} \right]$ with the help of this we will get,
$y{{\log }_{e}}(x)=(x-y){{\log }_{e}}e\,$
Here in this above equation we know that the value of $\,\left[ \because {{\log }_{e}}e=1 \right]$ we will put the value and then we will get,
$\Rightarrow y{{\log }_{e}}(x)=(x-y)\,$
On rearranging the above equation, we will get,

& \Rightarrow y\left( 1+{{\log }_{e}}x \right)=x\, \\\ & \Rightarrow y+y{{\log }_{e}}x=x\, \\\ \end{aligned} \right\\}...........\left( 1 \right)$$ On differentiating the above equation with respect to x on both sides we used the formula $$\,\,\left[ \because \dfrac{d(uv)}{dx}=u\dfrac{d(v)}{dx}+v\dfrac{d(u)}{dx} \right]$$ and then we will get, $$\dfrac{dy}{dx}+\left[ \left( y \right)\left( \dfrac{1}{x} \right)+\left( \dfrac{dy}{dx} \right)\left( {{\log }_{e}}x \right) \right]\,\,\,=1\,\,\,\,\,$$ On rearranging and simplifying the above equation we will get, $$\dfrac{dy}{dx}\left[ 1+{{\log }_{e}}x \right]+\dfrac{y}{x}=1$$ $$\Rightarrow \dfrac{dy}{dx}\left[ 1+{{\log }_{e}}x \right]=1-\dfrac{y}{x}$$ $$\Rightarrow \dfrac{dy}{dx}\left[ 1+{{\log }_{e}}x \right]=\dfrac{x-y}{x}$$ $$\Rightarrow \dfrac{dy}{dx}\left[ 1+{{\log }_{e}}x \right]=\dfrac{y+y{{\log }_{e}}x-y}{x}$$ $$\Rightarrow \dfrac{dy}{dx}\left[ 1+{{\log }_{e}}x \right]=\dfrac{y{{\log }_{e}}x}{x}=\dfrac{y}{x}{{\log }_{e}}x$$ Now again from the equation no.1 we can also write $\dfrac{y}{x}$ as $$\left( \dfrac{1}{1+{{\log }_{e}}x} \right)$$ $$\dfrac{dy}{dx}\left[ 1+{{\log }_{e}}x \right]=\left( \dfrac{1}{1+{{\log }_{e}}x} \right){{\log }_{e}}x$$ $$\Rightarrow \dfrac{dy}{dx}=\dfrac{\log x}{{{(1+\log x)}^{2}}}$$ Hence $$\dfrac{dy}{dx}=\dfrac{\log x}{{{(1+\log x)}^{2}}}$$ is proved. **Note:** For the exponent type of problems first try applying logarithms and simplify to get the answer and don’t bother about how we get the idea of applying logarithms, as the ideas will come by practice. More we practice more we can solve. In these types of question if you don’t take the logarithm then it not possible to solve the problem you have to know the formulas of derivation$$\left[ \because \dfrac{d(uv)}{dx}=u\dfrac{d(v)}{dx}+v\dfrac{d(u)}{dx} \right]$$ and also the logarithmic formulas like $$\,{{\log }_{e}}{{a}^{m}}=m{{\log }_{e}}a$$.