Question

Consider the function ${{y}^{x}}={{e}^{y-x}}$, then prove that $\dfrac{dy}{dx}=\dfrac{{{\left( 1+\log y \right)}^{2}}}{\log y}$

Answer 1

Proceed with the given equation, analyse and apply logarithm functions. Further simplify using logarithm properties like $\log f{{\left( x \right)}^{n}}=n\log f\left( x \right)$ and then start differentiating both the sides with respect to the respective variables using Power Rule $\left( \dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}} \right)$ and the Quotient Rule for differentiation when the function is given in $f\left( x \right)=\dfrac{g\left( x \right)}{h\left( x \right)}$ form, then the derivative is $f'\left( x \right)=\dfrac{g'\left( x \right)h\left( x \right)-h'\left( x \right)g\left( x \right)}{{{\left[ h\left( x \right) \right]}^{2}}}$.

Complete step-by-step solution:
We will start with the given equation in the question that: ${{y}^{x}}={{e}^{y-x}}$
Now, let’s apply log on both sides of the equation we get:
$\log {{y}^{x}}=\log {{e}^{y-x}}..............\text{ Equation 1}\text{.}$ ;
Now we know that according to one of the properties of logarithm: $\log f{{\left( x \right)}^{n}}=n\log f\left( x \right)$
Applying the above property in equation 1 on both sides, we get
$x\log y=(y-x)\log e..............\text{ Equation 2}\text{.}$
We know that $\log e=1$ , therefore equation 2 becomes:
$x\log y=y-x$
Taking x on the R.H.S.(Right Hand Side) , we will have:

& \log y=\dfrac{y-x}{x} \\\ & \log y=\dfrac{y}{x}-\dfrac{x}{x}\Rightarrow \log y=\dfrac{y}{x}-\dfrac{{x}}{{x}}\Rightarrow \log y=\dfrac{y}{x}-1 \\\ \end{aligned}$$ Taking 1 on the L.H.S. (Left Hand Side), we will get: $\log y+1=\dfrac{y}{x}$ ; Now to make differentiation easy we will move terms with y-variables on one side and those with x variables on another: $$x=\dfrac{y}{1+\log y}$$ Differentiating L.H.S. with respect to x and R.H.S. with respect to y. $$\begin{aligned} & \dfrac{d\left( x \right)}{dx}=\dfrac{d\left( \dfrac{y}{1+\log y} \right)}{dy}..............\text{ Equation 3}\text{.} \\\ & \\\ \end{aligned}$$ Differentiating L.H.S.: We will use the power rule here for differentiation, as we know that the power rule is : $\left( \dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}} \right)$ $$\dfrac{d\left( x \right)}{dx}=1.{{x}^{0}}\Rightarrow d(x)=1.dx$$ Differentiating R.H.S.: We will use the Quotient Rule here for the differentiation, as we know the Quotient rule is: If: $f\left( x \right)=\dfrac{g\left( x \right)}{h\left( x \right)}$ , then $$f'\left( x \right)=\dfrac{g'\left( x \right)h\left( x \right)-h'\left( x \right)g\left( x \right)}{{{\left[ h\left( x \right) \right]}^{2}}}$$ , Applying this to the RHS $$\begin{aligned} & \dfrac{d\left( \dfrac{y}{1+\log y} \right)}{dy} \\\ & =\dfrac{\dfrac{d}{dy}\left[ y \right].\left( 1+\log \left( y \right) \right)-y.\dfrac{d}{dy}\left[ 1+\log \left( y \right) \right]}{{{\left( 1+\log \left( y \right) \right)}^{2}}} \\\ & =\dfrac{1\left( 1+\log \left( y \right) \right)-y.\left[ \dfrac{d}{dy}\left[ 1 \right]+\dfrac{d}{dy}\left[ \log \left( y \right) \right] \right]}{{{\left( 1+\log \left( y \right) \right)}^{2}}} \\\ & =\dfrac{1+\log \left( y \right)-y.\left[ 0+\dfrac{1}{y} \right]}{{{\left( 1+\log \left( y \right) \right)}^{2}}} \\\ & =\dfrac{1+\log \left( y \right)-\left[ y.0+y.\dfrac{1}{y} \right]}{{{\left( 1+\log \left( y \right) \right)}^{2}}}=\dfrac{1+\log \left( y \right)-\left[ 0+{y}.\dfrac{1}{{y}} \right]}{{{\left( 1+\log \left( y \right) \right)}^{2}}}==\dfrac{1+\log \left( y \right)-0-1}{{{\left( 1+\log \left( y \right) \right)}^{2}}} \\\ & =\dfrac{\log y}{\left( 1+\log \left( y \right) \right)2} \\\ & \therefore d\left( \dfrac{y}{1+\log y} \right)==\dfrac{\log y}{{{\left( 1+\log \left( y \right) \right)}^{2}}}.dy \\\ \end{aligned}$$ Putting both the values obtained after differentiating in Equation 3: $$\begin{aligned} & 1.dx=\dfrac{\log y}{{{\left( 1+\log \left( y \right) \right)}^{2}}}.dy \end{aligned}$$ Taking $\dfrac{dy}{dx}$ on one side we will then have: $\dfrac{dy}{dx}=\dfrac{{{\left( 1+\log y \right)}^{2}}}{\log y}$ Hence Proved. **Note:** While solving (differentiating RHS) differential properties are applied like $\dfrac{d\left( \log y \right)}{dy}=\dfrac{1}{y}$ , students can make that mistake. In equation 2 log(e) can also be written as ln e. Student can make the mistake while changing the sign while performing logarithmic functions and hence calculation must be done with concentration.