Question

Prove the following equation, if \[{{x}^{y}}={{e}^{x-y}}\].
\[\dfrac{dy}{dx}=\dfrac{\ln x}{{{(1+\ln x)}^{2}}}\].

Answer 1

Hint: We know that \[\ln ({{x}^{y}})=y\ln x\] and \[\ln ({{e}^{x-y}})=x-y\]. Using these formulas we will get the right answer.
Complete step-by-step answer:
We are given \[{{x}^{y}}={{e}^{x-y}}\].
Now, we will apply natural logarithm on both sides.
On applying natural logarithm, we get
\[\ln ({{x}^{y}})=\ln ({{e}^{x-y}}).....\]equation\[(1)\]
Now, we know, \[\ln ({{a}^{b}})=b\times \ln (a)\]
So, \[\ln ({{x}^{y}})=y\ln (x)\] and \[\ln ({{e}^{x-y}})=(x-y)\ln (e)\]
Now, we will substitute the values of \[\ln ({{x}^{y}})\] and \[\ln ({{e}^{x-y}})\] in equation\[(1)\].
On substituting the values of \[\ln ({{x}^{y}})\] and \[\ln ({{e}^{x-y}})\] in equation\[(1)\], we will get
\[y\ln x=(x-y)\ln (e).....\]equation\[(2)\]
Now, we know \[\ln (e)=1\]
We will substitute \[\ln (e)=1\] in equation\[(2)\]
\[\Rightarrow y\ln x=x-y\]
Now, we will differentiate both sides with respect to \[x\].
On differentiating both sides with respect to \[x\], we get
\[\dfrac{d}{dx}(y\ln x)=1-\dfrac{dy}{dx}.......\] equation \[\left( 3 \right)\]
Now, \[y\ln (x)\] is of the form \[f(x).g(x)\] where \[f(x)=y\] and \[g(x)=\ln (x)\]
So, to find its derivative, we will apply the product rule of differentiation, which is given by \[\dfrac{d(u.v)}{dx}=uv'+vu'\].
Hence, we get \[\dfrac{d}{dx}(y\ln (x))=\dfrac{dy}{dx}.\ln (x)+y.\dfrac{1}{x}..........\]equation\[(4)\]
Now, we will substitute the value from equation \[(4)\] in equation \[(3)\]
On substituting equation \[(4)\] in equation \[(3)\], we get,
\[\dfrac{dy}{dx}.\ln (x)+\dfrac{y}{x}=1-\dfrac{dy}{dx}\]
Now, we will take all terms with \[\dfrac{dy}{dx}\] to LHS and remaining terms to RHS. We get
\[\dfrac{dy}{dx}(1+\ln x)=1-\dfrac{y}{x}.......\]equation\[(5)\]
Now, we need to find the value of \[\dfrac{y}{x}\].
From equation \[(2)\], we have
\[y\ln x=x-y\]
Taking \[y\] from LHS to RHS, we get
\[\ln x=\dfrac{x-y}{y}\]
Or,\[\ln x=\dfrac{x}{y}-1\]
Or, \[1+\ln x=\dfrac{x}{y}\]
Or, \[\dfrac{y}{x}=\dfrac{1}{1+\ln x}\]
Now, we will substitute \[\dfrac{y}{x}=\dfrac{1}{1+\ln x}\] in equation\[(5)\].
On substituting \[\dfrac{y}{x}=\dfrac{1}{1+\ln x}\] in equation\[(5)\] , we get ,
\[\dfrac{dy}{dx}(1+\ln x)=1-\dfrac{1}{1+\ln (x)}\]
\[\Rightarrow \dfrac{dy}{dx}(1+\ln x)=\dfrac{1+\ln (x)-1}{1+\ln (x)}\]
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{\ln (x)}{{{(1+\ln (x))}^{2}}}\]
Hence proved.
Note: Always remember that \[\dfrac{d}{dx}(\ln (x))=\dfrac{1}{x}\] and not \[\dfrac{-1}{x}\]. Students generally make this mistake. Also, questions with variables in exponents are generally solved using logarithms as applying logarithm changes exponents into product. This makes the differentiation easier. On applying log in exponents , they are converted into products . Then , to find their derivative , we can apply the product rule of differentiation , which is given as \[\dfrac{d(u.v)}{dx}=uv'+vu'\].