If $x^y=e^{x-y}$, find a general formula for $\displaystyle\frac{d^ny}{dx^n}$.

My Attempt:

$$x^y=e^{x-y}$$ $$\Rightarrow\ \ \ \ \ y\ln x=x-y$$ $$\Rightarrow\ \ \ \ \ y=\frac{x}{\ln x+1}$$ So $$\begin{align}\frac{dy}{dx}&=\frac{\ln x+1-1}{(\ln x+1)^2}\\ &=\frac{\ln x}{(\ln x+1)^2} \end{align}$$ Then $$\begin{align}\frac{d^2y}{dx^2}&=\frac{\frac{1}{x}(\ln x+1)^2+\frac{2\ln x}{x}(\ln x+1)}{(\ln x+1)^4} \end{align}$$ Now, I can see that the denominator satisfies a straight-forward pattern $(\ln x+1)^{2^n}$. So the general formula for $\displaystyle\frac{d^ny}{dx^n}$ is $$\frac{d^ny}{dx^n}=\frac{g(n)}{(\ln x+1)^{2^n}}$$But I can't derive any formula for the numerator $g(n)$, nor can I see any regular pattern in the structure. Should I differentiate the initial expression $x^y=e^{x-y}$ implicity? Can anyone help me in that?


Here is a formula of the $n$-th derivative of $\frac{f}{g}$ in terms of derivatives of $f$ and $g$ which might be helpful.

Let $D_x$ represent differentiation with respect to $x$. Hence $D^n_x f(x)$ is the $n$-th derivative of $f$ with respect to $x$. The following holds true for $n$ times differentiable functions $f$ and $g$ \begin{align*} D_x^n\left(\frac{f}{g}\right)=\sum_{k=0}^n\sum_{j=0}^{k} (-1)^j\binom{n}{k}\binom{k+1}{j+1}\frac{1}{g^{j+1}} D_x^{n-k}\left(f\right) D_x^{k}\left( g^j\right)\tag{1} \end{align*}

This formula is based upon the Hoppe Form of Generalized Chain Rule and a proof for it is given in this MSE post.

In the current situation we have $f(x)=x$ and $g(x)=\ln(x)+1$ and we obtain from (1) if we change the order of summation of the outer sum by replacing $k$ with $n-k$:

\begin{align*} \color{blue}{D_x^n\left(\frac{x}{\ln x+1}\right)} &=\sum_{k=0}^n\sum_{j=0}^{n-k}(-1)^j\binom{n}{k}\binom{n-k+1}{j+1} \frac{1}{(\ln x + 1)^{j+1}}D_x^k(x)D_x^{n-k}\left((\ln x + 1)^j\right)\\ &=\color{blue}{\sum_{j=0}^{n}(-1)^j\binom{n+1}{j+1} \frac{x}{(\ln x+1)^{j+1}}D_x^{n}\left((\ln x+1)^j\right)}\\ &\qquad\color{blue}{+\sum_{j=0}^{n-1}(-1)^jn\binom{n+1}{j+1} \frac{1}{(\ln x+1)^{j+1}}D_x^{n-1}\left((\ln x+1)^j\right)}\\ \end{align*} Since $D_x^k(x)=0$ if $k>1$ it is sufficient to consider the first two terms $k=0,1$ of the outer sum.

Let's look at a small example in order to see the formula in action

Example: $D^1_x\left(\frac{x}{\ln x + 1}\right)$

\begin{align*} \color{blue}{D_x^1\left(\frac{x}{\ln x +1}\right)} &=\sum_{j=0}^1(-1)^j\binom{2}{j+1}\frac{x}{(\ln x+1)^{j+1}}D_x^1\left((\ln x+1)^j\right)\\ &\qquad +\sum_{j=0}^0(-1)^j\binom{1}{j+1}\frac{1}{(\ln x+1)^{j+1}}D_x^0\left((\ln x+1)^j\right)\\ &=(-1)^0\binom{2}{1}\frac{x}{\ln x+1}D_x(1)+(-1)^1\binom{2}{2}\frac{x}{(\ln x+1)^2}D_x(\ln x + 1)\\ &\qquad +(-1)^0\binom{1}{1}\frac{1}{\ln x+1}D_x^0(1)\\ &=0-\frac{x}{(\ln x+1)^2}\cdot \frac{1}{x}+\frac{1}{\ln x+1}\\ &\color{blue}{=\frac{\ln x}{(\ln x+1)^2}} \end{align*}

in accordance with OPs result.


It looks that $$\frac{d^ny}{dx^n}=\frac{g_n(x) }{x^{n-1}(\log(x)+1)^{n+1}}$$ where $g_n(x)$ is a polynomial in $\log(x)$ of degree $(n-1)$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.