Proof by induction that $d^nf(x,y,z)=(adx+bdy+cdz)^n\cdot e^{ax+by+cz}$

Question

We have $f:\mathbb{R^3}\rightarrow\mathbb{R},f(x,y,z)=e^{ax+by+cz}$ and we need to proof by induction that $d^nf(x,y,z)=(adx+bdy+cdz)^n\cdot e^{ax+by+cz}$, where $d^nf(x,y,z)$ is n order differential of $f$, $\forall n\in\mathbb{N^*}$

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Here is all my steps:

• I noted with $\Gamma_{(n)}: d^nf(x,y,z)=(adx+bdy+cdz)^n\cdot e^{ax+by+cz}$.

Verification step: $$n=1\Rightarrow df(x,y,z)=\left[\frac{\partial dx}{\partial x}+\frac{\partial dy}{\partial y}+\frac{\partial dz}{\partial z}\right]\cdot f(x,y,z)=(adx+bdy+cdz)\cdot e^{ax+by+cz}$$

Demonstration step: We suppose that $\Gamma_{(k)}$ is true and we need to proof that it involve $\Gamma_{(k+1)}$ also true. Hence we need to prove that $\Gamma_{(k+1)}: d^{k+1}f(x,y,z)=(adx+bdy+cdz)^{k+1}\cdot e^{ax+by+cz},\forall x\in\mathbb{N^*}$

How can I prove it?

Answer 1

you showed it for n=1. Now suppose the for n=k the equality holds so:

\begin{equation} d^kf(x,y,z)=(adx+bdy+cdz)^k\cdot e^{ax+by+cz} \end{equation}

so for n=k+1 all we need to do is find the differential of the function \begin{equation} (adx+bdy+cdz)^k\cdot e^{ax+by+cz} \end{equation}

or calculate

\begin{equation} d\left \{ (adx+bdy+cdz)^k\cdot e^{ax+by+cz} \right \} = \\ = (adx+bdy+cdz)^{k} \cdot (adx+bdy+cdz) \cdot e^{ax+by+cz} = \\ = (adx+bdy+cdz)^{k+1}\cdot e^{ax+by+cz}, \end{equation}

which concludes the induction proof.