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Question Number 3370 by Filup last updated on 12/Dec/15

y=ax^n y+δy=a(x+δx)^n ⋮ (dy/dx)=alim_(δx→0) (((x+δx)^n −x^n )/(δx)) Show that (dy/dx)=anx^(n−1)

$${y}={ax}^{{n}} \\ $$$${y}+\delta{y}={a}\left({x}+\delta{x}\right)^{{n}} \\ $$$$\vdots \\ $$$$\frac{{dy}}{{dx}}={a}\underset{\delta{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left({x}+\delta{x}\right)^{{n}} −{x}^{{n}} }{\delta{x}} \\ $$$$ \\ $$$$\mathrm{Show}\:\mathrm{that}\:\frac{{dy}}{{dx}}={anx}^{{n}−\mathrm{1}} \\ $$

Answered by Yozzi last updated on 12/Dec/15

By the Binomial theorem, since n∈N+{0} (x+δx)^n =Σ_(r=0) ^n ((n),(r) ) x^(n−r) (δx)^r =x^n +Σ_(r=1) ^n ((n),(r) ) x^(n−r) (δx)^r ⇒(x+δx)^n −x^n =Σ_(r=1) ^n ((n),(r) )x^(n−r) (δx)^r ⇒(((x+δx)^n −x^n )/(δx))=Σ_(r=1) ^n ((n),(r) )x^(n−r) (δx)^(r−1) ∴ (dy/dx)=alim_(δx→0) {Σ_(r=1) ^n ((n),(r) )x^(n−r) (δx)^(r−1) } (dy/dx)=alim_(x→0) { ((n),(1) )x^(n−1) ×(δx)^0 +Σ_(r=2) ^n ((n),(r) )x^(n−r) (δx)^r } All terms in Σ_(r=2) ^n ((n),(r) )x^(n−r) (δx)^(r−1) is a product of at least one δx and another variable (which is of the form {constant}×{x^(n−r) }×{(δx)^r }). So lim_(δx→0) Σ_(r=2) ^n ((n),(r) )x^(n−r) (δx)^r =0+0+0+...+0 (n−1 times) =0 ⇒(dy/dx)=a×( ((n),(1) )x^(n−1) +0)=anx^(n−1) .

$${By}\:{the}\:{Binomial}\:{theorem},\:{since}\:{n}\in\mathbb{N}+\left\{\mathrm{0}\right\} \\ $$$$\left({x}+\delta{x}\right)^{{n}} =\underset{{r}=\mathrm{0}} {\overset{{n}} {\sum}}\begin{pmatrix}{{n}}\\{{r}}\end{pmatrix}\:{x}^{{n}−{r}} \left(\delta{x}\right)^{{r}} \:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={x}^{{n}} +\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\begin{pmatrix}{{n}}\\{{r}}\end{pmatrix}\:{x}^{{n}−{r}} \left(\delta{x}\right)^{{r}} \\ $$$$\Rightarrow\left({x}+\delta{x}\right)^{{n}} −{x}^{{n}} =\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\begin{pmatrix}{{n}}\\{{r}}\end{pmatrix}{x}^{{n}−{r}} \left(\delta{x}\right)^{{r}} \\ $$$$\Rightarrow\frac{\left({x}+\delta{x}\right)^{{n}} −{x}^{{n}} }{\delta{x}}=\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\begin{pmatrix}{{n}}\\{{r}}\end{pmatrix}{x}^{{n}−{r}} \left(\delta{x}\right)^{{r}−\mathrm{1}} \\ $$$$\therefore\:\frac{{dy}}{{dx}}={a}\underset{\delta{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left\{\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\begin{pmatrix}{{n}}\\{{r}}\end{pmatrix}{x}^{{n}−{r}} \left(\delta{x}\right)^{{r}−\mathrm{1}} \right\} \\ $$$$\frac{{dy}}{{dx}}={a}\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left\{\begin{pmatrix}{{n}}\\{\mathrm{1}}\end{pmatrix}{x}^{{n}−\mathrm{1}} ×\left(\delta{x}\right)^{\mathrm{0}} +\underset{{r}=\mathrm{2}} {\overset{{n}} {\sum}}\begin{pmatrix}{{n}}\\{{r}}\end{pmatrix}{x}^{{n}−{r}} \left(\delta{x}\right)^{{r}} \right\} \\ $$$${All}\:{terms}\:{in}\:\underset{{r}=\mathrm{2}} {\overset{{n}} {\sum}}\begin{pmatrix}{{n}}\\{{r}}\end{pmatrix}{x}^{{n}−{r}} \left(\delta{x}\right)^{{r}−\mathrm{1}} \\ $$$${is}\:{a}\:{product}\:{of}\:{at}\:{least}\:{one}\:\delta{x}\:{and}\:{another} \\ $$$${variable}\:\left({which}\:{is}\:{of}\:{the}\:{form}\:\right. \\ $$$$\left.\left\{{constant}\right\}×\left\{{x}^{{n}−{r}} \right\}×\left\{\left(\delta{x}\right)^{{r}} \right\}\right). \\ $$$${So}\:\underset{\delta{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\underset{{r}=\mathrm{2}} {\overset{{n}} {\sum}}\begin{pmatrix}{{n}}\\{{r}}\end{pmatrix}{x}^{{n}−{r}} \left(\delta{x}\right)^{{r}} =\mathrm{0}+\mathrm{0}+\mathrm{0}+...+\mathrm{0}\:\:\left({n}−\mathrm{1}\:{times}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{0} \\ $$$$\Rightarrow\frac{{dy}}{{dx}}={a}×\left(\begin{pmatrix}{{n}}\\{\mathrm{1}}\end{pmatrix}{x}^{{n}−\mathrm{1}} +\mathrm{0}\right)={anx}^{{n}−\mathrm{1}} . \\ $$$$ \\ $$

Commented by Filup last updated on 12/Dec/15

nice! i thought it was along that line of thought! thanks <3

$$\mathrm{nice}!\:\mathrm{i}\:\mathrm{thought}\:\mathrm{it}\:\mathrm{was}\:\mathrm{along}\:\mathrm{that}\:\mathrm{line} \\ $$$$\mathrm{of}\:\mathrm{thought}!\:\mathrm{thanks}\:<\mathrm{3} \\ $$

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