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Question Number 93064 by  M±th+et+s last updated on 10/May/20

prove that (d/dx)x^n =nx^(n−1)

$${prove}\:{that}\: \\ $$$$\frac{{d}}{{dx}}{x}^{{n}} ={nx}^{{n}−\mathrm{1}} \\ $$

Commented by mr W last updated on 10/May/20

(d/dx)x^n =lim_(Δx→0) (((x+Δx)^n −x^n )/(Δx)) =lim_(Δx→0) ((x^n (1+((Δx)/x))^n −x^n )/(Δx)) =lim_(Δx→0) ((x^n [1+n(((Δx)/x))+((n(n−1))/(2!))(((Δx)/x))^2 +((n(n−1)(n−2))/(3!))(((Δx)/x))^3 +...]−x^n )/(Δx)) =lim_(Δx→0) x^n [n((1/x))+((n(n−1))/(2!))×((Δx)/x^2 )+((n(n−1)(n−2))/(3!))×(((Δx)^2 )/x^3 )+...] =x^n n((1/x)) =nx^(n−1)

$$\frac{{d}}{{dx}}{x}^{{n}} \\ $$$$=\underset{\Delta{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left({x}+\Delta{x}\right)^{{n}} −{x}^{{n}} }{\Delta{x}} \\ $$$$=\underset{\Delta{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}^{{n}} \left(\mathrm{1}+\frac{\Delta{x}}{{x}}\right)^{{n}} −{x}^{{n}} }{\Delta{x}} \\ $$$$=\underset{\Delta{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}^{{n}} \left[\mathrm{1}+{n}\left(\frac{\Delta{x}}{{x}}\right)+\frac{{n}\left({n}−\mathrm{1}\right)}{\mathrm{2}!}\left(\frac{\Delta{x}}{{x}}\right)^{\mathrm{2}} +\frac{{n}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)}{\mathrm{3}!}\left(\frac{\Delta{x}}{{x}}\right)^{\mathrm{3}} +...\right]−{x}^{{n}} }{\Delta{x}} \\ $$$$=\underset{\Delta{x}\rightarrow\mathrm{0}} {\mathrm{lim}}{x}^{{n}} \left[{n}\left(\frac{\mathrm{1}}{{x}}\right)+\frac{{n}\left({n}−\mathrm{1}\right)}{\mathrm{2}!}×\frac{\Delta{x}}{{x}^{\mathrm{2}} }+\frac{{n}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)}{\mathrm{3}!}×\frac{\left(\Delta{x}\right)^{\mathrm{2}} }{{x}^{\mathrm{3}} }+...\right] \\ $$$$={x}^{{n}} {n}\left(\frac{\mathrm{1}}{{x}}\right) \\ $$$$={nx}^{{n}−\mathrm{1}} \\ $$

Commented by  M±th+et+s last updated on 10/May/20

thank you sir

$${thank}\:{you}\:{sir} \\ $$

Commented by mathmax by abdo last updated on 10/May/20

let f(x)=x^n we have f^′ (x)=lim_(h→0) ((f(x+h)−f(x))/h) =lim_(h→0) (((x+h)^n −x^n )/h) =lim_(h→0) (((x+h−x)((x+h)^(n−1) +(x+h)^(n−2) x+....(x+h)x^(n−2) +x^(n−1) )/h) =lim_(h→0) (x+h)^(n−1) +(x+h)^(n−2) x +....+(x+h)x^(n−2) +x^(n−1) =x^(n−1) +x^(n−1) +...+x^(n−1) (n time) =nx^(n−1) ⇒ f^′ (x) =nx^(n−1)

$${let}\:{f}\left({x}\right)={x}^{{n}} \:\:{we}\:{have}\:{f}^{'} \left({x}\right)={lim}_{{h}\rightarrow\mathrm{0}} \:\:\frac{{f}\left({x}+{h}\right)−{f}\left({x}\right)}{{h}} \\ $$$$={lim}_{{h}\rightarrow\mathrm{0}} \:\:\frac{\left({x}+{h}\right)^{{n}} −{x}^{{n}} }{{h}}\:={lim}_{{h}\rightarrow\mathrm{0}} \frac{\left({x}+{h}−{x}\right)\left(\left({x}+{h}\right)^{{n}−\mathrm{1}} \:+\left({x}+{h}\right)^{{n}−\mathrm{2}} {x}+....\left({x}+{h}\right){x}^{{n}−\mathrm{2}} +{x}^{{n}−\mathrm{1}} \right.}{{h}} \\ $$$$={lim}_{{h}\rightarrow\mathrm{0}} \:\:\:\left({x}+{h}\right)^{{n}−\mathrm{1}} \:+\left({x}+{h}\right)^{{n}−\mathrm{2}} {x}\:+....+\left({x}+{h}\right){x}^{{n}−\mathrm{2}} \:+{x}^{{n}−\mathrm{1}} \\ $$$$={x}^{{n}−\mathrm{1}} \:+{x}^{{n}−\mathrm{1}} +...+{x}^{{n}−\mathrm{1}} \:\left({n}\:{time}\right)\:={nx}^{{n}−\mathrm{1}} \:\:\Rightarrow \\ $$$${f}^{'} \left({x}\right)\:={nx}^{{n}−\mathrm{1}} \: \\ $$

Commented by  M±th+et+s last updated on 10/May/20

nice work. thank you sir

$${nice}\:{work}. \\ $$$${thank}\:{you}\:{sir} \\ $$

Commented by arcana last updated on 10/May/20

b the correct way to prove it is for induction ∗ prove it for the first cases. ∗assume that is true, then prove the case n+1.

$${b} \\ $$$$\mathrm{the}\:\mathrm{correct}\:\mathrm{way}\:\mathrm{to}\:\mathrm{prove}\:\mathrm{it}\:\mathrm{is}\:\mathrm{for} \\ $$$$\mathrm{induction} \\ $$$$\ast\:\mathrm{prove}\:\mathrm{it}\:\mathrm{for}\:\mathrm{the}\:\mathrm{first}\:\mathrm{cases}. \\ $$$$\ast\mathrm{assume}\:\mathrm{that}\:\mathrm{is}\:\mathrm{true},\:\mathrm{then}\:\mathrm{prove}\:\mathrm{the} \\ $$$$\mathrm{case}\:\mathrm{n}+\mathrm{1}. \\ $$

Commented by mathmax by abdo last updated on 11/May/20

you are welcome.

$${you}\:{are}\:{welcome}. \\ $$

Commented by Rio Michael last updated on 11/May/20

correct sir, but the method used above can also be used since the question didnt specify by induction.

$$\mathrm{correct}\:\mathrm{sir},\:\mathrm{but}\:\mathrm{the}\:\mathrm{method}\:\mathrm{used}\:\mathrm{above}\:\mathrm{can} \\ $$$$\mathrm{also}\:\mathrm{be}\:\mathrm{used}\:\mathrm{since}\:\mathrm{the}\:\mathrm{question}\:\mathrm{didnt}\:\mathrm{specify}\:\mathrm{by}\:\mathrm{induction}. \\ $$

Answered by Rio Michael last updated on 11/May/20

by mathematic induction, prove for n = 0 : (d/dx)(x^0 ) = 0 and n = 0 into nx^(n−1) = 0 prove for n = 1 , LHS = (d/dx)(x) = 1 RHS = 1(x)^(1−1) = 1 ⇒ the derivative formula is true for n=0,1 assume it is true for n = k ⇒ (d/dx) x^k = kx^(k−1) prove for n = k +1 (d/dx)(x)^(k+1) = (d/dx)(x^k × x) = x^k + x(d/dx)(x^k ) = x^k + x(kx^(k−1) ) = x^k + kx^k = (k+1)x^k hence the derivatve formula is true for n = k+1 hence ∀ n ∈Z , (d/dx)x^n = nx^(n−1) .

$$\mathrm{by}\:\mathrm{mathematic}\:\mathrm{induction}, \\ $$$$\mathrm{prove}\:\mathrm{for}\:{n}\:=\:\mathrm{0}\::\:\frac{{d}}{{dx}}\left({x}^{\mathrm{0}} \right)\:\:=\:\mathrm{0}\:\mathrm{and}\:{n}\:=\:\mathrm{0}\:\mathrm{into}\:{nx}^{{n}−\mathrm{1}} \:=\:\mathrm{0} \\ $$$$\mathrm{prove}\:\mathrm{for}\:{n}\:=\:\mathrm{1}\:,\:\mathrm{LHS}\:=\:\frac{{d}}{{dx}}\left({x}\right)\:=\:\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{RHS}\:=\:\mathrm{1}\left({x}\right)^{\mathrm{1}−\mathrm{1}} \:=\:\mathrm{1} \\ $$$$\Rightarrow\:\mathrm{the}\:\mathrm{derivative}\:\mathrm{formula}\:\mathrm{is}\:\mathrm{true}\:\mathrm{for}\:{n}=\mathrm{0},\mathrm{1} \\ $$$$\mathrm{assume}\:\mathrm{it}\:\mathrm{is}\:\mathrm{true}\:\mathrm{for}\:{n}\:=\:{k} \\ $$$$\Rightarrow\:\frac{{d}}{{dx}}\:{x}^{{k}} \:=\:{kx}^{{k}−\mathrm{1}} \\ $$$$\mathrm{prove}\:\mathrm{for}\:{n}\:=\:{k}\:+\mathrm{1}\: \\ $$$$\:\frac{{d}}{{dx}}\left({x}\right)^{{k}+\mathrm{1}} \:=\:\frac{{d}}{{dx}}\left({x}^{{k}} \:×\:{x}\right)\:=\:{x}^{{k}} \:+\:\:{x}\frac{{d}}{{dx}}\left({x}^{{k}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:{x}^{{k}} \:+\:{x}\left({kx}^{{k}−\mathrm{1}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:{x}^{{k}} \:+\:{kx}^{{k}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\left({k}+\mathrm{1}\right){x}^{{k}} \\ $$$$\mathrm{hence}\:\mathrm{the}\:\mathrm{derivatve}\:\mathrm{formula}\:\mathrm{is}\:\mathrm{true}\:\mathrm{for}\:{n}\:=\:{k}+\mathrm{1} \\ $$$$\mathrm{hence}\:\forall\:{n}\:\in\mathbb{Z}\:,\:\frac{{d}}{{dx}}{x}^{{n}} \:=\:{nx}^{{n}−\mathrm{1}} . \\ $$

Commented by  M±th+et+s last updated on 11/May/20

nice work. thanx

$${nice}\:{work}. \\ $$$${thanx} \\ $$

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