Question

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Using mathematical induction prove that :
$\bold{\frac{d}{dx} ( {x}^{n} ) = {nx}^{n - 1} \: for \: all \: n \in \: N}$
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$\red{\star \: \sf{chapter - derivation}}$
$\red{\star \: \sf{standard - 11 \: th}}$

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Answer 1

$\large\underline{\sf{Solution-}}$

$\rm :\longmapsto\:Let \: P(n) : \bold{\dfrac{d}{dx} ( {x}^{n} ) = {nx}^{n - 1} \: for \: all \: n \in \: N}$

Step :- 1 For n = 1

$\rm :\longmapsto\:\bold{\dfrac{d}{dx} ( {x}^{1} ) = {x}^{1 - 1} \:}$

$\rm :\longmapsto\:1 = {x}^{0}$

$\rm :\longmapsto\:1 = 1$

$\bf\implies \:P(n) \: is \: true \: for \: n \: = \: 1$

Step :- 2 Let assume that P(n) is true for n = k

$\rm :\longmapsto\: \bold{\dfrac{d}{dx} ( {x}^{k} ) = {kx}^{k - 1} \: for \: all \: k \in \: N - - - (1)}$

Step :- 3 Prove that P(n) is true for n = k + 1

$\rm :\longmapsto \: P(k + 1) : \bold{\dfrac{d}{dx} ( {x}^{k + 1} ) = {(k + 1)x}^{k} \: for \: all \: k \in \: N}$

Consider,

$\rm :\longmapsto \: \bold{\dfrac{d}{dx} ( {x}^{k + 1} ) }$

$\: \: \rm \: = \: \:\: \bold{\dfrac{d}{dx} ( {x}^{k} .x) }$

$\: \: \rm \: = \: \:\: \bold{x\dfrac{d}{dx} ( {x}^{k}) + {x}^{k}\dfrac{d}{dx} \: (x) }$

$\: \: \rm \: = \: \: \bf \: x \: (k {x}^{k - 1}) + {x}^{k} \times 1 \: \: \: \: \{ \: using \: (1) \: \}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \red{ \bigg\{ \sf \: \because \: \dfrac{d}{dx}x = 1 \: \bigg \}}$

$\: \: \rm \: = \: \: \bf \: k {x}^{k} + {x}^{k}$

$\: \: \rm \: = \: \: \bf \: {x}^{k}(k + 1)$

$\bf\implies \:P(n) \: is \: true \: for \: n \: = \: k + 1$

Hence,

By the process of Principal of Mathematical Induction,

$\rm :\longmapsto\: \bold{\dfrac{d}{dx} ( {x}^{n} ) = {nx}^{n - 1} \: for \: all \: n \in \: N}$

Answer 2

WE have to prove :⟹ $\bf{\frac{d}{dx}x^n=nx^{n-1}}$

for n = 1

• ⟹ $\bf{LHS =\frac{d}{dx}(x)=1}$

(x)=1

RHS = 1.x^{1-1} = 1

So, LHS = RHS

∴ P(1) is true.

• ∴ P(n) is true for n = 1

⟹ Let P(k) be true for some positive integer k.

Now, to prove that P(k + 1) is also true

⟹ RHS =$\bf{(k+1)x^{k+1-1}}(k+1)x$

⟹ $\tt LHS = \bf{\frac{d}{dx}(x^{k+1})=\frac{d}{dx}(x\times x^k)}$

⟹ $=\bf{x^k\frac{d}{dx}(x)+x\frac{d}{dx}(x^k)}x$

⟹ $=\bf{x^k.1+x.kx^{k-1}}x$

⟹ $=\bf{x^{k+1-1}+kx^{k+1-1}}x$

⟹ $=\bf{(k+1)x^{k+1-1}}(k+1)$

• ∴ LHS = RHS

Thus, P(k + 1) is true whenever P(k) is true.

Thus, P(k + 1) is true whenever P(k) is true.Therefore, by the principle of mathematical induction, the statement P(n) is true for every positive integer n.

★$\text{\underline{Hence proved}}$★