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1. Prove by mathematical induction that (frac{d}{d x}left(x^{n}right)=nleft(x^{n-1}right) )

Proof:

Consider the proposition P(n): (frac{d}{d x}left(x^{n}right)=nleft(x^{n-1}right) )

Let n = 1

LHS(=frac{d}{d x}left(x^{1}right)=1)

RHS(=1 times x^{1-1}=1)

Therefore the statement is true for (n=1)

Let us assume that the result is true for n = k,

So (frac{d}{d x}left(x^{k}right)=kleft(x^{k-1}right) ) be true

To prove P(n) is true for n = k+1 also, we need to show that (frac{d}{d x}left(x^{k+1}right)=(k+1)left(x^{k}right) )

To prove this lets us take LHS first

LHS =( frac{d}{d x}left(x^{k+1}right)=frac{d}{d x}left(k x^{k} cdot xright))

Use product rule differentiation, we get

(=x^{k}(1)+xcdot k cdot x^{k-1}),

(=x^{k}+k cdot x^{k}),

(=x^{k}(k+1))

=RHS

1. Prove by mathematical induction that (frac{d}{d x}left(x^{n}right)=nleft(x^{n-1}right) )

Proof:

Consider the proposition P(n): (frac{d}{d x}left(x^{n}right)=nleft(x^{n-1}right) )

Let n = 1

LHS(=frac{d}{d x}left(x^{1}right)=1)

RHS(=1 times x^{1-1}=1)

Therefore the statement is true for (n=1)

Let us assume that the result is true for n = k,

So (frac{d}{d x}left(x^{k}right)=kleft(x^{k-1}right) ) be true

To prove P(n) is true for n = k+1 also, we need to show that (frac{d}{d x}left(x^{k+1}right)=(k+1)left(x^{k}right) )

To prove this lets us take LHS first

LHS =( frac{d}{d x}left(x^{k+1}right)=frac{d}{d x}left(k x^{k} cdot xright))

Use product rule differentiation, we get

(=x^{k}(1)+xcdot k cdot x^{k-1}),

(=x^{k}+k cdot x^{k}),

(=x^{k}(k+1))

=RHS

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WordPress users may install and switch among different themes. Themes allow users to change the look and functionality of a WordPress website without altering the core code or site content. Every WordPress website requires at least one theme to be present and every theme should be designed using WordPress standards with structured PHP, valid HTML (HyperText Markup Language), and Cascading Style Sheets (CSS).