## Integration – Problme solving

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## Differentiation – Product rule

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## Differentiation by first principles

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

## Sample post

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

## Probability Density Function

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## IB Math Standard Level Revision

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## IB Math Higher Level Revision

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## International Mathematics Revision

You can access the IGCSE International Mathematics exam point of view revision here. This page contains topic wise revision materials, practice papers, and exam-style questions. Register free to access this page.

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## Volume of rotation

123

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