Hanumanth Srikar K.

£20 - £22 /hr

Maths and Economics (Bachelors) - LSE University

5.0

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#### Qualifications

MathematicsInternational Baccalaureate (IB) (HL)6
Further MathematicsInternational Baccalaureate (IB) (HL)6
ChemistryInternational Baccalaureate (IB) (HL)6
EconomicsInternational Baccalaureate (IB) (SL)7
GermanInternational Baccalaureate (IB) (SL)6
EnglishInternational Baccalaureate (IB) (SL)6

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#### Subjects offered

SubjectQualificationPrices
ChemistryGCSE£20 /hr
GermanGCSE£20 /hr
MathsGCSE£20 /hr
EconomicsIB£22 /hr
Further MathematicsIB£22 /hr
MathsIB£22 /hr

### Differentiation from first principles

Differentiaiton from principles requires the use of the following formula which is provided in the formula booklet:

f'(x) = limh->0 ((f(x+h) - f(x))/(h))

Consider a function:

f(x) = 6x2

Clearly we know that the function differentiates to:

f'(x) = 12x

by using the process of multiplying the coefficient by the power and then reducing the power by 1.

Using first principles however we must consider the formula mentioned previously.

f'(x) = limh->0 ((f(x+h) - f(x))/(h))

By computing the function for x+h and x we get:

f'(x) = limh->0 (6(x+h)2 - 6x2)/(h))

f'(x) = limh->0 (6(x2+2xh+h2) - 6x2)/(h))

f'(x) = limh->0 (6x2+12xh+6h2) - 6x2)/(h))

f'(x) = limh->0 (12xh+6h2)/(h))

We now cancel the h from above and below to get:

f'(x) = limh->0 12x+6h

Now consider the limit as h-> 0, clearly 12x remains unaffected but 6h will become 0 and is hence removed. Hence we are left with:

f'(x) = 12x

Which we know to be true from the trivial methods of differentiation considered earlier.

Differentiaiton from principles requires the use of the following formula which is provided in the formula booklet:

f'(x) = limh->0 ((f(x+h) - f(x))/(h))

Consider a function:

f(x) = 6x2

Clearly we know that the function differentiates to:

f'(x) = 12x

by using the process of multiplying the coefficient by the power and then reducing the power by 1.

Using first principles however we must consider the formula mentioned previously.

f'(x) = limh->0 ((f(x+h) - f(x))/(h))

By computing the function for x+h and x we get:

f'(x) = limh->0 (6(x+h)2 - 6x2)/(h))

f'(x) = limh->0 (6(x2+2xh+h2) - 6x2)/(h))

f'(x) = limh->0 (6x2+12xh+6h2) - 6x2)/(h))

f'(x) = limh->0 (12xh+6h2)/(h))

We now cancel the h from above and below to get:

f'(x) = limh->0 12x+6h

Now consider the limit as h-> 0, clearly 12x remains unaffected but 6h will become 0 and is hence removed. Hence we are left with:

f'(x) = 12x

Which we know to be true from the trivial methods of differentiation considered earlier.

2 years ago

786 views

Factorising a quadratic polynomial of the form,

ax2 + bx + c = 0

can be done in many depends depending on the values we have for a, b and c.

Some simple polynomial could be done simply by recognition. For example:

x2 + 4x + 3 = 0

In this scenario we would use the assumption that the factorisation is of the form:

(dx+e)(fx+g) = 0

and we would consider the values of d, e, f and g.

As the coefficient of the xterm is 1, d and f would both be 1. Hence:

(x+e)(x+g) = 0

Now we need numbers for e and g such that:

e*g = 3 and,

e+g = 4

We find that e = 3 and g = 1 or visa versa.

Hence we can factorise

x2 + 4x + 3 = 0

to (x+3)(x+1) = 0

Factorising a quadratic polynomial of the form,

ax2 + bx + c = 0

can be done in many depends depending on the values we have for a, b and c.

Some simple polynomial could be done simply by recognition. For example:

x2 + 4x + 3 = 0

In this scenario we would use the assumption that the factorisation is of the form:

(dx+e)(fx+g) = 0

and we would consider the values of d, e, f and g.

As the coefficient of the xterm is 1, d and f would both be 1. Hence:

(x+e)(x+g) = 0

Now we need numbers for e and g such that:

e*g = 3 and,

e+g = 4

We find that e = 3 and g = 1 or visa versa.

Hence we can factorise

x2 + 4x + 3 = 0

to (x+3)(x+1) = 0

2 years ago

837 views

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