Callum O. GCSE Maths tutor, A Level Maths tutor, 11 Plus Maths tutor,...

Callum O.

£18 - £22 /hr

Currently unavailable: for regular students

Studying: Computer Science (Masters) - Warwick University

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About me

With a strong passion for Mathematics and the Sciences, I never turn down oppurtunities to share these passions with keen individuals. Having had extensive tuition experience working in one-to-one tuition centres, I have been able to develop numerous tuition styles to suit you/your child's needs. 

I have a wealth of resources from numerous exam boards to facilitate in all aspects of learning. In addition, our tuition sessions would result in you not only being able to understand given topics, but more cruicially be able to define and tackle these topics in your exams.

With a strong passion for Mathematics and the Sciences, I never turn down oppurtunities to share these passions with keen individuals. Having had extensive tuition experience working in one-to-one tuition centres, I have been able to develop numerous tuition styles to suit you/your child's needs. 

I have a wealth of resources from numerous exam boards to facilitate in all aspects of learning. In addition, our tuition sessions would result in you not only being able to understand given topics, but more cruicially be able to define and tackle these topics in your exams.

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09/08/2016

Qualifications

SubjectQualificationGrade
Mathematics A-level (A2)A*
Further Mathematics A-level (A2)A
PhysicsA-level (A2)A
ChemistryA-level (A2)A

General Availability

Before 12pm12pm - 5pmAfter 5pm
mondays
tuesdays
wednesdays
thursdays
fridays
saturdays
sundays

Subjects offered

SubjectQualificationPrices
Further Mathematics A Level£20 /hr
MathsA Level£20 /hr
ChemistryGCSE£18 /hr
Further Mathematics GCSE£18 /hr
MathsGCSE£18 /hr
PhysicsGCSE£18 /hr
Maths13 Plus £18 /hr
Maths11 Plus£18 /hr

Questions Callum has answered

Differentiate with respect to X: x^2 + 2y^2+ 2xy = 2

Assuming the correct tools of differentiation have been taught, we can tackle each term seperately and then rearrange to have dy/dx as the subject.

Taking a look at the first term, x^2,  differentiating this term would become 2x (diffentiating x^n = nx^n-1)

Taking a look at the second term, 2y^2, it would appear we could differentiate it just like we did the first term. However this variable involves y and not x, meaning we must differentiate it implicitly.Therefore differentiating 2y^2 would become 4y(dy/dx)

Taking a look at the third term, 2xy, we immediately notice that it has both x terms and y terms involved; this should immediately hint to us that the product rule should be used. Therefore differentiating 2xy would become 2y + 2x(dy/dx) (Differentiating any term involving any other variable other than x with respect to x would require implicit differentiation).

Differentiating any constant (2) would = 0

Putting all these terms together would give:

2x + 4y(dy/dx) + 2y + 2x(dy/dx) = 0

With our basic GCSE knowledge of subject formula we can get:

2x + (dy/dx)(4y+2x) = 0

dy/dx = (-2x) / (4y+2x)

Assuming the correct tools of differentiation have been taught, we can tackle each term seperately and then rearrange to have dy/dx as the subject.

Taking a look at the first term, x^2,  differentiating this term would become 2x (diffentiating x^n = nx^n-1)

Taking a look at the second term, 2y^2, it would appear we could differentiate it just like we did the first term. However this variable involves y and not x, meaning we must differentiate it implicitly.Therefore differentiating 2y^2 would become 4y(dy/dx)

Taking a look at the third term, 2xy, we immediately notice that it has both x terms and y terms involved; this should immediately hint to us that the product rule should be used. Therefore differentiating 2xy would become 2y + 2x(dy/dx) (Differentiating any term involving any other variable other than x with respect to x would require implicit differentiation).

Differentiating any constant (2) would = 0

Putting all these terms together would give:

2x + 4y(dy/dx) + 2y + 2x(dy/dx) = 0

With our basic GCSE knowledge of subject formula we can get:

2x + (dy/dx)(4y+2x) = 0

dy/dx = (-2x) / (4y+2x)

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1 year ago

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