Manojhan S. GCSE Maths tutor, A Level Maths tutor, A Level Physics tutor
£18 - £20 /hr

Manojhan S.

Degree: Mechanical Engineering (Masters) - Bath University

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About me

About Me:

I am a Mechanical Engineering student at The University of Bath, where I am currently in my second year of the degree. I have found great enjoyment from learning about maths and science from my teachers and being able to apply this knowledge, which is why I’ve decided to study engineering and hope this passion for engineering and science can be passed down to you.

I am very friendly and understanding. I was part of a tutoring scheme in college where I helped out students completing their AS while I was studying for my A2 exams. I have also been part of the Sea Cadets for 3 years where I was often asked to instruct those new or younger about some of the courses on seamanship and communication.

Sessions:

During the sessions, I will be offering my knowledge and help onto topics that you wish to cover. I have done this previously, where I have researched into each topic before the session to ensure that you are getting the most understanding you do. I always go through many examples until you feel confident about the topic, and going through exam style questions.

I hope that during the sessions, they are fun. During the 55 mins, a lot of interesting content can be covered and by the end of each session I hope you have learned something new and intriguing.  

What next?

If you have any questions, you can email me or book a ‘Meet the Tutor Session’. Remember to inform me on the exam board and the topics you wish to cover.

I look forward to meeting you!

Subjects offered

SubjectLevelMy prices
Maths A Level £20 /hr
Maths GCSE £18 /hr

Qualifications

QualificationLevelGrade
MathematicsA-LevelA*
PhysicsA-LevelA*
ChemistryA-LevelA*
Disclosure and Barring Service

CRB/DBS Standard

No

CRB/DBS Enhanced

No

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Questions Manojhan has answered

Differentiate y = x^3 +x^2 - 4x +5 with respects to x.

When differentiating, you want to use the formula ax^n differentiates to (a*n)x^(n-1), so for the example above, x^3 where a is 1, and n is 3, the differentiation is (1*3)x^(3-1) which results to 3x^2. This is repeated for x^2 as 2x. For -4x, a = -4 and n = 1 so differentiating this becomes (...

When differentiating, you want to use the formula ax^n differentiates to (a*n)x^(n-1), so for the example above, x^3 where a is 1, and n is 3, the differentiation is (1*3)x^(3-1) which results to 3x^2.

This is repeated for x^2 as 2x. For -4x, a = -4 and n = 1 so differentiating this becomes (-4*1)x^(1-1)  which results to -4 as x^0 is 1. When the term does not have an x term like for instance +5, the term can be disregarded. It can be rewritten as 5x^0 and when differentiated, becomes (5*0)x^0-1. This results in the answer being 0.

The overall answer for this question is dy/dx = 3x^2 + 2x - 4.   

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