Nishit B. A Level Maths tutor, GCSE Maths tutor, A Level Further Math...
£18 - £20 /hr

Nishit B.

Degree: Engineering (Masters) - Oxford, New College University

MyTutor guarantee

Contact Nishit
Send a message

All contact details will be kept confidential.

To give you a few options, we can ask three similar tutors to get in touch. More info.

Contact Nishit

About me

Hello! I am a second year Engineering undergraduate at the University of Oxford. I am happy to help with A-level/GCSE Maths.

Subjects offered

SubjectLevelMy prices
Maths A Level £20 /hr
Maths GCSE £18 /hr
Physics GCSE £18 /hr
Maths 13 Plus £18 /hr
-Personal Statements- Mentoring £20 /hr

Qualifications

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

CRB/DBS Standard

No

CRB/DBS Enhanced

No

General Availability

Weeks availability
MonTueWedThuFriSatSun
Weeks availability
Before 12pm12pm - 5pmAfter 5pm
MONDAYMONDAY
TUESDAYTUESDAY
WEDNESDAYWEDNESDAY
THURSDAYTHURSDAY
FRIDAYFRIDAY
SATURDAYSATURDAY
SUNDAYSUNDAY

Please get in touch for more detailed availability

Questions Nishit has answered

If y = (1+3x)^2, what is dy/dx?

A good approach to solve this is to use the chain rule of differentiation. The chain rule states: dy/dx= (dy/du)*(du/dx). In this case let u = 1+3x, so y = u^2. Then dy/du = 2u and du/dx = 3, so dy/dx = (2u)*3 = (2(1+3x))*3 = 6+18x

A good approach to solve this is to use the chain rule of differentiation. The chain rule states: dy/dx= (dy/du)*(du/dx).

In this case let u = 1+3x, so y = u^2.

Then dy/du = 2u and du/dx = 3,

so dy/dx = (2u)*3 = (2(1+3x))*3 = 6+18x

5 months ago

176 views
Send a message

All contact details will be kept confidential.

To give you a few options, we can ask three similar tutors to get in touch. More info.

Contact Nishit

Still comparing tutors?

How do we connect with a tutor?

Where are they based?

How much does tuition cost?

How do tutorials work?

Cookies:

We use cookies to improve our service. By continuing to use this website, we'll assume that you're OK with this. Dismiss

mtw:mercury1:status:ok