Nat N. GCSE Maths tutor, A Level Maths tutor, GCSE Further Mathematic...

Nat N.

Currently unavailable: until 22/06/2016

Degree: Maths (Bachelors) - Warwick University

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About me

Me:

Hello! I’m a Maths student at Warwick university. I think Maths and it’s related subjects are exceptionally beautiful once you understand them. I hope to be able to help pupils reach that kind of understanding (as well as passing exams, of course).

Teaching Style:

Formulas and algorithms can be extremely useful time savers, but remembering one detail of them wrong can ruin an entire question. Whenever possible, I will try to give some insight into why a method is being used, as well as how to use it. Not only is this the only way to really enjoy the subject, but I also think it’s lot easier to recall how to do something if you have some understanding of exactly what you’re doing.

(This doesn’t mean I am against shortcuts and formulas: I think they’re great. But they are often worth examining quickly before learning them off).

I also think everyone needs to do more practise than they imagine necessary.

If this all sounds rather po-faced and serious, I apologise. I do possess something like a sense of humour, and will do my utmost to keep tutorials from becoming too boring.

Sessions:

I want to tailor the content of the sessions to fit whatever the pupil wants to work on. I would anticipate the three main areas I would be covering to be:

     – Going through concepts (ideally one or two per session)

     – Working through more difficult questions together

     – Working on past papers and exam technique

However, I'm more than happy to approach the sessions in a different way if the pupil requests it.

If you have any queries, I’ll happily answer any questions you message me with.

Subjects offered

SubjectLevelMy prices
Further Mathematics A Level £20 /hr
Maths A Level £20 /hr
Further Mathematics GCSE £18 /hr
Maths GCSE £18 /hr
Physics GCSE £18 /hr

Qualifications

QualificationLevelGrade
MathsA-LevelA*
Further MathsA-LevelA
PhysicsA-LevelA
PhilosophyA-LevelC
Disclosure and Barring Service

CRB/DBS Standard

No

CRB/DBS Enhanced

No

Currently unavailable: until

22/06/2016

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Questions Nat has answered

Differentiate y=x*ln(x^3-5)

We can immediately see that more than differentiation rule will be needed here. The expression in question is the product of two smaller expressions, so the product rule may be useful. But to apply the product rule, we need to be able to differentiate the two smaller expressions. ln(x3-5) is s...

We can immediately see that more than differentiation rule will be needed here. The expression in question is the product of two smaller expressions, so the product rule may be useful. But to apply the product rule, we need to be able to differentiate the two smaller expressions. ln(x3-5) is slightly more complicated to differentiate. However, notice it is the composition of two functions we know how to differentiate: x3-5 and ln(x). This suggest we may be able to apply the chain rule.

First, let u=ln(x3-5)

and v=x3-5

Then u=ln(v)

Differentiating u and v:

du/dv=1/v

dv/dx=3x2

Recall the formula for the chain rule, which in this case is du/dx=(du/dv)*(dv/dx)

Substituting into the chain rule:

du/dx=(du/dv)*(dv/dx)

=(1/v)*(3x2)

=3x2/v

=3x2/(x3-5)

So, d/dx(ln(x3-5))=3x2/(x3-5)

Note – In an exam, it may be faster simply to use the standard formula for differentiating ln: d/dx(ln(f(x)))=f'(x)/f(x) . You can use this formula whenever you spot you are differentiating ln of some function. You should be able to see how this would work in the above example. I have provided a full method for clarity, not because it is necessary to do so in your exam.

We are now in a position to apply the product rule. Recall that the formula for the product rule is d/dx(UV)=V*dU/dx+U*dV/dx   (U and V here used just to avoid confusion with u and v used earlier)

Let U=x

and V=ln(x3-5)

then dU/dx=1

and dV/dx=3x2/(x3-5)

Substituting into the product rule formula:

d/dx(x*ln(x3-5))=dx(UV)

=V*dU/dx+U*dV/dx

=ln(x3-5)*1+x*3x2/(x3-5)

=ln(x3-5)+3x3/(x3-5)

This gives us our answer:

dy/dx=ln(x3-5)+3x3/(x3-5)

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8 months ago

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