Hello, my name is Toby. I am a second year Mathematician at the University of Leeds, I am averaging a First Class degree at the moment and would love to be able to help out with whatever you need a hand with, GCSE and A-Level maths is where I would be most strong.

Hello, my name is Toby. I am a second year Mathematician at the University of Leeds, I am averaging a First Class degree at the moment and would love to be able to help out with whatever you need a hand with, GCSE and A-Level maths is where I would be most strong.

We only take tutor applications from candidates who are studying at the UK’s leading universities. Candidates who fulfil our grade criteria then pass to the interview stage, where a member of the MyTutor team will personally assess them for subject knowledge, communication skills and general tutoring approach. About 1 in 7 becomes a tutor on our site.

No DBS Check

We are looking for dy/dx, and in this question, the key rule to remember is the chain rule.

Therefore the first thing to do is the easy step, take the ^{3} down and minus 1 from it, this will give 3(6x-13)^{2}. But this is not the final answer as we have not also differentiated what is inside the bracket, hence we need to differentiate (6x-13), which is just 6, so we multiply this by the whole equation.

Therefore our answer will become 6*3(6x-13)^{2 }= 18(6x-13)^{2}

We are looking for dy/dx, and in this question, the key rule to remember is the chain rule.

Therefore the first thing to do is the easy step, take the ^{3} down and minus 1 from it, this will give 3(6x-13)^{2}. But this is not the final answer as we have not also differentiated what is inside the bracket, hence we need to differentiate (6x-13), which is just 6, so we multiply this by the whole equation.

Therefore our answer will become 6*3(6x-13)^{2 }= 18(6x-13)^{2}

To solve x^{2} - 8x + 15 = 0 we could try to use the Product, Sum, Numbers method (PSN).

The PSN method tells us that our numbers must multply to make the product of the first and last term, in this case 15. It also tells us that they must sum to the middle term, which is -8.

By thinking through all the factors of 15 (including negatives), we can work out which pair works:

15= 1 x 15, 3 x 5, -1 x -15 and -3 x -5

Of all these pairs, only the final pair add together to make -8, therefore these must be our two values of x.

We can then subsitute them in to say that:

x^{2 }- 3x + (-5x +15) = 0

Then we take the biggest factor out of both the first half and the second half:

x(x-3) -5(x-3) = 0

Which shows us that x^{2} - 8x + 15 = (x-3) (x-5)

So it can only be equal to zero when x = 3 or when x = 5

To solve x^{2} - 8x + 15 = 0 we could try to use the Product, Sum, Numbers method (PSN).

The PSN method tells us that our numbers must multply to make the product of the first and last term, in this case 15. It also tells us that they must sum to the middle term, which is -8.

By thinking through all the factors of 15 (including negatives), we can work out which pair works:

15= 1 x 15, 3 x 5, -1 x -15 and -3 x -5

Of all these pairs, only the final pair add together to make -8, therefore these must be our two values of x.

We can then subsitute them in to say that:

x^{2 }- 3x + (-5x +15) = 0

Then we take the biggest factor out of both the first half and the second half:

x(x-3) -5(x-3) = 0

Which shows us that x^{2} - 8x + 15 = (x-3) (x-5)

So it can only be equal to zero when x = 3 or when x = 5

The integration by parts rule looks like this:

∫ u * v' dx = u * v - ∫ ( v * u' ) dx

Hence in this example, we want to make our u = x and v' = sinx

So we now need to work out what u' and v are:

u' = 1 which is the easier of the two; to work out v, we should integrate v' = sinx, this will give us v = -cosx

Hence if we now subsititute these into the equations, we will find that:

∫ xsinx dx = -xcosx - ∫ (-cosx) dx

= -xcosx - (-sinx) + C (where C is the constant of integration)

= sinx - xcosx + C

The integration by parts rule looks like this:

∫ u * v' dx = u * v - ∫ ( v * u' ) dx

Hence in this example, we want to make our u = x and v' = sinx

So we now need to work out what u' and v are:

u' = 1 which is the easier of the two; to work out v, we should integrate v' = sinx, this will give us v = -cosx

Hence if we now subsititute these into the equations, we will find that:

∫ xsinx dx = -xcosx - ∫ (-cosx) dx

= -xcosx - (-sinx) + C (where C is the constant of integration)

= sinx - xcosx + C