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# differentiation help watch

1. im stuck on two questions, any help appreciated

1: given that y=8x^3 - 4x^0.5 + (3x^2 + 2)/x , x>0

i know how to differentiate but what does the x>0 mean, is it there to throw me off?

2: the curve c has equation y=(x+3)(x-8)/x , x>0
a: find dy/dx in its simplest form
b: find an equation of the tangent to c at the point where x=2

i started working it out but then got a ridiculous answer

thanks
2. (Original post by not_lucas)
im stuck on two questions, any help appreciated

1: given that y=8x^3 - 4x^0.5 + (3x^2 + 2)/x , x>0

i know how to differentiate but what does the x>0 mean, is it there to throw me off?

2: the curve c has equation y=(x+3)(x-8)/x , x>0
a: find dy/dx in its simplest form
b: find an equation of the tangent to c at the point where x=2

i started working it out but then got a ridiculous answer

thanks

if x = 0 then dividing by x will be a problem. It shouldn't matter for your working out, though.
3. i got dy/dx to = x-24x^-1 but i dont think it is right
4. (Original post by not_lucas)
i got dy/dx to = x-24x^-1 but i dont think it is right
Unfortunately not. I fear that you may have forgotten to apply the quotient rule properly, did you remember to
Spoiler:
Show
divide by the square of the denominator instead of just the denominator? Because your answer is nearly correct.
5. (Original post by not_lucas)
i got dy/dx to = x-24x^-1 but i dont think it is right
It's not correct. Can you post your working?
You don't need to use the quotient rule for this. Just simplify the fractions.
6. i multiplied out the brackets to get x^2-5x-24 which i then divided by x as shown in the curve c equation then from there i ended up with x-5-24x^-1 which i differentiated to x-24x^-1. I think i know where i went partly wrong, the x should become 1 due to it being 1x shouldn't it?

Also, im stuck on another question which asks me to find the area of a triangle, i have two coordinates but would i not need a third, i got those coordinates by subbing y=0 into the equations given as its says the normal and tangent meet the x axis at B and A respectively, do i get the third lot of coordinates by using simultaneous equations with the normal and tangent equations or something?
7. (Original post by not_lucas1)
i multiplied out the brackets to get x^2-5x-24 which i then divided by x as shown in the curve c equation then from there i ended up with x-5-24x^-1 which i differentiated to x-24x^-1. I think i know where i went partly wrong, the x should become 1 due to it being 1x shouldn't it?
Use the rule that , and , so using this on your function we have

You should get
Spoiler:
Show
8. (Original post by not_lucas1)
Also, im stuck on another question which asks me to find the area of a triangle, i have two coordinates but would i not need a third, i got those coordinates by subbing y=0 into the equations given as its says the normal and tangent meet the x axis at B and A respectively, do i get the third lot of coordinates by using simultaneous equations with the normal and tangent equations or something?
You would be better off posting a new thread entirely with the full question. :-)
9. I got 1+ 24x^-2. If you times the -24 by the power of -1 shouldn't the result be positive?

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