differentiation help

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

Reply 1

Kevin De Bruyne

21

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

Please post your working for part 2 :smile:

if x = 0 then dividing by x will be a problem. It shouldn't matter for your working out, though.

Reply 2

not_lucas

OP

4

i got dy/dx to = x-24x^-1 but i dont think it is right

Reply 3

Zacken

22

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

Reply 4

razzor

10

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.

Reply 5

username2250433

13

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?

Reply 6

Zacken

22

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

\frac{\mathrm{d}}{\mathrm{d}x} x^n = nx^{n-1}

, and

\frac{\mathrm{d}}{\mathrm{d}x} (f(x) + g(x)) = \frac{\mathrm{d}}{\mathrm{d}x}(f(x)) + \frac{\mathrm{d}}{\mathrm{d}x} (g(x))

, so using this on your function we have

\frac{\mathrm{d}}{\mathrm{d}x} (x - 24x^{-1}) = \frac{\mathrm{d}}{\mathrm{d}x} (x^1) - \frac{\mathrm{d}}{\mathrm{d}x} (24x^{-1}) = \cdots

You should get

Spoiler

(edited 8 years ago)

Reply 7

Zacken

22

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?