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# Find dy/dx when y = x / x^2 + 1 watch

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1. Hey,

For some reason I am unable to know how remove the denominator so if some kind person could go through it with me relatively slowly it would be great

Find dy / dx where y = x / x² + 1
2. (Original post by supreme)
Hey,

For some reason I am unable to know how remove the denominator so if some kind person could go through it with me relatively slowly it would be great

Find dy / dx where y = x / x² + 1
x/x²
= x^1 / x^2
= x^1 ÷ x^2

Using the laws of indices
x^a ÷ x^b = x^(a-b)

so
x^1 ÷ x^2
= x^(1-2)
=x^(-1)
3. (Original post by supreme)
Hey,

For some reason I am unable to know how remove the denominator so if some kind person could go through it with me relatively slowly it would be great

Find dy / dx where y = x / x² + 1
y = x / x² + 1

so x/x² = 1/x (think of it as x/x times x , so one cancels out on top an bottom)

so y = 1/x + 1

dy/dx = -1/x²

does that help?
4. Are you sure its not x/ (x^2 + 1)
5. (Original post by Vazzyb)
Are you sure its not x/ (x^2 + 1)
Yes

(Original post by moojoo)
y = x / x² + 1

so x/x² = 1/x (think of it as x/x times x , so one cancels out on top an bottom)

so y = 1/x + 1

dy/dx = -1/x²

does that help?
Thanks it mostly helps I understand up to y = 1/x + 1, I don't get where the -1 and x² comes from....though..?
6. (Original post by supreme)
Yes

Thanks it mostly helps I understand up to y = 1/x + 1, I don't get where the -1 and x² comes from....though..?
y= x-1 + 1
when you differentiate x-1 you times x by its power (i.e. -1) and then minus 1 from its original power (i.e. -1-1=-2) to give you its new power so...
dy/dx= -x-2
you could also write this as -1/x2
7. (Original post by *bobo*)
y= x-1 + 1
when you differentiate x-1 you times x by its power (i.e. -1) and then minus 1 from its original power (i.e. -1-1=-2) to give you its new power so...
dy/dx= -x-2
you could also write this as -1/x2
Thanks I get it now

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