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AS Level differentiation

jessica_anne_clu

I understand that 1/x would be dy/dx=-1/x^2
but
would xy=24 be dy/dx=x^-24 or x^-1 :s-smilie:

?

The q is: find the equation of the tangent on the curve at the point indicated:
xy=24 where x=2

Reply 1

TenOfThem

neither

you have

y = \frac{24}{x} = 24x^{-1}

Reply 2

TenOfThem

So when you differentiate you just multiply by the constant

Reply 3

jessica_anne_clu

Original post by TenOfThem

neither

you have

y = \frac{24}{x} = 24x^{-1}

Thankyou so much!!! :smile:

I accidentely clicked the thumbs down when I went to click the thumbs up for your quote, it isn't letting me change it :s-smilie:

?
Thankyou anyway :biggrin:

no worries :smile:

Original post by jessica_anne_clu

Thankyou so much!!! :smile:

I accidentely clicked the thumbs down when I went to click the thumbs up for your quote, it isn't letting me change it :s-smilie:

?
Thankyou anyway :biggrin:

So... when dy/dx=-24x^-2, m=6 so m of the normal would be -1/6

where would I go from therE?

Reply 6

TenOfThem

Did you need to find the equation of the tangent or of the normal?

Either way ... you know m

You know that x=2 ... what is y

Then you also have a point

Since you have a gradient and a point you should be able to find the equation

Reply 7

jessica_anne_clu

Original post by TenOfThem

trying to find the normal which is the reciprocal of the tangent changing the sign right? :smile:

Reply 8

TenOfThem

yes you have the gradient

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AS Level differentiation

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