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Differentiation

I'm aware this is probably rather easy but i just cant seem to get that answer and it's really irritating me, can someone talk us through it please.

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Reply 1

Slowbro93

Original post by jackpethick

I'm aware this is probably rather easy but i just cant seem to get that answer and it's really irritating me, can someone talk us through it please.

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What have you tried so far?

Reply 2

Zacken

Original post by jackpethick

I'm aware this is probably rather easy but i just cant seem to get that answer and it's really irritating me, can someone talk us through it please.

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Why don't you try differentiating with respect to

y

so that you get

\frac{\mathrm{d}x}{\mathrm{d}y}

and then flip them over (reciprocate) to get

\frac{\mathrm{d}y}{\mathrm{d}x}

You will need to use the chain and product rule to do the differentiation.

Reply 3

PlAtYpUsBoi97

Original post by Slowbro93

What have you tried so far?

I tried the chain rule and that didnt work so did the product rule. I'm gonna give it another go now ive left it for 5, literally couldnt seem to get to that answer haha

Reply 4

joostan

Original post by jackpethick

I'm aware this is probably rather easy but i just cant seem to get that answer and it's really irritating me, can someone talk us through it please.

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The easiest way to show this is probably using

\dfrac{dy}{dx}=1/\dfrac{dx}{dy}

.

EDIT: Oh my so very slow. :frown:

Reply 5

Slowbro93

Original post by Zacken

Why don't you try differentiating with respect to

y

so that you get

\frac{\mathrm{d}x}{\mathrm{d}y}

and then flip them over (reciprocate) to get

\frac{\mathrm{d}y}{\mathrm{d}x}

You will need to use the chain and product rule to do the differentiation.

It would be useful to see if he's already done that and whether it's the case that he's having an issue with algebra. Start with what they now and work from there.

Reply 6

Slowbro93

Original post by jackpethick

I tried the chain rule and that didnt work so did the product rule. I'm gonna give it another go now ive left it for 5, literally couldnt seem to get to that answer haha

Well you'll have to use the chain rule, but have you considered combining it with the product rule. Where is your working? :smile:

Reply 7

antebellum

I've worked it out for you, just trying to find out how to post a pic of my solution

Reply 8

joostan

Original post by antebellum

I've worked it out for you, just trying to find out how to post a pic of my solution

Please don't, it's best to try to guide the OP to the solution rather than do it for them.
Besides full solutions except as an absolute last resort are against forum guidelines.

(edited 8 years ago)

Reply 9

antebellum

Original post by joostan

Please don't, it's best to try to guide the OP to the solution rather than do it for them.

Well okay then, all they need to do is use a cheeky bit of product rule

Reply 10

antebellum

Original post by KINGYusuf

Yeah it's easy, leme send a pic

This is a maths forum, Yusuf. Unsolicited dick pics are largely frowned upon.

Reply 11

KINGYusuf

Original post by antebellum

This is a maths forum, Yusuf. Unsolicited dick pics are largely frowned upon.

LOOOL

Reply 12

Student403

Original post by KINGYusuf

Yeah it's easy, leme send a pic

Original post by joostan

Please don't, it's best to try to guide the OP to the solution rather than do it for them.
Besides full solutions except as an absolute last resort are against forum guidelines.

Reply 13

KINGYusuf

Original post by Student403

I didn't read that part, mb.

Reply 14

Student403

Original post by KINGYusuf

I didn't read that part, mb.

Thanks for deleting it :biggrin:

Reply 15

KINGYusuf

Original post by jackpethick

I tried the chain rule and that didnt work so did the product rule. I'm gonna give it another go now ive left it for 5, literally couldnt seem to get to that answer haha