C4 Implicit Differentiation

I've been having some trouble with implicit differentiation in C4, and just wondered if anyone knows of any useful sites that are good for learning this with examples.
Thanks all

Reply 1

Swayum

18

You basically just differentiate everything as normal, but when you differentiate anything involving y, you need to multiply by dy/dx as well. So if you have

y^2 + sinx - cosy + x^2y^3 = 10x - lny + e^x + y^2

If you differentiate without really thinking about anything, you get

2y + cosx + siny + 2xy^3 + 3x^2y^2 = 10 - 1/y + e^x + 2y

Right? (notice I differentiated x^2y^3 using the product rule)

Now remember what I said? I said to multiply by dy/dx whenever you differentiate y, so really the answer is:

2ydy/dx + cosx + sinydy/dx + 2xy^3 + 3x^2y^2dy/dx = 10 - 1/y * dy/dx + e^x + 2ydy/dx.

That's all.

Reply 2

Crazy Paving

17

What Swayum said. Simple as.

Then you just factor out dy/dx at the end to get the gradient if needed or whatever the question is.

Reply 3

CreativeBass

OP

Swayum

You basically just differentiate everything as normal, but when you differentiate anything involving y, you need to multiply by dy/dx as well. So if you have

y^2 + sinx - cosy + x^2y^3 = 10x - lny + e^x + y^2

If you differentiate without really thinking about anything, you get

2y + cosx + siny + 2xy^3 + 3x^2y^2 = 10 - 1/y + e^x + 2y

Right? (notice I differentiated x^2y^3 using the product rule)

Now remember what I said? I said to multiply by dy/dx whenever you differentiate y, so really the answer is:

2ydy/dx + cosx + sinydy/dx + 2xy^3 + 3x^2y^2dy/dx = 10 - 1/y * dy/dx + e^x + 2ydy/dx.

That's all.

Ahh thank you :smile:

I guess it just sounds harder then it actually is :smile:

Thanks again x

C4 Implicit Differentiation

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