Implicit Differentiation (1 Viewer)

[is2SWaNz]

I'm having some trouble understanding implicit differentiation. Here's the question:

I understand how they got 2x+2yy'=y... but i don't get how you can differentiate a '3'. The answer is xy' and i don't get how they got that.

One other question x^3+y^3=1

I know how to 'implicit differetiate' the LHS, but then the '1' is differentiated (is that a word??) into zero.. And that confuses me. Can anyone help me?

Thanks in advance! ><

 

[jyu]

I'm having some trouble understanding implicit differentiation. Here's the question:

I understand how they got 2x+2yy'=y... but i don't get how you can differentiate a '3'. The answer is xy' and i don't get how they got that.

One other question x^3+y^3=1

I know how to 'implicit differetiate' the LHS, but then the '1' is differentiated (is that a word??) into zero.. And that confuses me. Can anyone help me?

Thanks in advance! ><

Derivative of a constant with respect to any variable is zero. You can think of derivative of a function as the gradient function of the function. If the function is constant, then its gradient must be zero.

:wave:

 

[is2SWaNz]

I thought so too. About the constant being zero, but how come the '3' isn't zero? Its differentiated as xy'?

 

[jyu]

I thought so too. About the constant being zero, but how come the '3' isn't zero? Its differentiated as xy'?

Implicit diff. of xy with respect to x yields y + xy'.

:wave:

 

[pLuvia]

Basically where you differentiate a y with respect to x, it becomes y' so in your case since you want to implicitly differentiate xy it would be y (differentiating the x first) + xy' (differentiating the y), overall using the product rule

 

[is2SWaNz]

lol!!! thanks guys.

just last night i realised what a stupid stupid mistake i've done.... argh... how embarassing. I understand now.

thanks!

 

[dongypro]

d/dx (3) = 0