Calculus: Implicit derivatives and Natural logarithms

[Robert632]

Essentially, I don't really understand how to do those two things. Any help anyone could give is appreciated. Also, this is grade 12 calculus, so keep that in mind while answering.

In case anyone is confused, Implicit derivation is finding the derivative of something with two variables(I.E F(x)=x^2+xy+y^2) and Natural logarithms derivation is finding the derivative of something like x ln x.

 

[SckizoBoy]

Differentiation of a two variable system like what you've got simply means deriving a differential term within every part of the equation as appropriate. The example you've given would differentiate to:

dy/dx = 2x + y + x(dy/dx) + 2y(dy/dx)

Therefore dy/dx = (2x + y)/(1 - x - 2y)

Differentiate each term independently of each other, except for the fact that y-term also assumes the differential function as well as what it would normally become were it the only variable.

As for natural logarithms, I'll put it this way:

y = e[sup]x[/sup]
dy/dx = e[sup]x[/sup]

Therefore:

y = ln x
e[sup]y[/sup] = x
(dy/dx)e[sup]y[/sup] = 1
dy/dx = 1/(e[sup]y[/sup])
dy/dx = 1/x

 

[Robert632]

Differentiation of a two variable system like what you've got simply means deriving a differential term within every part of the equation as appropriate. The example you've given would differentiate to:

dy/dx = 2x + y + x(dy/dx) + 2y(dy/dx)

Therefore dy/dx = (2x + y)/(1 - x - 2y)

Differentiate each term independently of each other, except for the fact that y-term also assumes the differential function as well as what it would normally become were it the only variable.

As for natural logarithms, I'll put it this way:

y = e[sup]x[/sup]
dy/dx = e[sup]x[/sup]

Therefore:

y = ln x
e[sup]y[/sup] = x
(dy/dx)e[sup]y[/sup] = 1
dy/dx = 1/(e[sup]y[/sup])
dy/dx = 1/x

Thank you, That has helped me a bunch.

 

[SckizoBoy]

Thank you, That has helped me a bunch.

No sweat, just wish all the bullshit 'maths' question threads on Off-Topic were like this! -_-