Derivative of a composite function

[Tensel]

y =f(x), and y'=df(x)/dx, F(y)=y^3+(y')^2
how to deal with the (y')^2 when i calculate dF(y)/dy?
thanks.

 

[scurty]

Do you remember the chain rule? Are you having trouble differentiating that term?

 

[HallsofIvy]

If u(x) is a fuction of x, then, by the chain rule, the derivative of $u^2$ is 2u times the derivative of u.

If y is a function of x, then the derivative of $(y')^2$, with respect to x, is $2y'$ times the derivative of y' which is, of course, y''. That is, the derivative if $(y')^2$ is $2y' y''$.

 

[Tensel]

HallsofIvy, i want to calculate dF(y)/dy, not dF(y)/dx, but you remand me sth. thank you.

 

[HallsofIvy]

Well, that doesn't require any mention of x at all then! The derivative of $d(y)^2/dy= 2y$.