How do I compute the following differentiation by chain rule?

[gentsagree]

How do I compute the following differentiation by chain rule?

[tex]\frac{d}{d\lambda}(\lambda^{-1}\phi(\lambda^{-1}x))[/tex]

It is not a homework, but I can't figure out the exact way of getting the answer [itex]-\phi(x)-x^{s}\partial_{s}\phi(x)[/itex]

 

[HallsofIvy]

First, this is a product, [itex]\lambda^{-1}[/itex] times [itex]\phi(\lambda^{-1}x)[/itex] so you have to use the product rule. The derivative of [itex]\lambda^{-1}[/itex] with respect to [itex]\lambda[/itex] is [itex]-\lambda^{-2}[/itex] and the derivative of [itex]\phi(\lambda^{-1}x)[/itex] with respect to [itex]\lambda[/itex] is [itex]\phi'(\lambda^{-1}x) (-\lambda^{-2}x)[/itex].

So the derivative of [itex]\lambda^{-1}\phi(\lambda^{-1}x)[/itex] is [itex]-\lambda^{-2}\phi(\lambda^{-1}x)- \lambda^{-3}x\phi'(\lambda^{-1}x)[/itex]. That is NOT what you give but I don't know what [itex]-\phi(x)- x^s \partial_s\phi(x)[/itex] means because there is no "s" in your statement of the problem. Nor do I know where the "[itex]\lambda[/itex]" disappeared to! Are you sure you have stated the problem correctly?

Is it possible that this is the derivative at [itex]\lambda= 1[/itex]? But I still don't understand what "s" is.