Homogeneous function of degree n

[clairaut]

I can't see the proof clearly.

Gradient [f(tx,ty)] dot
<d/dt (tx) , d/dt (ty)> =
d/dt [f(tx,ty)] = n * t^(n-1) * [f(x,y)]

I don't see how the t^(n-1) term disappears.

HELP

 

[clairaut]

Help me!

 

[clairaut]

Echo!

 

[clairaut]

This proof is fundamental to our understanding of transcendental functions.

This proof is intimately connected to Euler's identity in that mass raised to the power of a rational number should equal e raised to the velocity function as fxn of mass.

Propulsion can also be designed in a cyclic process that is as stable as the circular motion that contains the constant centripetal acceleration.

Also, the circular wheel contains innate structural stability due to the forces pushing inward toward the center that causes tangential velocity.

Euler's number is so integral to the formation of a circle that there is an operator that brings out the power placed on the base as a multiple of the original function. This operator is known as the natural logarithm.

This proof must be understood
well!

Can somebody continue on with this incredible proof?

Prove inside this physics forum who is the real master of mathematical physics.

 

[pasmith]

I can't see the proof clearly.

Gradient [f(tx,ty)] dot
<d/dt (tx) , d/dt (ty)> =
d/dt [f(tx,ty)] = n * t^(n-1) * [f(x,y)]

I don't see how the t^(n-1) term disappears.

HELP

The result is true for all $t \in \mathbb{R}$, so in particular it is true for $t = 1$. Hence $$\mathbf{x} \cdot \nabla f = n 1^{n-1} f(\mathbf{x}) = n f(\mathbf{x})$$ as required.