Find Derivative of V(u): (a-pX/q)^B

[3.141592654]

Homework Statement

I need to find the derivative of V(u) with respect to x

V(u)=((a-pX)/q)^B

Homework Equations

f(x)=aX^b ==> f'(x)=baX^b-1

The Attempt at a Solution

I wanted to distribute the B throughout and then I would have got -BpX^B-1, but I know this isn't doable. What I really need to know is how to deal with the entire function raised to a power. Once I get the function by itself (not raised to power B) I can get through it. Thanks for your help!

 

[James R]

Do you know the chain rule?

 

[cmos]

Take q outside of the denominator and treat it as constant. This should help you visualize the rest.

 

[3.141592654]

I'm a little confused about the advice I got, let me write out the entire problem (I'm going to change the signs).

U(x,y)=(X^a)(Y^b).

Secondly, pX+cY=T

This is the given information.

First, I solved px+cy=T to get y in terms of x:

Y=(T-px)/c

NOW, should I use the chain rule to find U'(X,Y):

U'(x,y)=(ax)+(by)*(dy/dx)

=(ax)+(by)*(-px)

OR plug y(x) back into the U function like I did originally and solve:

U(x,y)=(x^a)((T-px)/c)^b?

If this second option is the case, how would I go about solving ((T-px)/c)^b? If I took c out of the denominator it'd be like this right: (T-px+(c^-1))^b. If this is the case, I am still having a hard time with what the derivative would be with the power b outside the equation. Thanks again.

 

[3.141592654]

I should probably note the objective is to find U'(x,y), set it equal to zero and solve for X and Y.

 

[HallsofIvy]

I'm a little confused about the advice I got, let me write out the entire problem (I'm going to change the signs).

U(x,y)=(X^a)(Y^b).

Secondly, pX+cY=T

This is the given information.

I don't see how this has anything to do with the original problem! And what exactly is the question here? You have given us two equations. What are we to do with them?

First, I solved px+cy=T to get y in terms of x:

Y=(T-px)/c

NOW, should I use the chain rule to find U'(X,Y):

U'(x,y)=(ax)+(by)*(dy/dx)

=(ax)+(by)*(-px)

OR plug y(x) back into the U function like I did originally and solve:

U(x,y)=(x^a)((T-px)/c)^b?

Oh, I see. Well, you "expand" that right part without knowing what b is. Use the product rule on the whole thing and the chain rule on the right. The derivative of ((T-px)/c)^b is b((T-px)/c)^(b-1) times the derivative of (T- px)/c.

If this second option is the case, how would I go about solving ((T-px)/c)^b? If I took c out of the denominator it'd be like this right: (T-px+(c^-1))^b. If this is the case, I am still having a hard time with what the derivative would be with the power b outside the equation. Thanks again.

 

[3.141592654]

Alright thanks. So just to be clear, say I had a function: f(x)=2/(x-1).

f(x)=2/(x-1) =2(x-1)^-1, so:

f'(x)=(-2(x-1)^-2)*1

Is this always true when x is part of a function and that function is to the power of any number, such as f(x)=2(x-1)^-1? If so, can anyone explain in basic terms the reason for needing to find the derivative of the entire function as well and of x? Thanks again for all your help!