# Differentiation of logarithms

1. Jun 15, 2015

### Rafiul Nakib

• Member warned about not using the template and showing work only in an image
the variables x and y are positive and related by x^a.y^b=(x+y)^(a+b) where a and b are positive constants. By taking logarithms of both sides, show that dy/dx=y/x. provided that bx not equal to ay.

#### Attached Files:

• ###### IMAG2067.jpg
File size:
16.6 KB
Views:
54
2. Jun 15, 2015

### ehild

You should have typed in your derivation. In that case, I would be able to show where your error is. The differentiation is not correct.

3. Jun 15, 2015

### Rafiul Nakib

x^a.y^b=(x+y)^(a+b)
=> ln(x^a.y^b) = ln((x+y)^(a+b))
=>a lnx + b lny = (a+b)ln(x+y)
=>a. d/dx lnx + b. d/dx lny = (a+b). d/dx ln(x+y)
=> a/x + b/y. dy/dx = (a+b)/(x+y). dy/dx

4. Jun 15, 2015

### ehild

Yes, d/dx [ln(x+y)]=1/(x+y) d/dx(x+y) . What is d/dx(x+y)? it is a sum, you have to differentiate both terms. What is dx/dx?

5. Jun 15, 2015

### Ray Vickson

I know the question tells you to use logarithms, but there is a much easier way: the constraint $x^a y^b = (x+y)^{a+b}$ implies that
$$\left(\frac{x}{x+y}\right)^a \left( \frac{y}{x+y} \right)^b = 1,$$
hence $r^a (1-r)^b = 1$, where $r = x/(x+y)$. For given $a,b$, that means that $r$ is the solution, or one of two solutions to an equation, so $x/(x+y)$ is a constant. Thus, $y = cx$ for some constant $c>0$.