Fairly basic natural logs question

Returning to differentiation after a fairly long break and presented with three questions. I'm sure there's a simpler way to do them, but I can't figure it out?

1. y(x) = ln[(f(x)+g(x))^2] => y(x) = ln[(u)^2]

y'(x) = 1*[2(f'(x)+g'(x))] / (f(x)+g(x))^2

2. y(x) = ln(x^a + zx)/(e^x*e^z)

y(x) = ln (u)/(v)

y'(x) = (V) / (U) * (V)(U') / (U)(V')

y'(x) = (V)(U)(V') / (U)(V)(U')

If that makes sense?

3. y(x) = f(g(x).u(z))

y'(x) = (g(x).u(z)).(g'(x))

Sorry if this seems a bit retarded, I'm finding the textbook a bit lacklustre in parts (Chiang / Wainwright) and statistics is much more my forte!

Re: Fairly basic natural logs question

1. $\displaystyle y = \ln{u^2}$

$\displaystyle y' = \frac{2u}{u'}$

2. $\displaystyle y = \frac{\ln{u}}{v}$

$\displaystyle y' = \frac{\frac{uv}{u'} - v' \cdot \ln{u}}{v^2}$

3. $\displaystyle y = f(u \cdot v)$

$\displaystyle y' = f'(u \cdot v) \cdot (uv' + vu')$

Re: Fairly basic natural logs question

Quote:

Originally Posted by

**skeeter** 1. $\displaystyle y = \ln{u^2}$

$\displaystyle y' = \frac{2u}{u'}$

I thought with natural logs, the derivative of e.g. $\displaystyle y= \ln{x^5}$ => $\displaystyle y' = 5x^4 / x^5$? I.e. it's the derivative over the original?

Re: #3, oops. it's the chain rule applying to the product rule, so I see why the answer is that.

Cheers!

Re: Fairly basic natural logs question

Quote:

Originally Posted by

**kedaha** I thought with natural logs, the derivative of e.g. $\displaystyle y= \ln{x^5}$ => $\displaystyle y' = 5x^4 / x^5$? I.e. it's the derivative over the original?

Re: #3, oops. it's the chain rule applying to the product rule, so I see why the answer is that.

Cheers!

$\displaystyle y' = \frac{2u'}{u}$ ... sorry, transposed the prime