Hello

I'd like some help on derive a function and simplifying the expression as much as possible.

$\displaystyle f(x) = \frac{ln(5x^7)}{ln(x^3)}$

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Here's my idea on how it ends up...

first, we can work a little bit with the function:

$\displaystyle f(x) = ln(5x^7) \cdot \frac{1}{ln(x^3)}$

Ok this looks a bit easier, we'll get the derivate of an product.

$\displaystyle D(u(x) \cdot v(x)) = u'(x) \cdot v(x) + u(x) \cdot v'(x)$

The general derivate of ln:

$\displaystyle D ln(kx^n) = \frac{1}{x^n}$

and $\displaystyle \frac{1}{x}$ simply ends up as $\displaystyle -\frac{1}{x^2}$

Let's move on...

Let's derivate $\displaystyle f(x) = ln(5x^7) \cdot \frac{1}{ln(x^3)}$

$\displaystyle \frac{1}{5x^7} \cdot 5 \cdot 7x^6 \cdot \frac{1}{ln(x^3)} + ln(5x^7) \cdot (- \frac{1}{ln(x^3)^2}) \cdot 3x^2$

I think it looks correct, can it be simplified?

$\displaystyle \frac {7}{x^7} \cdot x^4 \cdot \frac {1}{ln(x^3)} + (- \frac {ln(5x^7)}{ln(x^3)^2}) \cdot 3$ $\displaystyle |/x^2$

Anything further?

$\displaystyle \frac{7x^4}{x^7} - \frac {3ln(5x^7)}{ln(x^3)} |\cdot ln(x^3)$

$\displaystyle \frac{7}{x^3} - \frac{3ln(5x^7)}{ln(x^3)}$

Thanks for all replies.