# Math Help - Derivate

1. ## Derivate

Hello

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

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

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

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

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

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

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

The general derivate of ln:
$D ln(kx^n) = \frac{1}{x^n}$

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

Let's move on...

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

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

I think it looks correct, can it be simplified?

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

Anything further?

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

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

Thanks for all replies.

2. Hi

Originally Posted by λιεҗąиđ€ŗ
Hello

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

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

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

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

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

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

$D(u(x) \cdot v(x)) = u'(x) \cdot v(x) + u(x) \cdot v'(x)$
Until here it's ok

The general derivate of ln:
$D ln(kx^n) = \frac{1}{x^n}$
I don't agree.

$ln(k*x^n)=ln(k)+n ln(x)$

So the derivate should be n/x

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

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

I think it looks correct, can it be simplified?
It doesn't seem correct to me...

Let's see the derivate of 1/(ln(x^3)). This is a function similar to 1/u(x). But the derivate of such a function is -u'(x)/u²(x).
Here, u(x)=ln(x^3)
And the derivate of u(x) is not 3x² but -3/x.

Here is the mistake, try to write it down again

3. It could be easier to tackle this with the quotient rule;

$\frac{v\frac{du}{dx}~-~u\frac{dv}{dx}}{v^2}$

With a bit of re-arranging i got;

$\frac{-3\ln5}{x(\ln(x^3))^2}$

Sorry my latex is poor.

4. Originally Posted by Sean12345
It could be easier to tackle this with the quotient rule;
It might, but let's see what we get when we work a bit with the logarithms as Moo suggested.

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

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

$ln(5) + 7ln(x) \cdot \frac {1} {lnx^3}$

$ln(5) + 7ln(x) \cdot (ln(1) - 3ln(x))$ Error

$ln(5) + 7ln(x) \cdot (0 - 3ln(x))$

$ln(5) + 7ln(x) \cdot - 3ln(x)$

The above seems reasonable, I've kept it to the basics and I would be suprised if I made an error.

Let's derive that...

$\frac{7}{x} \cdot -3ln(x) + 7ln(x) \cdot - \frac{3}{x}$

can we divide by $ln(x)$? I think we can.

$\frac{7}{x} \cdot -3 + 7 \cdot - \frac{3}{x}$

$-\frac{21}{x} - \frac {21}{x}$

$-\frac{42}{x}$

Hmm not really sure if we can remove ln(x) just like that ... in that case

$-\frac{42ln(x)}{x}$

can $\frac{ln(x)}{x}$ be simplified somehow?

5. Originally Posted by λιεҗąиđ€ŗ
Hello

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

$f(x) = \frac{ln(5x^7)}{ln(x^3)}$
[snip]
$f(x) = \frac{\ln 5 + 7 \ln x}{3 \ln x}$.

Quotient Rule:

$f'(x) = \frac{(7/x) \, (3 \ln x) - (3/x) \, (\ln 5 + 7 \ln x)}{(3 \ln x)^2}$

$= \frac{21 \ln x - 3 \ln 5 - 21 \ln x}{9x \, (\ln x)^2}$

$= \frac{ -\ln 5}{3x \, (\ln x)^2}$.

6. ahhh i see the mistake λιεҗąиđ€ŗ

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

Goes to;

$ln(5) \cdot \frac {1} {lnx^3} + 7ln(x) \cdot \frac {1} {lnx^3}$

NOT

$ln(5) + 7ln(x) \cdot \frac {1} {lnx^3}$