# Did I use log differentiation right?

• Oct 28th 2012, 11:24 AM
AZach
Did I use log differentiation right?
$y = \frac{4}{3}x^{\frac{3}{4}-\pi}$

$ln(y) = ln(\frac{4}{3}x^{\frac{3}{4}-\pi})$

$ln(y) = ({\frac{3}{4}-\pi})ln(\frac{4}{3}x)$

$\frac{1}{y}\frac{dy}{dx} = (\frac{3}{4}-\pi(\frac{d}{dx}(ln(\frac{4}{3}x))))$

$\frac{1}{y}\frac{dy}{dx} = (\frac{3}{4}-\pi)(\frac{3}{4}x)*(\frac{4}{3})$

$\frac{dy}{dx} = y(\frac{3}{4}x-\pi(x))$

$\frac{dy}{dx} = \frac{4}{3}x^{\frac{3}{4}-\pi}(\frac{3}{4}x-\pi(x))$
• Oct 28th 2012, 12:26 PM
tom@ballooncalculus
Re: Did I use log differentiation right?
Quote:

Originally Posted by AZach
$\frac{1}{y}\frac{dy}{dx} = (\frac{3}{4}-\pi)(\frac{3}{4}x)*(\frac{4}{3})$

You mean,

$\frac{1}{y}\frac{dy}{dx} = (\frac{3}{4}-\pi)(\frac{3}{4x})*(\frac{4}{3})$
• Oct 28th 2012, 12:31 PM
MarkFL
Re: Did I use log differentiation right?
No, you incorrectly applied a certain log property, specifically:

$\log_a(bc^d)\ne d\cdot\log_a(bc)$

Since the exponent is a constant, you should just use the power rule to differentiate, unless specifically instructed to use logs.
• Oct 28th 2012, 01:28 PM
Soroban
Re: Did I use log differentiation right?
Hello, AZach!

Quote:

$\text{Differentiate: }\:y \:=\: \tfrac{4}{3}x^{\frac{3}{4}-\pi}$

Why are you using log differentiation?
The exponent is a constant.

Or did you mean: . $y \:=\:\tfrac{4}{3}x^{\frac{3}{4}-x}$
• Oct 28th 2012, 01:47 PM
AZach
Re: Did I use log differentiation right?
Quote:

Originally Posted by Soroban
Hello, AZach!

Why are you using log differentiation?
The exponent is a constant.

Or did you mean: . $y \:=\:\tfrac{4}{3}x^{\frac{3}{4}-x}$

For some reason I was thinking $\pi$ was a variable.

For a problem like this would I use log differentiation?

$y = \frac{4}{3}x^{\frac{3}{4}-x}$

$ln (y) = ln (\frac{4}{3}x^{\frac{3}{4}-x})$

$ln (y) = ln (\frac{4}{3}) + ({\frac{3}{4}-x})ln(x)$

$\frac{1}{y}\frac{dy}{dx} = \frac{3}{4} + (-1)ln(x) + (\frac{3}{4}-x)*(\frac{1}{x})$

$\frac{dy}{dx} = y( \frac{3}{4} -ln(x) + \frac{3}{4x}-1)$

$\frac{dy}{dx} = (\frac{4}{3}x^{\frac{3}{4}-x}) * ( \frac{3}{4} -ln(x) + \frac{3}{4x}-1)$

Does that look correct?
• Oct 28th 2012, 01:57 PM
Plato
Re: Did I use log differentiation right?
Quote:

Originally Posted by AZach
$y = \frac{4}{3}x^{\frac{3}{4}-\pi}$

I had the exact same question as Soroban.
$y' =\left(\frac{3}{4}-\pi\right)\left( \frac{4}{3}x^{-\frac{1}{4}-\pi}\right)$
• Oct 29th 2012, 05:54 AM
tom@ballooncalculus
Re: Did I use log differentiation right?
Quote:

Originally Posted by AZach
For a problem like this would I use log differentiation?

$y = \frac{4}{3}x^{\frac{3}{4}-x}$

$ln (y) = ln (\frac{4}{3}x^{\frac{3}{4}-x})$

$ln (y) = ln (\frac{4}{3}) + ({\frac{3}{4}-x})ln(x)$

$\frac{1}{y}\frac{dy}{dx} = \frac{3}{4} + (-1)ln(x) + (\frac{3}{4}-x)*(\frac{1}{x})$

...

Does that look correct?

Nearly. Just in case a picture helps...

http://www.ballooncalculus.org/draw/double/six.png

differentiate 4/3 x^(3/4 - x) - Wolfram|Alpha (Click on 'show steps'.)

Key in spoiler...

Spoiler:
http://www.ballooncalculus.org/asy/doubleChain.png

is the chain rule for two inner functions, i.e...

\frac{d}{dx}\ f(u(x), v(x)) = \frac{\partial f}{\partial u} \frac{du}{dx} + \frac{\partial f}{\partial v} \frac{dv}{dx}

As with...

http://www.ballooncalculus.org/asy/chain.png

... the ordinary chain rule, straight continuous lines differentiate downwards (integrate up) with respect to x, and the straight dashed line similarly but with respect to the (corresponding) dashed balloon expression which is (one of) the inner function(s) of the composite expression.

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