Did I use log differentiation right?

**AZach** · October 28th 2012, 11:24 AM

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

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

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

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

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

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

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

**tom@ballooncalculus** · October 28th 2012, 12:26 PM

Originally Posted by AZach

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

You mean,

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

**MarkFL** · October 28th 2012, 12:31 PM

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.

**Soroban** · October 28th 2012, 01:28 PM

Hello, AZach!

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

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

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

**AZach** · October 28th 2012, 01:47 PM

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?

**Plato** · October 28th 2012, 01:57 PM

Originally Posted by AZach

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

I had the exact same question as Soroban.
$y' =\left(\frac{3}{4}-\pi\right)\left( \frac{4}{3}x^{-\frac{1}{4}-\pi}\right)$

**tom@ballooncalculus** · October 29th 2012, 05:54 AM

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...

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

Key in spoiler...

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