Logarithmic Differentiation with binomial products

**ashleysmithd** · January 2nd 2011, 09:29 AM

Hi,

I've been asked to differentiate $y = 5x^2tan(6x)$ logarithmically.

I took the following steps:

$y = 5x^2tan(6x)$

$ln(y) = ln(5x^2tan6x) = (2ln5x)(lntan(6x))$

$f(x) = 2ln5x$
$g(x) = lntan(6x)$
$f'(x) = \frac {2}{x}$
$g'(x) = \frac {1}{6Sec^26x}$

$\frac {1}{y} \frac{dy}{dx} = (2ln5x)(\frac{1}{6Sec^26x})+(\frac{2}{x})(lntan(6x ))$

$\frac {dy}{dx} = y((2ln5x)(\frac{1}{6Sec^26x})+(\frac{2}{x})(lntan( 6x)))$

$\frac{dy}{dx} = 5x^2tan(6x) ((2ln5x)(\frac{1}{6Sec^26x})+(\frac{2}{x})(lntan(6 x)))$

This is my first attempt at differentiating binomial products logarithmically, can anyone please look at these steps and tell me if and how I've gone wrong?

Thanks.

(Please spare any factorising, I don't understand it very well right now)

**tom@ballooncalculus** · January 2nd 2011, 09:57 AM

Originally Posted by ashleysmithd

$y = 5x^2tan(6x)$

$ln(y) = ln(5x^2tan6x) = (2ln5x)(lntan(6x))$

The log of a product is a sum of two logs - not a product of 2 logs.

Originally Posted by ashleysmithd

$f(x) = 2ln5x$
$g(x) = lntan(6x)$
$f'(x) = \frac {2}{x}$
$g'(x) = \frac {1}{6Sec^26x}$

So you don't want the product rule, though you do want to differentiate f and g. Your f is fine, but g dodgy.

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**e^(i*pi)** · January 2nd 2011, 10:08 AM

$\ln(y) = \ln(5) + 2\ln(x) + \ln(\tan(6x))$

The first two terms are simple enough to differentiate but you will need to use the chain rule twice on the last term

Let $u = tan(6x) \implies \dfrac{du}{dx} = 6\sec^2(6x)$ and also gives $y = \ln(\tan(6x)) = \ln(u)$ and by the chain rule $\dfrac{dy}{dx} = \dfrac{dy}{du} \cdot \dfrac{du}{dx}$

(note that y here is used as an example to demonstrate the chain rule)

**ashleysmithd** · January 3rd 2011, 09:32 AM

I think I see what you mean, g(x) was a function of a function and needed the chain rule to differentiate it into: $(\frac{1}{tan(6x)}(6sec^26x))$ ?

I'm not quite what you meant by using the chain rule twice, I've equated it to this so far:
$\frac{dy}{dx} = 5x^2tan6x ((2ln5x)(\frac{1}{tan(6x)}(6sec^26x))+(lntan6x)(\f rac{2}{x}))$

Is this on the right lines?

**Plato** · January 3rd 2011, 09:37 AM

Originally Posted by ashleysmithd

$y = 5x^2tan(6x)$ logarithmically.

Why make it so complicated?
The derivative is simply: $10x\tan(6x)+30x^2\sec^2(6x)$ .

**ashleysmithd** · January 3rd 2011, 10:09 AM

I completely understand it can be factorised down, but I really don't understand factorising well yet so I'd rather work long hand and just get the rules of differentiation right first.
That said, all be it unnecessarily long, is the solution I posted above correct?

**Plato** · January 3rd 2011, 10:32 AM

It is a simple product: $5x^2$ times $\tan(6x)$ .

Use the product rule. Anything else is a waste of time and effort.

**ashleysmithd** · January 3rd 2011, 11:00 AM

I would do but the point of the exercise I've given is to learn how to differentiate logarithmically, and show working, so I can't use the product rule on its own annoyingly.

**tom@ballooncalculus** · January 3rd 2011, 11:29 AM

Originally Posted by ashleysmithd

I think I see what you mean, g(x) was a function of a function and needed the chain rule to differentiate it into: $(\frac{1}{tan(6x)}(6sec^26x))$ ?

Good.

I'm not quite what you meant by using the chain rule twice, I've equated it to this so far:
$\frac{dy}{dx} = 5x^2tan6x ((2ln5x)(\frac{1}{tan(6x)}(6sec^26x))+(lntan6x)(\f rac{2}{x}))$

Is this on the right lines?

No, it looks like you're still trying to involve the product rule even though the one advantage of doing it logarithmically (although, as Plato says, this example is not remotely a good case for that treatment) is that you got rid of the product so you don't need that rule.

Look at the bottom row of my diagram and multiply both sides by y, perhaps?

**ashleysmithd** · January 3rd 2011, 11:58 AM

I think the part where I'm getting confused is how
$ln(y) = ln(5x^2tan6x)$
becomes
$ln(y) = 2ln5x + lntan6x$

I don't understand why these functions are being added and not multiplied? Why does it become a sum, and not remain a product, is it to do with log properties?

**tom@ballooncalculus** · January 3rd 2011, 12:36 PM

Originally Posted by ashleysmithd

is it to do with log properties?

Yes - revise them.

$\ln(ab)\ =\ \ln(a) + \ln(b)\ \ \$

from

$a = e^x$ and $b = e^y$

and

$ab = e^{x+y}$

**e^(i*pi)** · January 3rd 2011, 12:55 PM

Originally Posted by ashleysmithd

I think the part where I'm getting confused is how
$ln(y) = ln(5x^2tan6x)$
becomes
$ln(y) = 2ln5x + lntan6x$

It doesn't - you can only take a power out when each term is raised to the power. You moved the 2 to the front but only the x was squared, not the 5

$\ln(y) = \ln(5x^2 \tan(6x)) = \ln(5x^2) + \ln(\tan(6x))$

You need to use the addition rule on $\ln(5x^2)$ before the power rule can be done on x^2 -> $\ln(5x^2) = \ln(5) + \ln(x^2) = \ln(5) + 2\ln(x)$

Hence $\ln(y) = \ln(5) + 2\ln(x) + \ln(\tan(6x))$

Use the chain rule (explained in posts 2 and 3) to differentiate $\ln(\tan(6x))$

**Plato** · January 3rd 2011, 01:08 PM

I you want to learn Logarithmic Differentiation then do right.
Suppose that each of $g,~h,~j,~\&~k$ is a differentiable function and $f(x)=g(x)\cdot h(x)\cdot j(x)\cdot k(x)\cdot$ .
Then the derivative is $f'(x) = f(x)\left[ {\frac{{h'(x)}}{{h(x)}} + \frac{{g'(x)}}{{g(x)}} + \frac{{j'(x)}}{{j(x)}} + \frac{{k'(x)}}{{k(x)}}} \right]$ .
Not that can be done for any numbers of factors.

So the derivative of $5x^2\tan(6x)$ is simply $5x^2 \tan (5x)\left[ {\frac{2}{{x^2 }} + \frac{{6\cos (6x)}}{{\sin (6x)}} + \frac{{6\sin (6x)}}{{\cos (6x)}}} \right]$ .

But this is a horrible example to teach that concept.

**ashleysmithd** · January 3rd 2011, 01:17 PM

Ok thanks, I think I just got caught out on the log rules.

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