differentiation simplifying

**dadon** · February 4th 2007, 07:43 AM

Hi all,

I have differentiated $y = \frac{e^{4x}sinx}{xcos2x}$ , using the quotient rule to get:

$\frac{dy}{dx}=\frac{xcos2x.e^{4x}(4sinx+cosx)-e^{4x}sinx(cos2x-2xsin2x)}{(xcos2x)^2}$

how can I simplify this further? please can you show step by step as I always make silly mistakes when trying to simplifly.

this is what the teacher said it is in simplest form:
$\frac{e^{4x}sinx(4+cotx-x^{-1}+2tan2x)}{xcos2x}$

thanks

**ticbol** · February 4th 2007, 10:30 AM

Originally Posted by dadon

Hi all,

I have differentiated $y = \frac{e^{4x}sinx}{xcos2x}$ , using the quotient rule to get:

$\frac{dy}{dx}=\frac{xcos2x.e^{4x}(4sinx+cosx)-e^{4x}sinx(cos2x-2xsin2x)}{(xcos2x)^2}$

how can I simplify this further? please can you show step by step as I always make silly mistakes when trying to simplifly.

this is what the teacher said it is in simplest form:
$\frac{e^{4x}sinx(4+cotx-x^{-1}+2tan2x)}{xcos2x}$

thanks

Well, the simplest form, or simplifying, depends on who is doing it. With Math he can continue playing until he gets what he thinks is the "simplest" form.

If your teacher said that is the simplest form, then let us get to your teacher's simplest form.
He seemed to make the denominator of yours into its first order. That is the key. So we need to get a factor of xcos2x from the numerator of yours.

{Xcos2X*e^(4X)[4sinX +cosX] -e^(4X)*sinX[cos2X -2Xsin2X]} / {(Xcos2X)^2} -------------------Yours (which is correct, btw.)

To make your denominator (Xcos2X)^2 into Xcos2X only, we need to get Xcos2X from your numerator.
The first part of the numerator has that already. The second part has none yet, so we introduce it, and maintain the original value of the 2nd part, by multiplying (Xcos2X / Xcos2X) to the 2nd part. Thus,

= [e^(4X) *Xcos2X]*{[4sinX +cosX] -(sinX / Xcos2X)[cos2X -2Xsin2X]} / {(Xcos2X)^2}

The Xcos2X from the numerator cancels one of those in the denominator. Simplifying further,
= [e^(4X)]*{[4sinX +cosX] -(sinXcos2X / Xcos2X) +(2XsinXsin2X / Xcos2X)} / {Xcos2X}

= [e^(4X)]*{4sinX +cosX -(sinX / X) +(2sinXtan2X)} / {Xcos2X}

Now we factor out the sinX in the numerator. The second term inside the curled bracket, the cosX, is the only one without a sinX, so we multiply (sinX /sinX) to it. Thus,

= [e^(4X)*sinX]*{4 +(cosX/sinX) -(1 / X) +(2tan2X)} / {Xcos2X}

= [e^(4X)*sinX]*{4 +cotX -x^(-1) +2tan2X} / {Xcos2X} ---------Your Teacher's simplest form.

**dadon** · February 4th 2007, 11:25 AM

Thank you, ticbol. Nicely explained. =)

**Soroban** · February 4th 2007, 12:05 PM

Hello, dadon!

Differentiate: . $y \:= \:\frac{e^{4x}\sin x}{x\cos2x}$

With a quotient of products, logarithmic differentiation may be in order.

Take logs: . $\ln(y)\;=\;\ln\left(\frac{e^{4x}\sin x}{x\cos2x}\right) \;=\;\ln(e^{4x}) + \ln(\sin x) - \ln(x) - \ln(\cos2x)$

And we have: . $\ln(y)\;=\;4x + \ln(\sin x) - \ln(x) - \ln(\cos2x)$

Differentiate implicitly: . $\frac{1}{y}\cdot\frac{dy}{dx}\;=\;4 + \frac{1}{\sin x}\cdot\cos x - \frac{1}{x} - \frac{1}{\cos2x}(-2\sin2x)$

We have: . $\frac{1}{y}\cdot\frac{dy}{dx}\;=\;4 + \cot x - \frac{1}{x} + 2\tan2x$

Then: . $\frac{dy}{dx}\;=\;y\left(4 + \cot x - x^{-1} + 2\tan2x\right)$
. . . . . . . . . . $\downarrow$
. . . . . . $= \;\frac{e^{4x}\sin x}{x\cos2x}\left(4 + \cot x - x^{-1} + 2\tan2x\right)$

Therefore: . $\frac{dy}{dx} \;=\;\frac{e^{4x}\sin x(4 + \cot x - x^{-1} + 2\tan2x)}{x\cos2x}$ . . . There!

**dadon** · February 4th 2007, 12:12 PM

Hi Soroban,

That's great. That way is much neater. Thank you

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