dy/dx if e^2x = sin(x+3y)


by lostintranslation
Tags: derivatives, dy or dx, sinx
lostintranslation
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#1
Nov20-05, 07:49 PM
P: 5
I need help with this problem.

Find dy/dx if e^2x = sin(x+3y)

Any help would be great!

I began by taking the natural log of both sides...which may not be correct??

Thanks!
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mathwonk
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#2
Nov20-05, 08:02 PM
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then what?
HallsofIvy
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#3
Nov20-05, 08:14 PM
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Quote Quote by lostintranslation
I need help with this problem.
Find dy/dx if e^2x = sin(x+3y)
Any help would be great!
I began by taking the natural log of both sides...which may not be correct??
Thanks!
That's certainly one way to do it: you get 2x= ln(sin(x+3y)).
Of course, you will have to use the chain rule on the right side when you differentiate now.
Couldn't you have used the chain rule on both sides of the equation the way it was written? What is [itex]\frac{de^u}{du}[/itex]? What is
[itex]\frac{de^u}{du}\frac{du}{dx}[/itex] if u= 2x?

courtrigrad
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#4
Nov20-05, 08:15 PM
P: 1,239

dy/dx if e^2x = sin(x+3y)


So [tex] e^{2x} = \sin(x+3y) [/tex] When you take the natural log of both sides you get: [tex] \ln e^{2x} = \ln\sin(x+3y) [/tex] or [tex] 2x = \ln\sin(x+3y) [/tex]. Now use implicit differentiation to find [tex] \frac{dy}{dx} [/tex]
lostintranslation
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#5
Nov20-05, 08:40 PM
P: 5
Ok, right.

So

2x = ln (sin(x+3y))

Then if you take the derivative...

2 = (1/sin(x+3y)) * cos (x+3y) * 3 * dy/dx

So dy/dx = 2(sin(x+3y))/ 3(cos(x+3y)) ?

Is that right?
courtrigrad
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#6
Nov20-05, 08:55 PM
P: 1,239
yes that is correct.
lostintranslation
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#7
Nov20-05, 09:08 PM
P: 5
Thanks so much! Does that simplify any farther?
courtrigrad
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#8
Nov20-05, 09:18 PM
P: 1,239
I guess you could write it as [tex] \frac{2\tan(x+3y)}{3} [/tex]
lostintranslation
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#9
Nov20-05, 09:23 PM
P: 5
Ok good! That is what I thought.

You have been so helpful! Thanks!

I have a couple more [shorter] questions, however don't feel obliged to respond if you don't want to...

1. Find dy/dx of y = x^5 * 7^x

2. Find dy/dx of y = (1/2) ^ x

3. Find dy/dx of y = (cosx) ^ x
courtrigrad
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#10
Nov20-05, 09:52 PM
P: 1,239
1. Use the chain rule. [tex] y = x^{5}7^{x} [/tex]. [tex] y' = x^{5}(7^{x}\ln 7) + 7^{x}(5x^{4}) [/tex]
2. [tex] (0.5)^{x} [/tex] is just [tex] (0.5)^{x}\ln 0.5 [/tex]
3. I'll let you do this one

In general, [tex] \frac{d}{dx} a^{u} = a^{u}\ln a \frac{du}{dx} [/tex]
lostintranslation
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#11
Nov20-05, 10:09 PM
P: 5
Right, right. That rule would make those very easy! Thanks bud!

I am still having trouble with the cosine one however. I apparently need to use multiple steps to find the derivative.
HallsofIvy
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#12
Nov21-05, 06:51 AM
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Quote Quote by courtrigrad
1. Use the chain rule. [tex] y = x^{5}7^{x} [/tex]. [tex] y' = x^{5}(7^{x}\ln 7) + 7^{x}(5x^{4}) [/tex]
2. [tex] (0.5)^{x} [/tex] is just [tex] (0.5)^{x}\ln 0.5 [/tex]
3. I'll let you do this one

In general, [tex] \frac{d}{dx} a^{u} = a^{u}\ln a \frac{du}{dx} [/tex]
Mistype in 2: [tex](0.5)^x[/tex] is just [tex]e^x\ln 0.5[/tex].

By the way, I don't see anything gained by taking the logarithm in the first problem you posted. It is just as easy to differentiate both sides of
e2x= sin(x+ 3y) the way it stands:
2e2x= cos(x+ 3y)(1+ y') so
[tex]1+ y'= \frac{2e^{2x}}{cos(x+3y}}[/tex]
and
[tex]y'= \frac{2e^{2x}}{cos(x+3y}}- 1[/tex]
VietDao29
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#13
Nov21-05, 07:23 AM
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Quote Quote by lostintranslation
2 = (1/sin(x+3y)) * cos (x+3y) * 3 * dy/dx
So dy/dx = 2(sin(x+3y))/ 3(cos(x+3y)) ?
Is that right?
Nope, that's {B]not[/B] correct!
The above line should read:
[tex]2 = \frac{\cos (x + 3y)}{\sin (x + 3y)} \left( 1 + 3 \frac{dy}{dx} \right)[/tex]
You forgot the (1 + 3 dy / dx)
Quote Quote by courtrigrad
yes that is correct.
???
How on Earth can that be called correct?
Quote Quote by HallsofIvy
Mistype in 2: [tex](0.5)^x[/tex] is just [tex]e^x\ln 0.5[/tex].
That's also a mistype.
0.5x = ex ln(0.5)
Quote Quote by HallsofIvy
e2x= sin(x+ 3y) the way it stands:
2e2x= cos(x+ 3y)(1+ y')
By the way, there's a typo there:
e2x= sin(x+ 3y)
Differentiate both sides with respect to x gives:
2e2x= cos(x+ 3y)(1+ 3y')



---------------------
To find dy / dx of y = cosxx. Just do the same:
cosxx = ex ln(cos(x))
---------------------
Or, you can do it a bit differently:
y = cosxx
Take the logarithm of both sides gives:
ln y = x ln(cos x)
Then you can differentiate both sides with respect to x, and see what you get.
courtrigrad
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#14
Nov21-05, 07:45 AM
P: 1,239
sorry for the typos

Thanks HallsofIvy and VietDao for correcting me.


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