Register to reply

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

by lostintranslation
Tags: derivatives, dy or dx, sinx
Share this thread:
lostintranslation
#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!
Phys.Org News Partner Science news on Phys.org
What lit up the universe?
Sheepdogs use just two simple rules to round up large herds of sheep
Animals first flex their muscles
mathwonk
#2
Nov20-05, 08:02 PM
Sci Advisor
HW Helper
mathwonk's Avatar
P: 9,480
then what?
HallsofIvy
#3
Nov20-05, 08:14 PM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,533
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
#4
Nov20-05, 08:15 PM
P: 1,236
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
#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
#6
Nov20-05, 08:55 PM
P: 1,236
yes that is correct.
lostintranslation
#7
Nov20-05, 09:08 PM
P: 5
Thanks so much! Does that simplify any farther?
courtrigrad
#8
Nov20-05, 09:18 PM
P: 1,236
I guess you could write it as [tex] \frac{2\tan(x+3y)}{3} [/tex]
lostintranslation
#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
#10
Nov20-05, 09:52 PM
P: 1,236
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
#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
#12
Nov21-05, 06:51 AM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,533
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
#13
Nov21-05, 07:23 AM
HW Helper
VietDao29's Avatar
P: 1,422
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
#14
Nov21-05, 07:45 AM
P: 1,236
sorry for the typos

Thanks HallsofIvy and VietDao for correcting me.


Register to reply

Related Discussions
Line Tangent ro a Parabola Precalculus Mathematics Homework 5
When to use derivatives? Calculus 8
Help with Derivatives Problem Calculus & Beyond Homework 0
3 different derivatives? Calculus & Beyond Homework 12
Find instantaneous rate of change of 7/3z^2 Calculus & Beyond Homework 2