Solving Integral of e^6x with U Substitution

[wiredmomar]

Hey All,

First post, hopefully it will be readable. I was going to try and word it correctly, but I might as well just post a problem I am having with a certain notation.

Take integral of e^6x. Easy enough question. Using U substitution:

u = 6x
du/dx = 6
du = 6 dx

Integral above now equals 1/6 e^6x + C (so the 1/6 cancels with our 6 in 6 dx).

Ok, now using dy/dx = dy/du * du/dx notation.

e^6x let u = 6x
y = e^u
u = 6x

dy/du = e^u
du/dx = 6

Since dy/dx = dy/du * du/dx, wouldn't the above equal e^6x * 6

The second notation confuses me a bit... Any help to explain would be appreciat
Thanks,

M

 

[Jerbearrrrrr]

dy/dx=6e^6x is correct,
if y=e^6x

You've differentiated y wrt x. It's correct, just not what you wanted to find.

 

[HallsofIvy]

Hey All,

First post, hopefully it will be readable. I was going to try and word it correctly, but I might as well just post a problem I am having with a certain notation.

Take integral of e^6x. Easy enough question. Using U substitution:

u = 6x
du/dx = 6
du = 6 dx

Integral above now equals 1/6 e^6x + C (so the 1/6 cancels with our 6 in 6 dx).

Ok, now using dy/dx = dy/du * du/dx notation.

e^6x let u = 6x
y = e^u
u = 6x

dy/du = e^u
du/dx = 6

Since dy/dx = dy/du * du/dx, wouldn't the above equal e^6x * 6

Yes, it would. One is integrating, the other differentiating. Since those are "opposite" operations, it shouldn't surprise you that one involves dividing by 6 and the other involves multiplying by 6.

The second notation confuses me a bit... Any help to explain would be appreciat
Thanks,

M