Implicit differentiation to change the dx in an integral into say, dy.?

Question

I was told to find an integral and given a substitution to use.
The substitution was let u = e^x. The next step that was given was
e^x dx = du. Don't understand this. If you derive both sides, shouldn't it be d/dx e^x = d/du u ??? Whats happened?

Answer 1

The mathematical object you are dealing with is called a differential.
dx = differential of x
dy = differential of y
etc.

The rules for dealing with differential are very similar to those of derivatives like df/dx, or dy/dz etc.

For example, if F is a function of x, i.e. F=F(x), then
dF = (dF/dx) * dx = (dF/dx) dx

[ * is the symbol for multiplication]


You can take the above as the difinition of a differential.

Now, consider y as a function of x i.e. y=y(x) and G as a function of y i.e. G=G(y)

Then, dG = (dG/dy) * dy
But dy = (dy/dx) * dx
Hence,
dG = (dG/dy) * (dy/dx) * dx .
This is called the chain rule for differentials.

Now for your problem
u = u(x) = e^(x) = exp(x) , i.e. u is a function of x.
Use chain rule:
du = (du/dx) * dx = [d{exp(x)}/dx ] * dx
= exp(x) * (dx/dx) * dx [ use chain rule for derivatives]
= exp(x) * dx
=e^(x) * dx
---------------------------------------------------------------------
NOTE: exp(x) is the symbol for exponential function. That is,
for example, exp(z) = e^z

Also, the chain rule for derivatives is:
f=f(z) and if z=z(y) then
df/dy = (df/dz) * (dz/dy) * dy/dy
= (df/dz) * (dz/dy) ,
[note that dp/dp=du/du=dx/dx=1 etc.]
----------------------------------------------------------

Cheers.

Answer 2

In fact you are wrong! If F is a function of X. By definition Df=d/dx(F)dx that is a definition. But infact if we are making a change of variables to comput an integral we need to multiply the neww integral to its Jacobian and it one variable integration Jacobiab is F^'(x)dx

Answer 3

You aren't substituting for a derivative, you're substituting for a differential.


Doug

Answer 4

u = e^x

Now differentiate both sides with respect to x.

(d/dx).(u) = (d/dx).(e^x)

du / dx = e^x