Question

f(x)=ln(e^(2x))
f'(x)= ?
can anyone help me with this?

Answer 1

2e^(2x)/e^(2x)

Answer 2

which simplifies to 2.

Answer 3

yes

Answer 4

thanks! so how would i find f(x)=ln[(e^(mx))+n] if m and n are constants, how would i find f'(x)= ?

Answer 5

f(x)=ln(e^(2x))

The exponent of a factor inside a logarithm can be expanded to the front of the expression using the third law of logarithms. The third law of logarithms states that the logarithm of a power of x is equal to the exponent of that power times the logarithm of x (e.g. log^b(x^(n))=nlog^b(x)).
((2xln(e)))

Remove the extra parentheses from the expression ((2xln(e))).
2xln(e)

When the base of a logarithm and the base of the exponent are the same, the result is the exponent of argument.
(2x(1))

Multiply 2x by 1 in the numerator.
(2x*1)

Multiply 2x by 1 to get 2x.
(2x)

Remove the parentheses from the numerator.
2x

The domain of an expression is all real numbers except for the regions where the expression is undefined. This can occur where the denominator equals 0, a square root is less than 0, or a logarithm is less than or equal to 0. All of these are undefined and therefore are not part of the domain.
e^(2x)<=0

Solve the equation to find where the original expression is undefined.
No Solution

The domain of the rational expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.
All real numbers

Answer 6

when you have ln all you have to do it put what ln is being multiply to at the bottom and derivative on top..

Answer 7

Since ln(a) and e^a are inverse operations, their composition is the identity. ln(e^a) = a. Also,\[\ln(a^{b}) = b(\ln a)\]so you can move the 2x out front that way and you're left with ln e = 1 (again, their inverses). In both cases you've simplified f(x) to 2x, which is easy to differentiate.

But, if you're supposed to be practicing the chain rule,\[\frac{d}{dx} f(g(x)) = f'(g(x)) g'(x)\]so\[\frac{d}{dx} \ln(e^{2x}) = \frac{1}{e^{2x}} \frac{d}{dx}e^{2x} = \frac{1}{e^{2x}} e^{2x} \frac{d}{dx} 2x = \frac{e^{2x}}{e^{2x}} 2 = 2\]

Answer 8

Using the facts that\[\frac{d}{dx} \ln x = \frac{1}{x}\]and\[\frac{d}{dx} e^{x} = e^{x}\]