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**Derivative Of ln x, Natural Logarithm –** The natural logarithm of a number x is the logarithm to the base e, where e is the mathematical constant approximately equal to 2.718. It is usually written using the shorthand notation ln x, instead of log ex as you might expect. For example, logarithms are used to solve for the half-life, decay constant, or unknown time in exponential decay problems. They are important in many branches of mathematics and the sciences and are used in finance to solve problems involving compound interest.

Exponential functions and their corresponding inverse functions, called logarithmic functions, have the following differentiation formulas: Note that the exponential function f( x) = e ^{x} has the special property that its derivative is the function itself, f′( x) = e ^{x} = f( x).

**The derivative of ln, ln x, ln(x) **

**Derivatives of logarithmic functions** are mainly based on the chain rule. However, we can generalize it for any differentiable function with a logarithmic function. The differentiation of log is only under the base $e,$ but we can differentiate under other bases, too.

Our task is to determine what is the derivative of the natural logarithm. We begin with the inverse definition. If

y = ln x

then

e^y = x

Now implicitly take the derivative of both sides with respect to x remembering to multiply by dy/dx on the left-hand side since it is given in terms of y, not x.

e^y dy/dx = 1

From the inverse definition, we can substitute x in for e^y to get

x dy/dx = 1

Finally, divide by x to get

dy/dx = 1/x

We have proven the following theorem

**The derivative of ln(2x)**

**Method 1**

You use the chain rule :

(f∘g)‘(x)=(f(g(x)))‘=f‘(g(x))⋅g‘(x).

In your case : (f∘g)(x)=ln(2x), f(x)=ln(x) and g(x)=2x.

Since f‘(x)=1/x and g‘(x)=2, we have :

(f∘g)‘(x)=(ln(2x))‘=1/2x⋅2=**1/x****.**

**Method 2**

We can use the chain rule here, naming u=2x and remembering that the chain rule states that

dy/dx=(dy/du)(du/dx)

So, now, for our function ln(u):

dy/du=1/u

And for the other part:

du/dx=2

Now, aggregating them:

dy/dx=1/u⋅2=(1/2x)⋅2=**1/x**

**Derivative of ln(x+1)**

(f∘g)‘(x)=(f(g(x)))‘=f‘(g(x))⋅g‘(x).

In your case : (f∘g)(x)=ln(x+1), f(x)=ln(x) and g(x)=x+1.

Since f‘(x)=1/x and g‘(x)=1, we have :

(f∘g)‘(x)=(ln(x+1))‘=1/x+1⋅1=**1/x+1****.**

**Derivative of ln(x^2)**

Applying the chain rule, along with the derivatives d/dxln(x)=1/x and d/dx(x²)=2x, we have

**Derivative of ln(3x)**

to find out the derivative of ln(3x) then suppose that

ln(3x)=y

e^y=3x

Now use implicit differentiation. Remember that:

dy/dy⋅dy/dx=dy/dx

If you use implicit differentiation…

e^y=3x

Should transform into…

e^y⋅dy/dx=3

Therefore:

dy/dx=3/e^y

dy/dx=3/3x

dy/dx=1/x

You could also differentiate it like this

**What is the derivative of ln 2 x?**

**How do you integrate Ln(x)?**

**Strategy: Use Integration by Parts.**

- ln(x) dx. set. u = ln(x), dv = dx. then we find. du = (1/x) dx, v = x.
- substitute. ln(x) dx = u dv.
- and use integration by parts. = uv – v du.
- substitute u=ln(x), v=x, and du=(1/x)dx.

**What is the derivative of a log?**

_{e}(x) (usually written “ln x”).

**What does Ln mean?**

**What are ln and log?**

**log**(x) means the base 10 logarithm; it can also be written as

**log**10(x).

**ln**(x) means the base e logarithm; it can also be written as

**log**e(x).

**ln**(x) tells you what power you must raise e to obtain the number x.

**How do you find the derivative of ln?**

**The steps are as follows:**

- Let y = ln(x).
- Use the definition of a logarithm to write y = ln(x) in logarithmic form. …
- Treat y as a function of x, and take the derivative of each side of the equation with respect to x.
- Use the chain rule on the left-hand side of the equation to find the derivative.

**How do you add LN?**

**ln(x/y) = ln(x) – ln(y)**

- ln(x/y) = ln(x) – ln(y)
- The natural log of the division of x and y is the difference of the ln of x and ln of y.
- Example: ln(7/4) = ln(7) – ln(4)

**How do you differentiate ln3x?**

**What is the derivative of an exponential function?**

^{x}has the special property that its derivative is the function itself, f′( x) = e

^{x}= f( x).