# The Derivative Of Lnx And Examples

# The Derivative Of Lnx And Examples

**Derivative Of Lnx: **Part of calculus is mastering the basic derivative rules like the production rule, the power rule, or the chain rule. One of the rules you will see come up often is the rule for the outgrowth of lnx. In the accompanying lesson, we will look at some illustrations of how to apply this rule to finding extraordinary types of by-products. We will also see how using the laws of logarithms can help make taking these kinds of by-products even easier.

The aggressive function has an inverse gathering, which is called the **natural logarithm**, and is stood for ln(x). Our goal on this page is to verify that the derivative of the natural logarithm is a sensible function. Precisely, we show that the following is true.

ln(x) = 1/x

## What Is The Derivative Of Lnx

Our next task is to resolve what is the derivative of the genuine logarithm. We begin with the contrary definition. If

y = ln x

then

e^{y} = x

Now essentially take the derivative of both sides with relation to x reminding to multiply by dy/dx on the left hand side from the time of 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

If f(x) = ln x, then

f ‘(x) = 1/x

## Derivative Of Lnx^2

**ln(x) = **lim_{(d->0)} [ ln(x+d) – ln(x) ] / d = lim ln((x+d)/x) / d

= lim (1/d) ln(1 + d/x) = lim [ ln (1 + d/x)^{^(1/d)} ].

*Set u=d/x and substitute:*

lim_{(u->0)} [ ln (1 + u)^{^(1/(ux))} ] = 1/x ln [ lim_{(u->0)} (1 + u)^{^(1/u)} ]

= 1/x ln (e) (Definition of e)

**= 1/x**.

## Anti Derivative Of Lnx

Remember that when taking the outgrowth, you can break the by-product up over addition/subduction, and you can take out guideline. This allows us to find the following.

### Example

(3ln(x))′=3(1x)=3x(3ln(x))′=3(1x)=3x

### Example

(ln(x)5)′=15(ln(x))′=(15)(1x)=15x(ln(x)5)′=15(ln(x))′=(15)(1x)=15x

### Example

(2x2–ln(x))′=4x–1x(2×2–ln(x))′=4x–1x

These show you the more straightforward types of derivatives you can find using this rule. But, if we combine this with the laws of logarithms we can do even more.

## Derivative Of 1/Lnx **Example**

**Example**

**Example**

find the derivative of

f(x) = ln[(1 + x)(1 + x^{2})^{2}(1 + x^{3})^{3 }]

**Solution**

The last thing that we want to do is to use the product rule and chain rule multiple times. Instead, we first simplify with properties of the natural logarithm. We have

ln[(1 + x)(1 + x^{2})^{2}(1 + x^{3})^{3 }] = ln(1 + x) + ln(1 + x^{2})^{2} + ln(1 + x^{3})^{3 }

= ln(1 + x) + 2 ln(1 + x^{2}) + 3 ln(1 + x^{3})

Now the derivative is not so daunting. We have use the chain rule to get

1 4x 9x^{2}

f ‘(x) = + +

1 + x 1 + x^{2} 1 + x^{3}

## Derivative Of Lnx

Find the derivative of the function:

y=ln(x2)y=ln(x2)

### Solution

Before applying any calculus rules, first expand the expression using the laws of logarithms. Here, we can use rule (1). This step is all algebra; no calculus is done until after we expand the expression.

y=ln(x2)=2ln(x)y=ln(x2)=2ln(x)

Now, take the derivative. This is the calculus step.

y′=(2ln(x))′=2(ln(x))′=2(1x)=2xy′=(2ln(x))′=2(ln(x))′=2(1x)=2x

In the example above, only one rule was needed to fully expand the expression. The next example shows you how to apply more than one rule.

### Example

Find the derivative of the function.

y=ln(5x4)y=ln(5×4)

Before taking the derivative, we will expand this expression. Since the exponent is only on the x, we will need to first break this up as a product, using rule (2) above. Then, we can apply rule (1).

y=ln(5x4)=ln(5)+ln(x4)=ln(5)+4ln(x)y=ln(5×4)=ln(5)+ln(x4)=ln(5)+4ln(x)

Now take the derivative of the expanded form of the function, and then simplify.

y′=(ln(5)+4ln(x))′=(ln(5))′+4(ln(x))′=4(1x)=4xy′=(ln(5)+4ln(x))′=(ln(5))′+4(ln(x))′=4(1x)=4x

You may be wondering what happened to ln(5)ln(5). Remember – this is a constant. If you check your calculator, you will find that ln(5)≈1.61ln(5)≈1.61. The derivative of a constant is zero.

None of these examples have used rule (3), so let’s look at one more example to see how that might be applied.

### Example

Find the derivative of the function.

y=ln(6x2)y=ln(6×2)

Here we have a fraction, which we can expand with rule (3), and then a power, which we can expand with rule (1). Remember that this is just algebra – no calculus is involved just yet.

y=ln(6x2)=ln(6)–ln(x2)=ln(6)–2ln(x)y=ln(6×2)=ln(6)–ln(x2)=ln(6)–2ln(x)

Now that we have ln(x)ln(x) by itself, we can apply the derivative rule for the natural log.

y′=(ln(6)–2ln(x))′=(ln(6))′–2(ln(x))′=−2(1x)=−2x

## What is derivative of natural log?

**DERIVATIVE** OF THE **NATURAL LOGARITHM**. y = ln(x) The exponential function has an inverse function, which is called the **natural logarithm**, and is denoted ln(x).

## What is the derivative of Lnx 2?

Thus, the **derivative of ln x2** is **2**/x.

## How do you differentiate LN?

**Solution**

- We use the formula. ln x. f(x) = ln 4. so that. f ‘(x) = x ln 4.
- We again use the formula. ln(3x + 4) f(x) = ln 10. now use the chain rule to get. f ‘(x) = (3x + 4) ln 10.
- Use the product rule to get. f ‘(x) = log(2x) + x(log(2x))’ Now use the formula to get. ln(2x) log (2x) = ln 10. The chain rule gives. 2 1.