## How to calculate the derivative the ln(x+1)

The chain dominion is valuable for recognize the derivative of one expression which can have been distinguished had it remained in terms that x, but it is in the form of an additional expression which could also be identified if that stood top top its own.

You are watching: Derivative of 1/ln(x)

In this case:

We know just how to distinguish x+1 (the prize is 1)We know exactly how to distinguish ln(x) (the prize is 1/x)This way the chain preeminence will enable us to do the differentiation that the role ln(x+1).

To do the differentiation, the chain rule says us must distinguish the expression as if it were just in terms of x as long as we then multiply that result by the derivative that what the expression was in reality in terms of (in this instance the derivative that x+1).

Using the chain rule to find the derivative the ln(x+1)ln(x+1) is in the kind of the standard natural log function ln(x), except it go not have x as an argument, instead it has actually another role of x (x+1).

Let’s speak to the role in the discussion g(x), i m sorry means:

g(x) = x+1

From this it complies with that:

ln(x+1) = ln(g(x))

So if the role f(x) = ln(x) and also the duty g(x) = x+1, climate the function ln(x+1) deserve to be created as a composite function.

f(x) = ln(x)f(g(x)) = ln(g(x)) (but g(x) = x+1)f(g(x)) = ln(x+1)

Let’s specify this composite role as F(x):

F(x) = f(g(x)) = ln(x+1)

We can find the derivative that ln(x+1) (F"(x)) by making use of the chain rule.

**The Chain Rule:****For two differentiable attributes f(x) and also g(x)****If F(x) = f(g(x))****Then the derivative of F(x) is F"(x) = f’(g(x)).g’(x)**

Now we have the right to just plug f(x) and also g(x) into the chain rule. But before we execute that, simply a recap top top the derivative the the natural logarithm.

The derivative of ln(x) with respect come x is (1/x)The derivative the ln(s) v respect to s is (1/s)

In a similar way, the derivative that ln(x+1) with respect to **x+1** is 1/(x+1). We will usage this reality as component of the chain dominance to discover the derivative the ln(x+1) through respect come **x**.

**How to find the derivative the ln(x+1) making use of the Chain Rule:**

F"(x) | = f"(g(x)).g"(x) | Chain dominance Definition |

= f"(g(x)).(1) | g(x) = x+1 ⇒ g"(x) = 1 | |

= (1/(x+1)).1 | f(g(x)) = ln(x+1) ⇒ f"(g(x)) = 1/(x+1)(The derivative of ln(x+1) with respect come x+1 is 1/(x+1) | |

= 1/(x+1) |

Using the chain rule, we uncover that the derivative of ln(x+1) is 1/(x+1)

Finally, just a note on syntax and notation: ln(x+1) is periodically written in the forms listed below (with the derivative together per the calculations above). Simply be aware that not all of the forms listed below are mathematically correct.

lnx+1 | ► Derivative of lnx+1 =1/(x+1) |

ln x+1 | ► Derivative of ln x+1 = 1/(x+1) |

ln x + 1 | ► Derivative that ln x +1 = 1/(x+1) |

## The second Derivative the ln(x+1)

To calculate the second derivative of a function, you just differentiate the first derivative.

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From above, we discovered that the first derivative of ln(x+1) = 1/(x+1). So to uncover the 2nd derivative the ln(x+1), we just need to distinguish 1/(x+1).

We can use the quotient ascendancy to discover the derivative that 1/(x+1), and we get response of -1/(x+1)2