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### Course: Differential Calculus > Unit 2

Lesson 9: Derivatives of cos(x), sin(x), 𝑒ˣ, and ln(x)- Derivatives of sin(x) and cos(x)
- Worked example: Derivatives of sin(x) and cos(x)
- Derivatives of sin(x) and cos(x)
- Proving the derivatives of sin(x) and cos(x)
- Derivative of 𝑒ˣ
- Derivative of ln(x)
- Derivatives of 𝑒ˣ and ln(x)
- Proof: The derivative of 𝑒ˣ is 𝑒ˣ
- Proof: the derivative of ln(x) is 1/x

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# Derivative of ln(x)

The derivative of ln(x) is 1/x. We show why it is so in a different video, but you can get some intuition here.

## Want to join the conversation?

- But ln(x) is a logarithmic function defined only for x-values greater than zero, while 1/x is a rational function defined for
**all**non-zero x's. So would it be more accurate to say: the derivative of ln(x) is 1/x such that x is greater than zero?(13 votes)- Really good thinking here, but since the domain is already limited with ln(x) when we start, we don't need to carry that over, since we already know x can't be 0 or less. if this were the other way around , where we started with a larger domain we would have to do something to the domain of the derivative.

Spoiler alert, this happens when you get to integrals.(20 votes)

- what happens when you have an equation like 3ln(x) would it be 3/x ?(4 votes)
- That's correct, because we leave scaling constants alone when taking derivatives.(8 votes)

- Can we say that if x--> inf

the slope is 0 and vice- versa. This might seem foolish but Im just curious(4 votes) - What is the derivative of 2x?(0 votes)
- The derivative of a function is its slope.

y=2x is a line of slope 2.

So the derivative of 2x is 2.(5 votes)

- Hi, what happens when we have to do ln 4x - is it 1/4x?(1 vote)
- No, actually. To find the derivative of ln(4x), you have to use the chain rule.

ln(4x) = 1/(4x) * 4 = 1/x

Hope this helps!(3 votes)

- Hear me out, what if x < 0(1 vote)
- since the function isnt defined at x < 0 then it isnt differentiable or the derivative doesnt exist at x < 0 (there are no derivatives/tangent lines for x values that dont exist)

so basically the derivative of a function has the same domain as the function itself. Therefore the derivative of the function f(x)= ln(x), which is defined only of x > 0, is also defined only for x > 0 (f'(x) = 1/x where x > 0).

i hope this makes sense(2 votes)

- I got a question asking "d/dx ln(2-e^x)". I thought the answer was "1/(2-e^x)" but the answer was

"(1/(2-e^x))(-e^x)". Can someone explain?(1 vote)- You need to apply the chain rule (https://www.khanacademy.org/math/ap-calculus-ab/ab-differentiation-2-new/ab-3-1a/v/chain-rule-introduction). The chain rule tells us how to find the derivative of a composite function, and ln(2-e^x) is a composite function [f(g(x))] where f(x) = ln(x) and g(x) = 2 - e^x.(2 votes)

- "Slope of a tangent line" and the derivative are the same thing, right?

It's like if you take a chunk from the original function's line at a certain point, same sized bit on either side of the line and "straighten it out", you get the tangent line and the derivative of that point, an m value. Right?(1 vote)- Be careful with your wording. However, I think you got right idea(1 vote)

- What is the derivative of ln(f(x))?(0 votes)
- This is an example of a composite function. A composite function like g(f(x)).

The differentiation of composite functions is done using the chain rule. This will be covered in the next modules but for now the differentiation of

*d/dx(ln(f(x)))*=*1/f(x)*f'(x)*(2 votes)

- how does Sal draw the tangent line to make the slope equal to some specific value ?(0 votes)
- The slope of a tangent is equal to the gradient of the curve.(1 vote)

## Video transcript

- [Instructor] In this
video, we're going to think about what the derivative
with respect to x of the natural log of x's. And I'm gonna go straight
to the punch line. It is equal to one over x. In a future video, I'm
actually going to prove this. It's a little bit involved. But in this one, we're
just going to appreciate that this seems like it is actually true. So right here is the graph of y is equal to the natural log of x. And just to feel good about the statement, let's try to approximate what the slope of the tangent
line is at different points. So let's say right over here, when x is equal to one, what does the slope of the
tangent line look like? Well, it looks like here, the slope looks like it is equal, pretty close to being equal to one, which is consistent with the statement. If x is equal to one, one
over one is still one, and that seems like what
we see right over there. What about when x is equal to two? Well, this point right over
here is the natural log of two, but more interestingly,
what's the slope here? Well, it looks like, let's see, if I try to
draw a tangent line, the slop of the tangent line
looks pretty close to 1/2. Well, once again, that is one over x. One over two is 1/2. Let's keep doing this. If I go right over here,
when x is equal to four, this point is four comma
natural log of four, but the slope of the tangent line here looks pretty close to 1/4 and if you accept this, it is exactly 1/4, and you could even go
to values less than one. Right over here, when x is equal to 1/2, one over 1/2, the slope should be two. And it does indeed, let me do this in a slightly different color, it does indeed look like
the slope is two over there. So once again, you take the derivative with respect to x of the natural
log of x, it is one over x. And hopefully, you get a sense that that is actually true here. In a future video,
we'll actually prove it.