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Derivatives of sin(x), cos(x), tan(x), eˣ & ln(x)

Learn the derivatives of several common functions. Created by Sal Khan.

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  • leaf green style avatar for user Aronyo Mitra
    Lets say I have an equation sin x = 1/2. Then clearly, x = 30 degrees or pi/6 radians.
    Now if I differentiate both sides of the equation with respect to x (because both are equal, their derivatives should also be equal), I will have cos x = 0, right?
    But that means x = 90 degrees, which is obviously not the solution! Could somebody please explain why this is happening?
    (14 votes)
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    • leaf green style avatar for user SteveSargentJr
      Once you say that "x = some number", x is no longer a variable but rather it is a constant -- and you must then treat it like a constant.

      Since x = π/6, both "x" and "sin(x)" are constants (because sin(π/6) = 1/2) and the derivative of a constant is zero.

      Does this clear things up for you?
      (42 votes)
  • leaf green style avatar for user avneesh.muralitharan
    What is e?
    (10 votes)
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    • piceratops seed style avatar for user Michael
      Think of e like pi - it's just an accepted name for a number that's useful in a bunch of formulas. It's equal to around 2.71828, so plug that into wherever you see e and you'll be good, or you can often leave things in terms of e.
      (18 votes)
  • aqualine ultimate style avatar for user SD
    What is the explanation of why the derivative of sec (x) is sec (x) tan(x)?

    It comes up in the questions for this topic on Khan Academy in the Special derivatives exercise.
    (11 votes)
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    • piceratops ultimate style avatar for user Just Keith
      If we accept that d/dx (cos x) = − sin x, and the power rule then:
      sec x ≡ 1/cos x
      Let u = cos x, thus du = − sin x dx
      sec x = 1/u
      (1/u) = (u⁻¹)
      By the power rule:
      derivative of (u⁻¹) = −u⁻² du
      Back substituting:
      = −(cos x)⁻² ( − sin x) ∙ dx
      = [sin x / (cos x)²] ∙ dx
      = [(sin x / cos x) ∙ (1/cos x)] ∙ dx
      = [tan (x) ∙ sec (x)] ∙ dx
      (10 votes)
  • aqualine seed style avatar for user Joseph Mandes
    I think I understand how to do the derivatives of the trig functions. But What if instead of Sin(x,
    there was Sin( and equation. Like, Sin(x^2+2). How would you take the derivative when it's something like that, and not just Sin or Cos?
    (5 votes)
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    • male robot hal style avatar for user Sid
      You always have to multiply the outer derivative with the inner derivative. That's true even for sin(x), it's just that the inner derivative is 1. (d/dx x = 1)

      d/dx sin(x) = cos(x) * 1 = cos(x)

      d/dx sin(2x) = cos(2x) * 2 = 2 cos(2x)

      d/dx sin(x^2) = cos(x^2) * 2x = 2x cos(x^2)

      d/dx sin(x^2 + 2) = cos(x^2 + 2) * 2x = 2x cos(x^2 + 2)
      (9 votes)
  • stelly blue style avatar for user Julian Delgadillo Marin
    Where i can find the proof of these derivatives?
    (5 votes)
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  • leaf green style avatar for user TeraVolt
    Is there a way of proving the derivative of e^x without using the derivative of ln(x)? If yes, can you please provide a link to it?
    (4 votes)
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    • piceratops ultimate style avatar for user Just Keith
      Yes, I can prove it:
      derivative of e^x , by definition of derivative:
      = lim h→0 { e^(x+h) - e^x} / h
      = lim h → 0 {e^x(e^h) -e^x}/h
      = lim h → 0 e^x{(e^h) -1}/h
      = (e^x) lim h → 0 {(e^h) -1}/h
      By definition of e:
      e = lim h→ 0 (1+h)^(1/h)
      And so,
      e^h = lim h→ 0 (1+h)^(h/h)
      = lim h→ 0 (1+h)
      Substituting that in:
      = (e^x) lim h → 0 {(1+h) -1}/h
      = (e^x) lim h → 0 { h/h}
      = (e^x) lim h → 0 { 1 }
      = e^x * (1)
      = e^x
      (6 votes)
  • aqualine tree style avatar for user CodeLoader
    What are the proofs of sin, cos & tan derivatives? Are there any videos proving them?
    (4 votes)
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  • female robot grace style avatar for user Tyler Barnes-Diana
    Is there a video that goes over the intuition behind integrals of trig functions (sec(x) and cot(x) included ideally)?
    (3 votes)
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  • purple pi purple style avatar for user Darko
    Are there videos of proving the derivations of trig functions? I've only seen the sine limit approaching zero proof where he proves it using sandwhich / squeeze theorem, (that is proof that limit of sinx/x is equal to 1 as x approaches zero). Also, could I use squeeze function and 2 parabolas to prove sine function instead of the overly complicated way he's proving it? And another thing, how can derivative of function tanx be cosecant when he's clearly written 1/cos^2(x) which is cosecant squared?
    (1 vote)
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    • aqualine ultimate style avatar for user Kyler Kathan
      Unfortunately there's no proof currently on Khan of the derivatives of sine, cosine, or tangent.
      Also, the derivative of tangent is secant squared.
      1/cos x = sec x
      d/dx (tan x) = 1/cos^2 x = sec^2 x

      As for proofs, here's a good proof of the derivative of sine:
      https://proofwiki.org/wiki/Derivative_of_Sine_Function/Proof_2

      Using the proof for sine, you can easily prove cosine using the equality
      cos(x) = sin(x+π/2) and the chain rule.

      Using the derivative of sine and the derivative of cosine, you can use the definition of tangent
      tan(x) = sin(x)/cos(x) and the quotient rule to prove the derivative of tangent.
      (5 votes)
  • piceratops tree style avatar for user ethan brito
    what about the derivatives for arccos, arcsin, and arctan?
    (2 votes)
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Video transcript

Let's get some exposure to the derivatives of some of the most common functions. We're not going to prove them in this video, but at least understand what the derivatives are. So first, let's start with the trig functions. If I want to take the derivative with respect to x of sine of x, this is going to be equal to cosine of x. And if you look at their graphs, it'll make intuitive sense. Once again I have not proved it here, but this is a good thing to know, that the derivative of sine of x is cosine of x. Now what about the derivative of cosine of x? What about the derivative with respect to x of cosine of x? Well, this one's going to be negative sine of x. So the derivative of sine is cosine, and the derivative cosine is negative sine. And then finally, the derivative of tangent of x is equal to 1 over cosine squared of x, which is equal to the secant squared of x. Once again, these are all very good things to know. Now let's talk a little bit about exponentials and logarithms. So the derivative-- and actually, this is one of the coolest results, and it once again speaks to how cool e is as a number, the derivative with respect to x of e to the x-- we need a drum roll for this one. This is one of the coolest things in mathematics. The derivative of e to the x is e to the x. Now what does that tell us? And I have to take a little pause here, because this is just so exciting. So let me graph e to the x. So that's my y-axis. Let's say that this right over here is my x-axis. So if I have very negative values of x, e to a very negative value, we are approaching zero. And then e to the 0 is 1, so that's going to be 1 right over there. So it's going to look something like this. And then it's an exponential. It's going to go, it's going to start increasing really, really, really, really, really fast. So let's say that's the graph of y is equal to e to the x. What this tells us is that at any point-- so let's say I go right over here. I say when x is equal to 0, e to the 0 is 1, what's the slope of the tangent line here? Turns out that is also 1. Amazing. If I go to x is equal to 1 right over here, the function evaluated here gets us e to the 1 power or just e. And what's the slope of the tangent line right over here? It is also e. At any point right over here, the slope of the tangent line is equal to the value of the function at that point. This is amazing. This is what is so cool about e. Anyway, that's not the point of this video. This video is to give you a catalog of all of the derivatives that you might really need. So then finally, if we're thinking about the derivative with respect to x of the natural log of x, this is going to be equal to-- and this is also fascinating. This is equal to 1 over x or x to the negative 1. So somehow, we have our natural log has kind of inserted itself into-- when you take the derivative, as filling in the gap that the power rule left vacant, which is, is there some function whose derivative is equal to x to the negative 1? The power rule gave us functions whose derivatives might be x to the negative 2, x to the negative 3, or x to the squared or x to the fifth. But it left the x to the negative 1 vacant, and it's filled by the natural log of x. Now I haven't proved it here. I've just catalogued these for you. And then we can use these in future videos, and we'll prove them in future videos.