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## Calculus, all content (2017 edition)

### Course: Calculus, all content (2017 edition) > Unit 2

Lesson 28: Logarithmic functions differentiation- Derivatives of sin(x), cos(x), tan(x), eˣ & ln(x)
- Derivative of logₐx (for any positive base a≠1)
- Worked example: Derivative of log₄(x²+x) using the chain rule
- Differentiate logarithmic functions
- Differentiating logarithmic functions using log properties
- Derivative of logarithm for any base (old)
- Differentiating logarithmic functions review

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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.

## Want to join the conversation?

- 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)- 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?(43 votes)

- 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)

- 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)- 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)

- 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)- 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)

- Where i can find the proof of these derivatives?(5 votes)
- The website https://proofwiki.org has proofs for many mathematical identities and theorems, including various common derivatives.(7 votes)

- 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)
- 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)

- What are the proofs of sin, cos & tan derivatives? Are there any videos proving them?(4 votes)
- There are quite a few ways to do the proofs, and quite a few web pages with them.

This page has proofs for quite a few derivatives, including the trig functions.

https://www.wyzant.com/resources/lessons/math/calculus/derivative_proofs(4 votes)

- Is there a video that goes over the intuition behind integrals of trig functions (sec(x) and cot(x) included ideally)?(3 votes)
- 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)
- 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)

- what about the derivatives for arccos, arcsin, and arctan?(2 votes)
- D[arcsin](x) = 1/√(1 - x²).

D[arccos](x) = -1 / √(1 - x²).

D[arctan](x) = 1 / (1 + x²).(3 votes)

## 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.