Question 1

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?

Accepted Answer

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?

Question 2

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.

Accepted Answer

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

Question 3

What is e?

Accepted Answer

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.

Question 4

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?

Accepted Answer

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)

Question 5

Where i can find the proof of these derivatives?

Accepted Answer

The website https://proofwiki.org  has proofs for many mathematical identities and theorems, including various common derivatives.

Question 6

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?

Accepted Answer

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

Question 7

What are the proofs of sin, cos & tan derivatives? Are there any videos proving them?

Accepted Answer

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

Question 8

what about the derivatives for arccos, arcsin, and arctan?

Accepted Answer

D[arcsin](x)  =  1/√(1 - x²).
D[arccos](x) = -1 / √(1 - x²).
D[arctan](x)   = 1 / (1 + x²).

Question 9

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?

Accepted Answer

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.

Class 12 math (India)

Course: Class 12 math (India) > Unit 5

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

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