Derivatives of Trig Functions

Memorize all of these.

Simple Harmonic Motion

Any back-and-forth motion

Pendulum, spring, almost any vibration, �waves…

When you isolate one dimension of a �two-dimensional orbit, each dimension �could be described as simple harmonic motion.

Where is the derivative the highest? The lowest?

Let’s have a look at sin(x)

sin(x) is in blue. Never a straight line.�Derivative will keep changing.

When sin(x) reaches a max or min,�its derivative is zero.

When sin(x) reaches a node, its derivative is at its highest value.

What function does the red line (the derivative) look like?

^d/_dx sin(x) = cos(x)

The derivative of sin(x) is cos(x)�cos(x) shown in blue this time

But...what’s the derivative of cos(x)?

Same deal as sin(x): extrema produce a derivative of zero, nodes produce extrema

Shown in red, the derivative is -sin(x)

Phase Shift (sin(x) in black)

^d/_dx sin(x) = cos(x)�^d/_dx cos(x) = -sin(x)�^d/_dx -sin(x) = -cos(x)�^d/_dx -cos(x) = sin(x)

Engineers would call this a phase shift - the function is basically the same, it’s simply in a different phase of its cycle.

sin(x) = cos(x - 𝜋/2)

The other trig functions

tan(x) = sin(x) sec(x) = __1__� cos(x) cos(x)

cot(x) = cos(x) csc(x) = __1__� sin(x) sin(x)

You could do the quotient rule to find out the derivatives of all these.

Or...

Trig Derivatives

^d/_dx tan(x) = sec²(x) ^d/_dx sec(x) = sec(x)tan(x)

^d/_dx cot(x) = -csc²(x) ^d/_dx csc(x) = -csc(x)cot(x)

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Yes, you need to memorize these!

In Practice

What’s the second derivative of csc(x)?

First derivative - we have a formula for that!�d/dx csc(x) = -csc(x)cot(x)

Second derivative - Product rule!�[-csc(x) ∙ -csc²(x)] + [csc(x)cot(x) ∙ cot(x)]

= csc³(x) + csc(x)cot²(x)�= 1 + cos²(x)� sin³(x)

A step further

Find the equation for the line tangent to the derivative of tan(x) at the point where x = 𝜋/3

If it’s tangent to the derivative, we need the slope of the derivative (we need the second derivative).

^d/_dx tan(x) = sec²(x)

...continued

d/dx sec²(x) Product rule!

[sec(x) ∙ sec(x)tan(x)] + [sec(x)tan(x) ∙ sec(x)]�= 2sec²(x)tan(x)

Since sec(x) = 1/cos(x) and tan = sin(x)/cos(x)�= 2sin(x)/cos³(x)

Now we need the slope at x = 𝜋/3 2sin(𝜋/3)� (cos(𝜋/3))³

...still going!

sin(𝜋/3) = √(3)/2�cos(𝜋/3) = ½

2√(3)/2 = √3 = 8√3� (½)³ ⅛

WE FINALLY HAVE THE SLOPE!!!

Now we need the y-intercept

y = mx + b

m = 8√3

Keep in mind, we’re focused on the derivative of tan(x), or sec²(x)

sec²(𝜋/3) = 4 Our point is (𝜋/3, 4)

4 = [8√(3)] ∙ (𝜋/3) + ???

y = 8x√(3) + 3/2𝜋√3 FINALLY SOLVED!!!

Chain Rule!

7cos(9x³ - 7x)

f(x) = 7cos(x)�g(x) = 9x³ - 7x

(f ० g)’(x) = f’(g(x)) ∙ g’(x)

-7sin(9x³ - 7x) ∙ (27x² - 7) = -189x²sin(9x³ - 7x) + 49sin(9x³ - 7x)

U-Substitution

Another way of doing the Chain Rule

Define the “inside” function as a single variable: u�Now take the derivative u’ like normal�Multiply the result by u’

Example:�-4cos(7x^-2) u = 7x^-2 u’ = -14x^-3�-4cos(u) → 4sin(u) ∙ u’�4sin(7x^-2) ∙ -14x^-3 → -56x^-3sin(7x^-2)

And Another Thing

^d/_dx cos³(x)

Keep in mind, cos³(x) = [cos(x)]³

3[cos(x)]² ∙ -sin(x)

-3cos²(x)sin(x)

4.2 #10

If y = sin(3x), then ^dy/_dx =

Extra Links

Find the derivative of 7cot³(4x¹² - 5x) same thing as 7[cot(4x¹²)]³

^d/_dx of 4x¹² = 48x¹¹ - 5�^d/_dx of cot(4x¹²) = -csc²(4x¹²) · (48x¹¹ - 5)�^d/_dx of 7[cot(4x¹²)]³ = 21[cot(4x¹²)]² · -csc²(4x¹²) · (48x¹¹ - 5)

Start inside and work your way out! Find the derivative each time, and keep applying the Chain Rule:�(f ० g)’(x) = f’(g(x)) ∙ g’(x)

4.2 #25

∫₀^𝜋/4 sin(x) dx =

Recap

Derivatives of sin(x) and cos(x) cause a phase shift - memorize the sequence.�Memorize all your other trig derivatives too!

Trig derivatives often involve the Power and Quotient Rules�Know your trig definitions (like sec(x) = 1/cos(x)) - it’ll save you a lot of headaches.

Today’s lesson had a lot of equations to memorize.

Derivative of e^x

Everyone’s favorite derivative

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