Review: Basic Trig Derivatives You should already know these:
d d x [ sin x ] = cos x d d x [ cos x ] = − sin x \frac{d}{dx}[\sin x] = \cos x \qquad \frac{d}{dx}[\cos x] = -\sin x d x d [ sin x ] = cos x d x d [ cos x ] = − sin x They’re the foundation for everything else here.
The Four Remaining Trig Derivatives Memory Aid: The “C” Rule When a trig function starts with “c ” (cosine, cotangent, cosecant), its derivative has a negative sign in front.
Group 1: Tangent and Cotangent d d x [ tan x ] = sec 2 x \frac{d}{dx}[\tan x] = \sec^2 x d x d [ tan x ] = sec 2 x d d x [ cot x ] = − csc 2 x \frac{d}{dx}[\cot x] = -\csc^2 x d x d [ cot x ] = − csc 2 x Cotangent starts with “c,” so it gets the negative.
Group 2: Secant and Cosecant d d x [ sec x ] = sec x tan x \frac{d}{dx}[\sec x] = \sec x \tan x d x d [ sec x ] = sec x tan x d d x [ csc x ] = − csc x cot x \frac{d}{dx}[\csc x] = -\csc x \cot x d x d [ csc x ] = − csc x cot x Secant becomes “secant tangent,” and cosecant becomes “negative cosecant cotangent.”
Another way to see it: the c-count matches. If the original function has two c’s (like c sec ant), the answer has two c’s too.
Where d d x [ tan x ] = sec 2 x \frac{d}{dx}[\tan x] = \sec^2 x d x d [ tan x ] = sec 2 x You should just memorize it, but here’s the derivation. Start with
tan x = sin x cos x \tan x = \frac{\sin x}{\cos x} tan x = c o s x s i n x and apply the quotient rule:
d d x [ sin x cos x ] = ( cos x ) ( cos x ) − ( sin x ) ( − sin x ) cos 2 x \frac{d}{dx}\left[\frac{\sin x}{\cos x}\right] = \frac{(\cos x)(\cos x) - (\sin x)(-\sin x)}{\cos^2 x} d x d [ cos x sin x ] = cos 2 x ( cos x ) ( cos x ) − ( sin x ) ( − sin x ) = cos 2 x + sin 2 x cos 2 x = 1 cos 2 x = sec 2 x = \frac{\cos^2 x + \sin^2 x}{\cos^2 x} = \frac{1}{\cos^2 x} = \sec^2 x = cos 2 x cos 2 x + sin 2 x = cos 2 x 1 = sec 2 x Where the Other Three Come From Same move each time: rewrite the function with sine and cosine, then use the quotient rule.
Cotangent: cot x = cos x sin x \cot x = \frac{\cos x}{\sin x} cot x = s i n x c o s x d d x [ cos x sin x ] = ( − sin x ) ( sin x ) − ( cos x ) ( cos x ) sin 2 x = − ( sin 2 x + cos 2 x ) sin 2 x = − 1 sin 2 x = − csc 2 x \frac{d}{dx}\left[\frac{\cos x}{\sin x}\right] = \frac{(-\sin x)(\sin x) - (\cos x)(\cos x)}{\sin^2 x} = \frac{-(\sin^2 x + \cos^2 x)}{\sin^2 x} = -\frac{1}{\sin^2 x} = -\csc^2 x d x d [ sin x cos x ] = sin 2 x ( − sin x ) ( sin x ) − ( cos x ) ( cos x ) = sin 2 x − ( sin 2 x + cos 2 x ) = − sin 2 x 1 = − csc 2 x Secant: sec x = 1 cos x \sec x = \frac{1}{\cos x} sec x = c o s x 1 d d x [ 1 cos x ] = ( 0 ) ( cos x ) − ( 1 ) ( − sin x ) cos 2 x = sin x cos 2 x = 1 cos x ⋅ sin x cos x = sec x tan x \frac{d}{dx}\left[\frac{1}{\cos x}\right] = \frac{(0)(\cos x) - (1)(-\sin x)}{\cos^2 x} = \frac{\sin x}{\cos^2 x} = \frac{1}{\cos x} \cdot \frac{\sin x}{\cos x} = \sec x \tan x d x d [ cos x 1 ] = cos 2 x ( 0 ) ( cos x ) − ( 1 ) ( − sin x ) = cos 2 x sin x = cos x 1 ⋅ cos x sin x = sec x tan x Cosecant: csc x = 1 sin x \csc x = \frac{1}{\sin x} csc x = s i n x 1 d d x [ 1 sin x ] = ( 0 ) ( sin x ) − ( 1 ) ( cos x ) sin 2 x = − cos x sin 2 x = − 1 sin x ⋅ cos x sin x = − csc x cot x \frac{d}{dx}\left[\frac{1}{\sin x}\right] = \frac{(0)(\sin x) - (1)(\cos x)}{\sin^2 x} = -\frac{\cos x}{\sin^2 x} = -\frac{1}{\sin x} \cdot \frac{\cos x}{\sin x} = -\csc x \cot x d x d [ sin x 1 ] = sin 2 x ( 0 ) ( sin x ) − ( 1 ) ( cos x ) = − sin 2 x cos x = − sin x 1 ⋅ sin x cos x = − csc x cot x If you ever forget one of the four, this is how to rebuild it in under a minute.
Common Challenges Challenge 1: Memorization You have to commit all six derivatives to memory. Sine and cosine show up the most, but the others are essential too.
Challenge 2: Unit Circle Values You need to be able to recall things like:
csc ( π 6 ) = 1 sin ( π 6 ) = 1 1 / 2 = 2 \csc\left(\frac{\pi}{6}\right) = \frac{1}{\sin\left(\frac{\pi}{6}\right)} = \frac{1}{1/2} = 2 csc ( 6 π ) = s i n ( 6 π ) 1 = 1/2 1 = 2 cos ( π 4 ) = 2 2 \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} cos ( 4 π ) = 2 2 tan ( π 3 ) = 3 \tan\left(\frac{\pi}{3}\right) = \sqrt{3} tan ( 3 π ) = 3 This is precalc knowledge but it’s critical here.
Challenge 3: Simplifying Trig Expressions You have to recognize equivalent forms:
sin x ⋅ 1 cos x = sin x cos x = tan x \sin x \cdot \frac{1}{\cos x} = \frac{\sin x}{\cos x} = \tan x sin x ⋅ cos x 1 = cos x sin x = tan x Multiple choice answers are usually simplified, so you’ll need to manipulate your result to match.
Challenge 4: Calculator Input for Reciprocal Functions Most calculators don’t have cotangent, secant, or cosecant buttons. Enter them as reciprocals:
Cotangent: 1 ÷ tan x 1 \div \tan x 1 ÷ tan x Secant: 1 ÷ cos x 1 \div \cos x 1 ÷ cos x Cosecant: 1 ÷ sin x 1 \div \sin x 1 ÷ sin x Warning: the
sin − 1 \sin^{-1} sin − 1 button is
inverse sine (arcsine), NOT
1 sin x \frac{1}{\sin x} s i n x 1 .
Practice Problems Example 1: Product Rule with Trig Functions Find
d y d x \frac{dy}{dx} d x d y if
y = cos x ⋅ cot x y = \cos x \cdot \cot x y = cos x ⋅ cot x .
Two functions multiplied, so use the product rule:
d y d x = ( − sin x ) ( cot x ) + ( cos x ) ( − csc 2 x ) = − sin x cot x − cos x csc 2 x \frac{dy}{dx} = (-\sin x)(\cot x) + (\cos x)(-\csc^2 x) = -\sin x \cot x - \cos x \csc^2 x d x d y = ( − sin x ) ( cot x ) + ( cos x ) ( − csc 2 x ) = − sin x cot x − cos x csc 2 x Simplify using
cot x = cos x sin x \cot x = \frac{\cos x}{\sin x} cot x = s i n x c o s x :
= − sin x ⋅ cos x sin x − cos x csc 2 x = − cos x − cos x csc 2 x = -\sin x \cdot \frac{\cos x}{\sin x} - \cos x \csc^2 x = -\cos x - \cos x \csc^2 x = − sin x ⋅ sin x cos x − cos x csc 2 x = − cos x − cos x csc 2 x = − cos x ( 1 + csc 2 x ) = -\cos x(1 + \csc^2 x) = − cos x ( 1 + csc 2 x ) Example 2: Quotient Rule and Evaluation Find
f ′ ( π 4 ) f'\left(\frac{\pi}{4}\right) f ′ ( 4 π ) if
f ( x ) = x tan x f(x) = \dfrac{x}{\tan x} f ( x ) = tan x x .
Step 1: Quotient rule.
f ′ ( x ) = ( 1 ) ( tan x ) − ( x ) ( sec 2 x ) tan 2 x = tan x − x sec 2 x tan 2 x f'(x) = \frac{(1)(\tan x) - (x)(\sec^2 x)}{\tan^2 x} = \frac{\tan x - x\sec^2 x}{\tan^2 x} f ′ ( x ) = tan 2 x ( 1 ) ( tan x ) − ( x ) ( sec 2 x ) = tan 2 x tan x − x sec 2 x Step 2: Evaluate at
x = π 4 x = \frac{\pi}{4} x = 4 π . From the unit circle,
tan ( π 4 ) = 1 \tan\left(\frac{\pi}{4}\right) = 1 tan ( 4 π ) = 1 and
sec ( π 4 ) = 1 2 / 2 = 2 \sec\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}/2} = \sqrt{2} sec ( 4 π ) = 2 /2 1 = 2 .
f ′ ( π 4 ) = 1 − π 4 ( 2 ) 2 1 2 = 1 − π 4 ( 2 ) 1 = 1 − π 2 = 2 − π 2 ≈ − 0.571 f'\left(\frac{\pi}{4}\right) = \frac{1 - \frac{\pi}{4}(\sqrt{2})^2}{1^2} = \frac{1 - \frac{\pi}{4}(2)}{1} = 1 - \frac{\pi}{2} = \frac{2 - \pi}{2} \approx -0.571 f ′ ( 4 π ) = 1 2 1 − 4 π ( 2 ) 2 = 1 1 − 4 π ( 2 ) = 1 − 2 π = 2 2 − π ≈ − 0.571 Example 3: Calculator Usage Find the derivative of
f ( x ) = csc 2 ( 3 x ) f(x) = \csc^2(3x) f ( x ) = csc 2 ( 3 x ) at
x = 0.5 x = 0.5 x = 0.5 using a calculator.
Calculator steps (TI-83/84):
Press MATH, then select option 8 (nDeriv) Enter the function as (1/sin(3x))^2. Type (1 ÷ sin(3X)) then press x 2 x^2 x 2 1 sin ( 3 x ) \frac{1}{\sin(3x)} s i n ( 3 x ) 1 Enter the variable: X Enter the evaluation point: 0.5 Press ENTER Make sure your calculator is in RADIAN mode (check the MODE menu).
The result should be about
f ′ ( 0.5 ) ≈ − 0.428 f'(0.5) \approx -0.428 f ′ ( 0.5 ) ≈ − 0.428 .
Note on squaring:
Correct: (1/sin(3x))^2 squares the whole reciprocal Incorrect: 1/sin((3x)^2) only squares the argument Example 4: Rewrite with Identities First Find
f ′ ( x ) f'(x) f ′ ( x ) if
f ( x ) = 1 + sin x cos x f(x) = \dfrac{1 + \sin x}{\cos x} f ( x ) = cos x 1 + sin x .
You could use the quotient rule, but splitting the fraction and using identities is faster:
f ( x ) = 1 cos x + sin x cos x = sec x + tan x f(x) = \frac{1}{\cos x} + \frac{\sin x}{\cos x} = \sec x + \tan x f ( x ) = cos x 1 + cos x sin x = sec x + tan x f ′ ( x ) = sec x tan x + sec 2 x f'(x) = \sec x \tan x + \sec^2 x f ′ ( x ) = sec x tan x + sec 2 x Check with the quotient rule:
f ′ ( x ) = ( cos x ) ( cos x ) − ( 1 + sin x ) ( − sin x ) cos 2 x = cos 2 x + sin x + sin 2 x cos 2 x = 1 + sin x cos 2 x f'(x) = \frac{(\cos x)(\cos x) - (1 + \sin x)(-\sin x)}{\cos^2 x} = \frac{\cos^2 x + \sin x + \sin^2 x}{\cos^2 x} = \frac{1 + \sin x}{\cos^2 x} f ′ ( x ) = cos 2 x ( cos x ) ( cos x ) − ( 1 + sin x ) ( − sin x ) = cos 2 x cos 2 x + sin x + sin 2 x = cos 2 x 1 + sin x Those match, since
sec x tan x + sec 2 x = sin x cos 2 x + 1 cos 2 x = 1 + sin x cos 2 x \sec x \tan x + \sec^2 x = \frac{\sin x}{\cos^2 x} + \frac{1}{\cos^2 x} = \frac{1 + \sin x}{\cos^2 x} sec x tan x + sec 2 x = c o s 2 x s i n x + c o s 2 x 1 = c o s 2 x 1 + s i n x . Both forms are correct, which is why you may need to rearrange your answer to match a multiple-choice option.
Key Takeaways C-Rule: derivatives of functions starting with “c” have a negative sign in frontAll six trig derivatives: d d x [ sin x ] = cos x \frac{d}{dx}[\sin x] = \cos x d x d [ sin x ] = cos x d d x [ cos x ] = − sin x \frac{d}{dx}[\cos x] = -\sin x d x d [ cos x ] = − sin x d d x [ tan x ] = sec 2 x \frac{d}{dx}[\tan x] = \sec^2 x d x d [ tan x ] = sec 2 x d d x [ cot x ] = − csc 2 x \frac{d}{dx}[\cot x] = -\csc^2 x d x d [ cot x ] = − csc 2 x d d x [ sec x ] = sec x tan x \frac{d}{dx}[\sec x] = \sec x \tan x d x d [ sec x ] = sec x tan x d d x [ csc x ] = − csc x cot x \frac{d}{dx}[\csc x] = -\csc x \cot x d x d [ csc x ] = − csc x cot x Pattern: tangent/cotangent derivatives involve squared functions, secant/cosecant derivatives are products Calculator reciprocals: use 1 ÷ sin x 1 \div \sin x 1 ÷ sin x sin − 1 \sin^{-1} sin − 1 Your derivative may need algebraic manipulation to match the answer choices Forgot one? Rewrite it with sine and cosine and use the quotient rule 
