Subjects

Pre-Calculus

Trigonometry

Sinusoidal Equations

•

Dec 7, 2025

•

4 pages

Understanding the Derivatives of Trigonometric Functions

Drizzle Hinata

@drizzlehinata

Derivatives of trigonometric functions allow us to find rates of... Show more

1 / 4

ENGGMA TH2
DERIVATIVE OF TRIGONOMETRIC
FUNCTIONS
1-3 y
=
sin 3x
*
U = 3x
CALCULUS
dy
dx
d
;
=
dx
dx
dx
dx
dx
dx
sin u = cos u
3 cos3x
cos u

Derivative Formulas for Trigonometric Functions

When finding derivatives of trig functions, we need to remember six key formulas:

The derivative of sine: $\frac{d}{dx}\sin u = \cos u \frac{du}{dx}$
The derivative of cosine: $\frac{d}{dx}\cos u = -\sin u \frac{du}{dx}$
The derivative of tangent: $\frac{d}{dx}\tan u = \sec^2 u \frac{du}{dx}$
The derivative of cosecant: $\frac{d}{dx}\csc u = -\csc u \cot u \frac{du}{dx}$
The derivative of secant: $\frac{d}{dx}\sec u = \sec u \tan u \frac{du}{dx}$
The derivative of cotangent: $\frac{d}{dx}\cot u = -\csc^2 u \frac{du}{dx}$

Let's practice with an example: $y = \sin 3x$ . To find the derivative, we identify $u = 3x$ which means $\frac{du}{dx} = 3$ . Applying the sine formula: $\frac{dy}{dx} = \cos 3x (3) = 3 \cos 3x$ .

Pro Tip: Always start by identifying what's inside the trig function (the "u") and find its derivative $\frac{du}{dx}$ before applying the formula. This chain rule application is crucial for correctly differentiating composite trig functions.

More Complex Trigonometric Derivatives

When working with more complex expressions, we still follow the same pattern. For $y = 2 \csc(1 - 3x)$ , we first identify $u = 1 - 3x$ , giving us $\frac{du}{dx} = -3$ .

Using the cosecant formula and applying the chain rule: $\frac{dy}{dx} = 2-\csc $1 - 3x$ \cot $1 - 3x$ = 6 \csc $1 - 3x$ \cot $1 - 3x$ $

For nested functions like $y = \tan(x \sin x)$ , we need to find the derivative of the inside function first. With $u = x \sin x$ , we get $\frac{du}{dx} = x \cos x + \sin x$ using the product rule.

Then we apply the tangent formula: $\frac{dy}{dx} = \sec^2(x \sin x)[x \cos x + \sin x]$

Remember that these problems require careful attention to the chain rule. Break down each step methodically, and you'll see the pattern emerge.

Remember: The chain rule tells us to multiply by the derivative of the inner function, which is why we need $\frac{du}{dx}$ in all these formulas.

Product Rule with Trigonometric Functions

When two trig functions are multiplied together, we need the product rule: $\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}$

For $y = \tan x \sin x$ , we set $u = \tan x$ and $v = \sin x$ : $\frac{dy}{dx} = \tan x \frac{d}{dx}(\sin x) + \sin x \frac{d}{dx}(\tan x)$ $= \tan x \cos x + \sin x \sec^2 x$ $= \frac{\sin x}{\cos x} \cdot \cos x + \sin x \sec^2 x$ $= \sin x + \sin x \sec^2 x = \sin x(1 + \sec^2 x)$

For powers of trig functions like $y = \cos^4 t - \sin^4 t$ , we use the power rule along with the trig derivative formulas: $\frac{dy}{dx} = 4\cos^3 t (-\sin t) - 4\sin^3 t (\cos t)$ $= -4\cos^3 t \sin t - 4\sin^3 t \cos t$

We can factor out common terms: $= -4 \cos t \sin t (\cos^2 t + \sin^2 t) = -4 \cos t \sin t (1) = -2(2\sin t \cos t) = -2\sin 2t$

Simplify when possible: Notice how we used the identity $\sin 2t = 2\sin t \cos t$ to make our final answer more elegant. Always look for ways to simplify your final expressions.

Practice Problems and Special Cases

Let's examine some additional examples:

For $y = \sin(\cos x)$ , we identify the nested functions and apply the chain rule:

Outer function: sine
Inner function: $\cos x$ with derivative $-\sin x$

Using $\frac{d}{dx}\sin u = \cos u \frac{du}{dx}$ : $\frac{dy}{dx} = \cos(\cos x) \cdot (-\sin x) = -\sin x \cdot \cos(\cos x)$ The answer is $y' = -\cos(\cos x) \cdot \sin x$

For $r = \frac{\cos 2\theta}{1 - \sin 2\theta}$ , we need the quotient rule along with derivatives of trig functions. This gives us $\frac{dr}{d\theta} = \frac{2}{1 - \sin 2\theta}$

These problems demonstrate the importance of recognizing patterns and applying multiple rules in combination. With practice, you'll develop an intuition for approaching even the most complex trig derivatives.

Test yourself: Try creating your own trig derivative problems and solving them step by step. Teaching yourself is one of the best ways to master these concepts!

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Subjects

Pre-Calculus

Trigonometry

Sinusoidal Equations

Pre-Calculus

•

Dec 7, 2025

•

4 pages

Understanding the Derivatives of Trigonometric Functions

Drizzle Hinata

@drizzlehinata

Derivatives of trigonometric functions allow us to find rates of change for wave-like patterns and cycles. This topic is essential in calculus and has many real-world applications in physics, engineering, and signal processing. Let's break down how to find these... Show more

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Derivative Formulas for Trigonometric Functions

When finding derivatives of trig functions, we need to remember six key formulas:

The derivative of sine: $\frac{d}{dx}\sin u = \cos u \frac{du}{dx}$
The derivative of cosine: $\frac{d}{dx}\cos u = -\sin u \frac{du}{dx}$
The derivative of tangent: $\frac{d}{dx}\tan u = \sec^2 u \frac{du}{dx}$
The derivative of cosecant: $\frac{d}{dx}\csc u = -\csc u \cot u \frac{du}{dx}$
The derivative of secant: $\frac{d}{dx}\sec u = \sec u \tan u \frac{du}{dx}$
The derivative of cotangent: $\frac{d}{dx}\cot u = -\csc^2 u \frac{du}{dx}$

Let's practice with an example: $y = \sin 3x$ . To find the derivative, we identify $u = 3x$ which means $\frac{du}{dx} = 3$ . Applying the sine formula: $\frac{dy}{dx} = \cos 3x (3) = 3 \cos 3x$ .

Pro Tip: Always start by identifying what's inside the trig function (the "u") and find its derivative $\frac{du}{dx}$ before applying the formula. This chain rule application is crucial for correctly differentiating composite trig functions.

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More Complex Trigonometric Derivatives

When working with more complex expressions, we still follow the same pattern. For $y = 2 \csc(1 - 3x)$ , we first identify $u = 1 - 3x$ , giving us $\frac{du}{dx} = -3$ .

Using the cosecant formula and applying the chain rule: $\frac{dy}{dx} = 2-\csc $1 - 3x$ \cot $1 - 3x$ = 6 \csc $1 - 3x$ \cot $1 - 3x$ $

For nested functions like $y = \tan(x \sin x)$ , we need to find the derivative of the inside function first. With $u = x \sin x$ , we get $\frac{du}{dx} = x \cos x + \sin x$ using the product rule.

Then we apply the tangent formula: $\frac{dy}{dx} = \sec^2(x \sin x)[x \cos x + \sin x]$

Remember that these problems require careful attention to the chain rule. Break down each step methodically, and you'll see the pattern emerge.

Remember: The chain rule tells us to multiply by the derivative of the inner function, which is why we need $\frac{du}{dx}$ in all these formulas.

Sign up to see the contentIt's free!

Access to all documents

Improve your grades

Join milions of students

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Product Rule with Trigonometric Functions

When two trig functions are multiplied together, we need the product rule: $\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}$

We can factor out common terms: $= -4 \cos t \sin t (\cos^2 t + \sin^2 t) = -4 \cos t \sin t (1) = -2(2\sin t \cos t) = -2\sin 2t$

Simplify when possible: Notice how we used the identity $\sin 2t = 2\sin t \cos t$ to make our final answer more elegant. Always look for ways to simplify your final expressions.

Sign up to see the contentIt's free!

Access to all documents

Improve your grades

Join milions of students

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Practice Problems and Special Cases

Let's examine some additional examples:

For $y = \sin(\cos x)$ , we identify the nested functions and apply the chain rule:

Outer function: sine
Inner function: $\cos x$ with derivative $-\sin x$

Using $\frac{d}{dx}\sin u = \cos u \frac{du}{dx}$ : $\frac{dy}{dx} = \cos(\cos x) \cdot (-\sin x) = -\sin x \cdot \cos(\cos x)$ The answer is $y' = -\cos(\cos x) \cdot \sin x$

For $r = \frac{\cos 2\theta}{1 - \sin 2\theta}$ , we need the quotient rule along with derivatives of trig functions. This gives us $\frac{dr}{d\theta} = \frac{2}{1 - \sin 2\theta}$

Test yourself: Try creating your own trig derivative problems and solving them step by step. Teaching yourself is one of the best ways to master these concepts!