•

Feb 3, 2026

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4 pages

Mastering Trigonometric Function Derivatives

Ella Nadine

@nadenden

Trigonometric derivatives are essential tools for calculus that allow us... Show more

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$DERIVATIVE OF TRIGONOMETRIC FUNCTIONS $ \frac{d}{dx} \sin u = \cos u \frac{du}{dx} $ $ \frac{d}{dx} \cos u = -\sin u \frac{du}{dx} $ $ \frac$

Derivatives of Trigonometric Functions

When working with trigonometric functions, you need to memorize six key derivative formulas:

$\frac{d}{dx} \sin u = \cos u \frac{du}{dx}$
$\frac{d}{dx} \cos u = -\sin u \frac{du}{dx}$
$\frac{d}{dx} \tan u = \sec^2 u \frac{du}{dx}$
$\frac{d}{dx} \csc u = -\csc u \cot u \frac{du}{dx}$
$\frac{d}{dx} \sec u = \sec u \tan u \frac{du}{dx}$
$\frac{d}{dx} \cot u = -\csc^2 u \frac{du}{dx}$

Let's see this in action with $y = \sin 3x$ . Identify that $u = 3x$ , which means $\frac{du}{dx} = 3$ . Applying the sine derivative formula: $\frac{dy}{dx} = \cos u \cdot \frac{du}{dx} = \cos 3x \cdot 3 = 3\cos 3x$ .

Pro Tip: The chain rule is crucial here - always identify your inner function (u) and find its derivative $\frac{du}{dx}$ before applying the trig formula!

$DERIVATIVE OF TRIGONOMETRIC FUNCTIONS $ \frac{d}{dx} \sin u = \cos u \frac{du}{dx} $ $ \frac{d}{dx} \cos u = -\sin u \frac{du}{dx} $ $ \frac$

More Complex Trig Derivatives

For $y = 2\csc(1-3x)$ , we need the formula $\frac{d}{dx} \csc u = -\csc u \cot u \frac{du}{dx}$ .

First, identify $u = 1-3x$ , which gives $\frac{du}{dx} = -3$ . Then apply the formula: $\frac{dy}{dx} = 2[-\csc(1-3x) \cot(1-3x)] \cdot (-3) = 6\csc(1-3x)\cot(1-3x)$

With $y = \tan(x\sin x)$ , we face a more complex situation. Here, $u = x\sin x$ and we need to find $\frac{du}{dx}$ using the product rule: $\frac{du}{dx} = x\cos x + \sin x$

Now we can apply the tangent derivative formula: $\frac{dy}{dx} = \sec^2(x\sin x) \cdot (x\cos x + \sin x)$

Remember: When your inner function gets complicated, break down its derivative step by step before plugging into the main formula!

$DERIVATIVE OF TRIGONOMETRIC FUNCTIONS $ \frac{d}{dx} \sin u = \cos u \frac{du}{dx} $ $ \frac{d}{dx} \cos u = -\sin u \frac{du}{dx} $ $ \frac$

Product Rule with Trig Functions

For $y = \tan x \sin x$ , we need to use the product rule since we have a product of two functions: $\frac{dy}{dx} = \tan x \cdot \frac{d}{dx}(\sin x) + \sin x \cdot \frac{d}{dx}(\tan x)$

Substituting the derivatives: $\frac{dy}{dx} = \tan x \cdot \cos x + \sin x \cdot \sec^2 x$

Simplifying: $= \frac{\sin x}{\cos x} \cdot \cos x + \sin x \cdot \sec^2 x = \sin x + \sin x \sec^2 x = \sin x(1 + \sec^2 x)$

For $y = \cos^4 t - \sin^4 t$ , we use the power rule for each term: $\frac{dy}{dx} = 4\cos^3 t \cdot (-\sin t) - 4\sin^3 t \cdot \cos t = -4\cos^3 t \sin t - 4\sin^3 t \cos t$

Factoring out common terms and using trig identities: $= -4\cos t \sin t(\cos^2 t + \sin^2 t) = -4\cos t \sin t = -2(2\sin t \cos t) = -2\sin 2t$

Simplify smartly: Trig identities like $\sin^2 t + \cos^2 t = 1$ and $\sin 2t = 2\sin t \cos t$ can dramatically simplify your answers!

$DERIVATIVE OF TRIGONOMETRIC FUNCTIONS $ \frac{d}{dx} \sin u = \cos u \frac{du}{dx} $ $ \frac{d}{dx} \cos u = -\sin u \frac{du}{dx} $ $ \frac$

Practice Problems

When tackling a problem like $y = \sin(\cos x)$ , identify the nested structure. Here $u = \cos x$ is inside a sine function.

Apply the chain rule: $\frac{dy}{dx} = \cos(\cos x) \cdot \frac{d}{dx}(\cos x) = \cos(\cos x) \cdot (-\sin x) = -\sin x \cdot \cos(\cos x)$

For $r = \frac{\cos 2\theta}{1-\sin 2\theta}$ , you'll need the quotient rule along with the derivatives of $\cos 2\theta$ and $\sin 2\theta$ .

These examples show how trig derivatives combine multiple calculus techniques - the chain rule, product rule, quotient rule, and simplification using trig identities.

Build your confidence: Start with simpler problems and work your way up. With practice, even the most complex trig derivatives will become manageable!

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Calculus 1

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Feb 3, 2026

•

4 pages

Mastering Trigonometric Function Derivatives

Ella Nadine

@nadenden

Trigonometric derivatives are essential tools for calculus that allow us to find rates of change for oscillating functions. In this guide, we'll explore how to find derivatives of various trig functions using standard formulas and the chain rule.

$DERIVATIVE OF TRIGONOMETRIC FUNCTIONS $ \frac{d}{dx} \sin u = \cos u \frac{du}{dx} $ $ \frac{d}{dx} \cos u = -\sin u \frac{du}{dx} $ $ \frac$

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Derivatives of Trigonometric Functions

When working with trigonometric functions, you need to memorize six key derivative formulas:

$\frac{d}{dx} \sin u = \cos u \frac{du}{dx}$
$\frac{d}{dx} \cos u = -\sin u \frac{du}{dx}$
$\frac{d}{dx} \tan u = \sec^2 u \frac{du}{dx}$
$\frac{d}{dx} \csc u = -\csc u \cot u \frac{du}{dx}$
$\frac{d}{dx} \sec u = \sec u \tan u \frac{du}{dx}$
$\frac{d}{dx} \cot u = -\csc^2 u \frac{du}{dx}$

Pro Tip: The chain rule is crucial here - always identify your inner function (u) and find its derivative $\frac{du}{dx}$ before applying the trig formula!

$DERIVATIVE OF TRIGONOMETRIC FUNCTIONS $ \frac{d}{dx} \sin u = \cos u \frac{du}{dx} $ $ \frac{d}{dx} \cos u = -\sin u \frac{du}{dx} $ $ \frac$

Sign up to see the contentIt's free!

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Join milions of students

More Complex Trig Derivatives

For $y = 2\csc(1-3x)$ , we need the formula $\frac{d}{dx} \csc u = -\csc u \cot u \frac{du}{dx}$ .

First, identify $u = 1-3x$ , which gives $\frac{du}{dx} = -3$ . Then apply the formula: $\frac{dy}{dx} = 2[-\csc(1-3x) \cot(1-3x)] \cdot (-3) = 6\csc(1-3x)\cot(1-3x)$

With $y = \tan(x\sin x)$ , we face a more complex situation. Here, $u = x\sin x$ and we need to find $\frac{du}{dx}$ using the product rule: $\frac{du}{dx} = x\cos x + \sin x$

Now we can apply the tangent derivative formula: $\frac{dy}{dx} = \sec^2(x\sin x) \cdot (x\cos x + \sin x)$

Remember: When your inner function gets complicated, break down its derivative step by step before plugging into the main formula!

$DERIVATIVE OF TRIGONOMETRIC FUNCTIONS $ \frac{d}{dx} \sin u = \cos u \frac{du}{dx} $ $ \frac{d}{dx} \cos u = -\sin u \frac{du}{dx} $ $ \frac$

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Product Rule with Trig Functions

For $y = \tan x \sin x$ , we need to use the product rule since we have a product of two functions: $\frac{dy}{dx} = \tan x \cdot \frac{d}{dx}(\sin x) + \sin x \cdot \frac{d}{dx}(\tan x)$

Substituting the derivatives: $\frac{dy}{dx} = \tan x \cdot \cos x + \sin x \cdot \sec^2 x$

Simplifying: $= \frac{\sin x}{\cos x} \cdot \cos x + \sin x \cdot \sec^2 x = \sin x + \sin x \sec^2 x = \sin x(1 + \sec^2 x)$

For $y = \cos^4 t - \sin^4 t$ , we use the power rule for each term: $\frac{dy}{dx} = 4\cos^3 t \cdot (-\sin t) - 4\sin^3 t \cdot \cos t = -4\cos^3 t \sin t - 4\sin^3 t \cos t$

Factoring out common terms and using trig identities: $= -4\cos t \sin t(\cos^2 t + \sin^2 t) = -4\cos t \sin t = -2(2\sin t \cos t) = -2\sin 2t$

Simplify smartly: Trig identities like $\sin^2 t + \cos^2 t = 1$ and $\sin 2t = 2\sin t \cos t$ can dramatically simplify your answers!

$DERIVATIVE OF TRIGONOMETRIC FUNCTIONS $ \frac{d}{dx} \sin u = \cos u \frac{du}{dx} $ $ \frac{d}{dx} \cos u = -\sin u \frac{du}{dx} $ $ \frac$

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Improve your grades

Join milions of students

Practice Problems

When tackling a problem like $y = \sin(\cos x)$ , identify the nested structure. Here $u = \cos x$ is inside a sine function.

Apply the chain rule: $\frac{dy}{dx} = \cos(\cos x) \cdot \frac{d}{dx}(\cos x) = \cos(\cos x) \cdot (-\sin x) = -\sin x \cdot \cos(\cos x)$

For $r = \frac{\cos 2\theta}{1-\sin 2\theta}$ , you'll need the quotient rule along with the derivatives of $\cos 2\theta$ and $\sin 2\theta$ .

These examples show how trig derivatives combine multiple calculus techniques - the chain rule, product rule, quotient rule, and simplification using trig identities.

Build your confidence: Start with simpler problems and work your way up. With practice, even the most complex trig derivatives will become manageable!