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!

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!

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!

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!