Derivatives of trigonometric functions worksheet with problems 43-64.

To solve the derivatives of the given functions, we will use various differentiation rules, including the chain rule, product rule, quotient rule, and trigonometric differentiation formulas. Let's go through a few examples step by step.

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Example 1: Problem 43

Function: \( y = \cos(4x) \)

Solution:
We use the chain rule here. The chain rule states:
\[
\frac{d}{dx}[\cos(u)] = -\sin(u) \cdot \frac{du}{dx}
\]
Here, \( u = 4x \), so \( \frac{du}{dx} = 4 \).

Thus,
\[
\frac{dy}{dx} = -\sin(4x) \cdot 4 = -4\sin(4x)
\]

Answer:
\[
\boxed{-4\sin(4x)}
\]

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Example 2: Problem 45

Function: \( g(x) = 5\tan(3x) \)

Solution:
We use the chain rule. The derivative of \( \tan(u) \) is \( \sec^2(u) \cdot \frac{du}{dx} \). Here, \( u = 3x \), so \( \frac{du}{dx} = 3 \).

Thus,
\[
g'(x) = 5 \cdot \sec^2(3x) \cdot 3 = 15\sec^2(3x)
\]

Answer:
\[
\boxed{15\sec^2(3x)}
\]

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Example 3: Problem 47

Function: \( y = \sin(\pi x)^2 \)

Solution:
This can be interpreted as \( y = [\sin(\pi x)]^2 \). We use the chain rule twice. Let \( u = \sin(\pi x) \), so \( y = u^2 \).

First, differentiate \( y = u^2 \):
\[
\frac{dy}{du} = 2u
\]

Next, differentiate \( u = \sin(\pi x) \):
\[
\frac{du}{dx} = \cos(\pi x) \cdot \pi = \pi \cos(\pi x)
\]

Using the chain rule:
\[
\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = 2u \cdot \pi \cos(\pi x)
\]

Substitute back \( u = \sin(\pi x) \):
\[
\frac{dy}{dx} = 2\sin(\pi x) \cdot \pi \cos(\pi x) = 2\pi \sin(\pi x) \cos(\pi x)
\]

Using the double-angle identity \( \sin(2\theta) = 2\sin(\theta)\cos(\theta) \):
\[
\frac{dy}{dx} = \pi \sin(2\pi x)
\]

Answer:
\[
\boxed{\pi \sin(2\pi x)}
\]

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Example 4: Problem 51

Function: \( f(x) = \frac{\cot x}{\sin x} \)

Solution:
We use the quotient rule, which states:
\[
\left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2}
\]
Here, \( u = \cot x \) and \( v = \sin x \).

First, find \( u' \) and \( v' \):
\[
u' = \frac{d}{dx}[\cot x] = -\csc^2 x
\]
\[
v' = \frac{d}{dx}[\sin x] = \cos x
\]

Now apply the quotient rule:
\[
f'(x) = \frac{(-\csc^2 x)(\sin x) - (\cot x)(\cos x)}{\sin^2 x}
\]

Simplify the numerator:
\[
-\csc^2 x \cdot \sin x = -\frac{1}{\sin^2 x} \cdot \sin x = -\frac{1}{\sin x} = -\csc x
\]
\[
\cot x \cdot \cos x = \frac{\cos x}{\sin x} \cdot \cos x = \frac{\cos^2 x}{\sin x}
\]

So the numerator becomes:
\[
-\csc x - \frac{\cos^2 x}{\sin x} = -\frac{1}{\sin x} - \frac{\cos^2 x}{\sin x} = -\frac{1 + \cos^2 x}{\sin x}
\]

Thus,
\[
f'(x) = \frac{-\frac{1 + \cos^2 x}{\sin x}}{\sin^2 x} = -\frac{1 + \cos^2 x}{\sin^3 x}
\]

Using the Pythagorean identity \( 1 + \cos^2 x = 2 - \sin^2 x \):
\[
f'(x) = -\frac{2 - \sin^2 x}{\sin^3 x}
\]

Answer:
\[
\boxed{-\frac{2 - \sin^2 x}{\sin^3 x}}
\]

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Example 5: Problem 60

Function: \( y = 3x - 5\cos(\pi x)^2 \)

Solution:
Differentiate term by term. The first term \( 3x \) is straightforward:
\[
\frac{d}{dx}[3x] = 3
\]

For the second term \( -5\cos(\pi x)^2 \), let \( u = \cos(\pi x) \), so \( y = -5u^2 \).

First, differentiate \( y = -5u^2 \):
\[
\frac{dy}{du} = -10u
\]

Next, differentiate \( u = \cos(\pi x) \):
\[
\frac{du}{dx} = -\sin(\pi x) \cdot \pi = -\pi \sin(\pi x)
\]

Using the chain rule:
\[
\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = -10u \cdot (-\pi \sin(\pi x)) = 10\cos(\pi x) \cdot \pi \sin(\pi x)
\]

Using the double-angle identity \( \sin(2\theta) = 2\sin(\theta)\cos(\theta) \):
\[
10\cos(\pi x) \cdot \pi \sin(\pi x) = 5\pi \sin(2\pi x)
\]

Thus, the derivative of the entire function is:
\[
\frac{dy}{dx} = 3 + 5\pi \sin(2\pi x)
\]

Answer:
\[
\boxed{3 + 5\pi \sin(2\pi x)}
\]

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Summary

The solutions to the problems involve applying the chain rule, product rule, quotient rule, and trigonometric differentiation formulas. Each problem requires careful identification of the appropriate rule and step-by-step application.

If you need solutions for other specific problems, please let me know!