https://file.aiquickdraw.com/mathbot/file/1727242551175gncn59ee.jpg

The image presents several functions to differentiate. Let's go through each one systematically and find the derivatives.

### (a) \( f(x) = x - 3 \sin x \)
The derivative is:
\[
f'(x) = \frac{d}{dx} \left( x \right) - \frac{d}{dx} \left( 3 \sin x \right) = 1 - 3 \cos x
\]

### (b) \( g(t) = t^3 \cos t \)
Using the product rule:
\[
g'(t) = \frac{d}{dt} \left( t^3 \right) \cdot \cos t + t^3 \cdot \frac{d}{dt} \left( \cos t \right)
\]
\[
g'(t) = 3t^2 \cos t - t^3 \sin t
\]

### (c) \( f(x) = x \sin x \)
Using the product rule:
\[
f'(x) = \frac{d}{dx} \left( x \right) \cdot \sin x + x \cdot \frac{d}{dx} \left( \sin x \right)
\]
\[
f'(x) = \sin x + x \cos x
\]

### (d) \( y = x \cos x \sin x \)
This is another product of functions, and we can treat \( \cos x \sin x \) as a single term for differentiation:
\[
y' = \frac{d}{dx} \left( x \right) \cdot \cos x \sin x + x \cdot \frac{d}{dx} \left( \cos x \sin x \right)
\]
Using the product rule for \( \cos x \sin x \):
\[
\frac{d}{dx} \left( \cos x \sin x \right) = -\sin^2 x + \cos^2 x
\]
Thus:
\[
y' = \cos x \sin x + x (-\sin^2 x + \cos^2 x)
\]

### (e) \( y = \frac{1 + \sin x}{x + \cos x} \)
This is a quotient, so we use the quotient rule:
\[
y' = \frac{(x + \cos x) \cdot \frac{d}{dx}(1 + \sin x) - (1 + \sin x) \cdot \frac{d}{dx}(x + \cos x)}{(x + \cos x)^2}
\]
Simplifying each term:
\[
y' = \frac{(x + \cos x) \cos x - (1 + \sin x) (1 - \sin x)}{(x + \cos x)^2}
\]

### (f) \( y = \frac{x}{\cos x} \)
This is another quotient, so we apply the quotient rule:
\[
y' = \frac{\cos x \cdot 1 - x (-\sin x)}{\cos^2 x}
\]
\[
y' = \frac{\cos x + x \sin x}{\cos^2 x}
\]

### (g) \( y = \sin 4x \)
Using the chain rule:
\[
y' = \cos 4x \cdot 4 = 4 \cos 4x
\]

### (h) \( y = \cos(\sin x) \)
Using the chain rule:
\[
y' = -\sin(\sin x) \cdot \cos x
\]

### (i) \( y = \sqrt{\sin x} \)
Rewrite as \( y = (\sin x)^{1/2} \), then use the chain rule:
\[
y' = \frac{1}{2} (\sin x)^{-1/2} \cdot \cos x = \frac{\cos x}{2 \sqrt{\sin x}}
\]

### (j) \( y = \cos \sqrt{x} \)
Using the chain rule:
\[
y' = -\sin \sqrt{x} \cdot \frac{1}{2\sqrt{x}} = \frac{-\sin \sqrt{x}}{2\sqrt{x}}
\]

### (k) \( f(x) = \sin^5(3x) \)
This requires both the chain rule and the power rule:
\[
f'(x) = 5 \sin^4(3x) \cdot \cos(3x) \cdot 3 = 15 \sin^4(3x) \cos(3x)
\]

### (l) \( y = 4 \sin^2 x + 5 \cos^2 x \)
Differentiate each term separately:
\[
y' = 4 \cdot 2 \sin x \cos x + 5 \cdot 2 \cos x (-\sin x) = 8 \sin x \cos x - 10 \sin x \cos x
\]
\[
y' = -2 \sin x \cos x
\]

Would you like any further clarifications on these derivatives?

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### 5 Relative Questions:
1. How does the product rule work in differentiating composite functions?
2. Can you explain the chain rule used in \( y = \sin 4x \) and \( y = \cos \sqrt{x} \)?
3. What is the quotient rule and how is it applied to rational functions like in (e)?
4. How would you differentiate a function with a trigonometric term raised to a power, as in (k)?
5. What is the significance of simplifying the derivative after applying the quotient rule?

### Tip:
When differentiating composite functions, always consider the order of operations to apply the chain rule or product rule accurately.

Differentiate several functions including f(x) = x - 3 sin x, g(t) = t^3 cos t, y = sin 4x, and others.

Explore the differentiation of complex functions, including product and quotient of trigonometric functions, using rules such as the product rule, quotient rule, and chain rule. Functions include f(x) = x - 3 sin x, y = sin 4x, and more, with step-by-step solutions.

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