Let’s solve these derivatives one by one.

(a) $f(x) = (4x^3 - 2x^2)^3$

To differentiate this, we’ll use the chain rule. Let $u = 4x^3 - 2x^2$, so $f(x) = u^3$.

1. Differentiate $f(x) = u^3$ with respect to $u$: $\frac{df}{du} = 3u^2$

2. Differentiate $u = 4x^3 - 2x^2$ with respect to $x$: $\frac{du}{dx} = 12x^2 - 4x$

3. Now apply the chain rule: $f'(x) = \frac{df}{du} \cdot \frac{du}{dx} = 3(4x^3 - 2x^2)^2 \cdot (12x^2 - 4x)$ Simplifying this gives: $f'(x) = 3(4x^3 - 2x^2)^2 \cdot (12x^2 - 4x)$

(b) $g(x) = \cos(4x + 5)$

To differentiate this, we’ll use the chain rule.

1. Differentiate the outer function $\cos(u)$ where $u = 4x + 5$: $\frac{d}{dx} \cos(u) = -\sin(u)$

2. Differentiate $u = 4x + 5$ with respect to $x$: $\frac{du}{dx} = 4$

3. Now apply the chain rule: $g'(x) = -\sin(4x + 5) \cdot 4 = -4\sin(4x + 5)$

(c) $h(x) = \tan^2(x)$

For this, we’ll use both the chain rule and the derivative of $\tan(x)$.

1. Let $u = \tan(x)$, so $h(x) = u^2$.

2. Differentiate $u^2$ with respect to $u$: $\frac{dh}{du} = 2u$

3. Differentiate $u = \tan(x)$ with respect to $x$: $\frac{du}{dx} = \sec^2(x)$

4. Apply the chain rule: $h'(x) = 2\tan(x) \cdot \sec^2(x)$ So, $h'(x) = 2\tan(x) \sec^2(x)$

(d) $k(x) = e^{4x^2 + 5x}$

For this, we’ll use the chain rule with the exponential function.

1. Differentiate the outer function $e^u$ where $u = 4x^2 + 5x$: $\frac{d}{dx} e^u = e^u$

2. Differentiate $u = 4x^2 + 5x$ with respect to $x$: $\frac{du}{dx} = 8x + 5$

3. Apply the chain rule: $k'(x) = e^{4x^2 + 5x} \cdot (8x + 5)$ So, $k'(x) = (8x + 5)e^{4x^2 + 5x}$

If you have any questions or would like further details, please let me know!

Here are some related questions to expand your understanding:

1. How does the chain rule apply to nested functions with multiple layers?
2. What are some applications of the derivative of exponential functions in real-world scenarios?
3. How can trigonometric derivatives be used in physics, especially in wave and oscillation problems?
4. What is the significance of the derivative of the tangent function in calculus?
5. How can we differentiate more complex composite functions, such as trigonometric functions raised to powers?

Tip: When applying the chain rule, always identify the "inner" and "outer" functions first; this makes differentiation clearer and more structured.