To find the nth derivatives of the given functions, we apply specific differentiation techniques such as product rule, chain rule, and higher-order derivative formulas.

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i) $y = \frac{1}{\cos(4x)\cos(6x)}$

We rewrite the function: $y = \sec(4x)\sec(6x)$

First derivative:

Using the product rule and the derivative of $\sec(kx)$: $\frac{dy}{dx} = \sec(4x)\tan(4x) \cdot 4 \sec(6x) + \sec(4x) \cdot 6\sec(6x)\tan(6x)$

For the nth derivative: The higher-order derivatives involve products of terms like powers of $\sec(kx)$ and $\tan(kx)$, with recursive patterns due to their nature.

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ii) $y = 2e^{\sin(x)}\cos(3x)$

This function involves the product of $e^{\sin(x)}$ and $\cos(3x)$.

First derivative:

$\frac{dy}{dx} = 2 \left[ e^{\sin(x)} \cos(3x) \cos(x) - 3e^{\sin(x)}\sin(3x) \right]$

nth derivative:

Using Leibniz's rule for nth derivatives of a product and the fact that derivatives of $\cos(3x)$ cycle through $\cos(3x)$, $-\sin(3x)$, $-\cos(3x)$, and $\sin(3x)$, the nth derivative can be expressed as: $\frac{d^n y}{dx^n} = 2 e^{\sin(x)} \cdot P_n(x)$ where $P_n(x)$ involves terms of $\cos(x)$, $\sin(x)$, and their derivatives combined with $\cos(3x)$ or $\sin(3x)$.

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iii) $y = \frac{x}{(x+1)^5}$

Rewrite: $y = x (x+1)^{-5}$

First derivative:

Using the product rule: $\frac{dy}{dx} = (x+1)^{-5} - 5x(x+1)^{-6}$

nth derivative:

The nth derivative will involve terms from the generalized Leibniz rule for products and the nth derivatives of $(x+1)^{-5}$, which follow the binomial pattern: $\frac{d^n y}{dx^n} = C(x+1)^{-5-n} + \text{other terms}$ where $C$ is a coefficient dependent on $n$.

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Would you like a deeper explanation or step-by-step solutions for any of these?

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Questions for Further Exploration:

1. How do higher-order derivatives behave for trigonometric functions like $\sec(4x)\sec(6x)$?
2. What are the patterns in nth derivatives of exponential functions involving trigonometric terms?
3. How can Leibniz's rule simplify computing nth derivatives of products?
4. What is the role of the binomial theorem in nth derivatives of rational functions?
5. How do we handle mixed terms when finding nth derivatives of products?

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Tip:

For nth derivatives, recognizing recursive patterns can simplify computations significantly.