Find the derivative of y with respect to x for the given functions and evaluate the limit.

$$\frac{d}{dx}(x^2) = 2x$$ $$\frac{d}{dx}(3x) = 3$$ $$\frac{d}{dx}(-11) = 0$$

$$\frac{dy}{dx} = 2x + 3 + 0$$ $$\frac{dy}{dx} = 2x + 3$$

The derivative is $\boxed{\text{2x + 3}}$

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إذا كانت $ص = (س - 3)^4$، فإن $\frac{د ص}{د س}$
To find the derivative $\frac{dy}{dx}$, we use the chain rule.

$$\frac{dy}{dx} = 4(x - 3)^3 \cdot 1$$ $$\frac{dy}{dx} = 4(x - 3)^3$$

The derivative is \boxed{\text{4(x - 3)^3}}

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نها $س \to 4 \frac{س - 4}{س^2 - 16}$
To evaluate the limit, we first try direct substitution, which gives $\frac{0}{0}$, an indeterminate form. We need to simplify the expression.

$$\frac{1}{4 + 4} = \frac{1}{8}$$

The limit is $\boxed{\text{}\frac{1}{8}\text{}}$

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إذا كانت $د(س) = 3س - 11$، فإن $د'(س)$
To find the derivative $f'(x)$, we apply the power rule for differentiation.

$$\frac{d}{dx}(3x) = 3$$ $$\frac{d}{dx}(-11) = 0$$

$$f'(x) = 3 - 0$$ $$f'(x) = 3$$

The derivative is $\boxed{\text{3}}$

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إذا كانت $د(س) = 10$، فإن $د'(س)$
To find the derivative $f'(x)$, we differentiate the constant function.

$$f'(x) = 0$$

The derivative is $\boxed{\text{0}}$

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إذا كانت $د(س) = \frac{16}{س}$، فإن $د'(س)$
To find the derivative $f'(x)$, we first rewrite the function using negative exponents.

$$f'(x) = 16 \cdot (-1)x^{-1-1}$$ $$f'(x) = -16x^{-2}$$

$$f'(x) = -\frac{16}{x^2}$$

The derivative is $\boxed{\text{}-\frac{16}{x^2}\text{}}$

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إذا كانت $د(س) = س^3 + 9س + 1$، فإن $د'(س)$
To find the derivative $f'(x)$, we apply the power rule for differentiation to each term.

$$\frac{d}{dx}(x^3) = 3x^2$$ $$\frac{d}{dx}(9x) = 9$$ $$\frac{d}{dx}(1) = 0$$

$$f'(x) = 3x^2 + 9 + 0$$ $$f'(x) = 3x^2 + 9$$

The derivative is \boxed{\text{3x^2 + 9}}

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إذا كانت $د(س) = (س - 3)(س