i. $y = 4\sqrt{x}$

We rewrite the square root in exponent form: $y = 4x^{\frac{1}{2}}$ To differentiate: $\frac{dy}{dx} = 4 \cdot \frac{1}{2} x^{-\frac{1}{2}} = 2x^{-\frac{1}{2}} = \frac{2}{\sqrt{x}}$

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ii. $y = (5x - 9x^3)(8 + x^2)$

Here, we apply the product rule:
If $y = u(x)v(x)$, then $\frac{dy}{dx} = u'v + uv'$.

Let:

• $u(x) = 5x - 9x^3$
• $v(x) = 8 + x^2$

First, find the derivatives of $u(x)$ and $v(x)$: $u'(x) = 5 - 27x^2$ $v'(x) = 2x$

Now apply the product rule: $\frac{dy}{dx} = (5 - 27x^2)(8 + x^2) + (5x - 9x^3)(2x)$ Expanding both terms: $= (40 - 27x^2 \cdot 8 + 5x^2 - 27x^4) + (10x - 18x^4)$ $= 40 - 216x^2 + 5x^2 - 27x^4 + 10x - 18x^4$ Combine like terms: $= 40 + 10x - 211x^2 - 45x^4$

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iii. $y = \frac{7x+4}{5 + x^2}$

This is a quotient rule problem. If $y = \frac{u(x)}{v(x)}$, then $\frac{dy}{dx} = \frac{v(x)u'(x) - u(x)v'(x)}{(v(x))^2}$

Let:

• $u(x) = 7x + 4$
• $v(x) = 5 + x^2$

First, find the derivatives of $u(x)$ and $v(x)$: $u'(x) = 7$ $v'(x) = 2x$

Now apply the quotient rule: $\frac{dy}{dx} = \frac{(5 + x^2)(7) - (7x + 4)(2x)}{(5 + x^2)^2}$ Simplify the numerator:

= \frac{35 + 7x^2 - 14x^2 - 8x}{(5 + x^2)^2} = \frac{35 - 7x^2 - 8x}{(5 + x^2)^2}$$ --- ### iv. $$\int \sqrt{x^3} \, dx$$ First, rewrite $$\sqrt{x^3}$$ as $$x^{\frac{3}{2}}$$. Now apply the power rule for integration: $$\int x^{\frac{3}{2}} \, dx = \frac{x^{\frac{5}{2}}}{\frac{5}{2}} + C = \frac{2}{5}x^{\frac{5}{2}} + C$$ --- Would you like further details or clarifications? Here are five related questions to explore: 1. How does the product rule apply when differentiating a product of functions? 2. What is the quotient rule and how does it differ from the product rule? 3. How do you simplify expressions when applying the chain rule? 4. What are common applications of integration involving power functions? 5. How does differentiating a square root function compare to differentiating a polynomial? **Tip**: Always rewrite square roots and radicals in exponent form for easier differentiation and integration.