Let's solve the differentiation problems given in the image step by step.

Problem 11:

$y = (1 + x^3)(3 - 2x)^{5/2}$

To differentiate this, we will use the product rule and chain rule. The product rule states:

$\frac{d}{dx}[u \cdot v] = u'v + uv'$

Here, let:

• $u = 1 + x^3$ and
• $v = (3 - 2x)^{5/2}$.

Step 1: Differentiate $u = 1 + x^3$

$u' = \frac{d}{dx}(1 + x^3) = 3x^2$

Step 2: Differentiate $v = (3 - 2x)^{5/2}$ using the chain rule

First, differentiate the outer function $(3 - 2x)^{5/2}$: $v' = \frac{5}{2}(3 - 2x)^{3/2} \cdot \frac{d}{dx}(3 - 2x)$ Now, differentiate the inner function $3 - 2x$: $\frac{d}{dx}(3 - 2x) = -2$ So, $v' = \frac{5}{2}(3 - 2x)^{3/2} \cdot (-2) = -5(3 - 2x)^{3/2}$

Step 3: Apply the product rule

Now, apply the product rule to combine the derivatives: $\frac{dy}{dx} = u'v + uv' = 3x^2(3 - 2x)^{5/2} + (1 + x^3)(-5(3 - 2x)^{3/2})$ Simplified, the derivative is: $\frac{dy}{dx} = 3x^2(3 - 2x)^{5/2} - 5(1 + x^3)(3 - 2x)^{3/2}$

Problem 12:

$y = \sqrt[3]{x^2 - 4x^{-3}}$

We rewrite the expression as: $y = (x^2 - 4x^{-3})^{1/3}$

We use the chain rule again for differentiation.

Step 1: Differentiate the outer function

First, differentiate the outer function $(x^2 - 4x^{-3})^{1/3}$: $\frac{dy}{dx} = \frac{1}{3}(x^2 - 4x^{-3})^{-2/3} \cdot \frac{d}{dx}(x^2 - 4x^{-3})$

Step 2: Differentiate the inner function

Now, differentiate the inner function $x^2 - 4x^{-3}$: $\frac{d}{dx}(x^2) = 2x$ $\frac{d}{dx}(-4x^{-3}) = 12x^{-4}$ So, the derivative of the inner function is: $\frac{d}{dx}(x^2 - 4x^{-3}) = 2x + 12x^{-4}$

Step 3: Combine the results

Now, plug the inner derivative into the outer function derivative: $\frac{dy}{dx} = \frac{1}{3}(x^2 - 4x^{-3})^{-2/3}(2x + 12x^{-4})$

This is the derivative for problem 12.

Would you like me to explain any part of the solution further?

Here are 5 related questions for further exploration:

1. How do we use the product rule for more than two functions?
2. What is the chain rule, and when is it applied?
3. How can we simplify the result of these derivatives for specific values of $x$?
4. How do we handle negative exponents in differentiation?
5. What is the physical meaning of taking a derivative in such problems?

Tip: Always simplify the expression before applying the product or chain rules—it can make the differentiation process easier and clearer.