Let's go through the given problems one by one and solve them:

a)

$\frac{d}{dx} \left[ (3x^2 + 6)\left(2x - \frac{1}{4}\right) \right]$

This expression involves the product of two functions of $x$, so we will use the product rule, which states: $\frac{d}{dx}[f(x)g(x)] = f'(x)g(x) + f(x)g'(x)$ Let:

• $f(x) = 3x^2 + 6$
• $g(x) = 2x - \frac{1}{4}$

Now, differentiate each function:

• $f'(x) = 6x$
• $g'(x) = 2$

Apply the product rule: $\frac{d}{dx} \left[ (3x^2 + 6)(2x - \frac{1}{4}) \right] = (6x)(2x - \frac{1}{4}) + (3x^2 + 6)(2)$

Simplifying: $= 6x \left(2x - \frac{1}{4}\right) + 2(3x^2 + 6)$ $= 12x^2 - \frac{6x}{4} + 6x^2 + 12$ $= 18x^2 - \frac{6x}{4} + 12$ $= 18x^2 - \frac{3x}{2} + 12$

b)

$\frac{d}{dx} \left[ 2y^3 t + t^3 y \right] = 1$ This is an implicit differentiation problem. Differentiate both sides with respect to $x$, treating $y$ as a function of $x$, meaning you need to apply the chain rule when differentiating terms involving $y$.

First, differentiate the left side: $\frac{d}{dx} \left( 2y^3 t + t^3 y \right)$ Using the product and chain rules: $= 2 \left( 3y^2 \frac{dy}{dx} \cdot t \right) + 2y^3 \cdot \frac{dt}{dx} + \left( 3t^2 \frac{dt}{dx} \cdot y \right) + t^3 \frac{dy}{dx}$

Now, set the derivative equal to 1 (as given on the right side of the equation): $6y^2 t \frac{dy}{dx} + 2y^3 \frac{dt}{dx} + 3t^2 y \frac{dt}{dx} + t^3 \frac{dy}{dx} = 1$

c)

$\frac{d}{dx} \left( \frac{\sin x \sec x}{1 + x \tan x} \right)$ This is a quotient of two functions, so we will apply the quotient rule: $\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$ Let:

• $u = \sin x \sec x = 1$ (since $\sin x \cdot \sec x = 1$)
• $v = 1 + x \tan x$

Now, differentiate $u$ and $v$:

• $\frac{du}{dx} = 0$
• $\frac{dv}{dx} = \tan x + x \sec^2 x$

Using the quotient rule: $\frac{d}{dx} \left( \frac{\sin x \sec x}{1 + x \tan x} \right) = \frac{(1 + x \tan x)(0) - (1)(\tan x + x \sec^2 x)}{(1 + x \tan x)^2}$ $= \frac{-(\tan x + x \sec^2 x)}{(1 + x \tan x)^2}$

d)

$\frac{d}{dx} \left[ \frac{(x^2 - 8)^{\frac{1}{3}} \sqrt{x^3 + 1}}{x^6 - 7x} \right]$ Here, we have a quotient, so again we'll use the quotient rule, and both the numerator and denominator involve functions that require the chain rule.

Let:

• $u = (x^2 - 8)^{\frac{1}{3}} \sqrt{x^3 + 1}$
• $v = x^6 - 7x$

First, differentiate $u$ using the product rule and chain rule: $u = (x^2 - 8)^{\frac{1}{3}} \cdot (x^3 + 1)^{\frac{1}{2}}$ Applying the product rule: $\frac{du}{dx} = \frac{1}{3}(x^2 - 8)^{-\frac{2}{3}} \cdot 2x \cdot (x^3 + 1)^{\frac{1}{2}} + (x^2 - 8)^{\frac{1}{3}} \cdot \frac{1}{2}(x^3 + 1)^{-\frac{1}{2}} \cdot 3x^2$

For the denominator: $\frac{dv}{dx} = 6x^5 - 7$

Now apply the quotient rule: $\frac{d}{dx} \left[ \frac{u}{v} \right] = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$

Would you like further details on any of these steps? Here are some related questions to deepen your understanding:

1. How do you apply the product rule for differentiating functions?
2. What is the chain rule, and when is it used in differentiation?
3. How do you differentiate implicit equations?
4. How do you simplify the result of a quotient rule application?
5. What is the geometric meaning of the derivative of a quotient?

Tip: Always simplify your derivative before applying rules like the product or quotient rule—it can make the calculation much easier!