Let's analyze and solve the problems from the image step by step:

(e) $f(t) = \frac{t^4}{6 \sin(t)}$

To find $f'(t)$, apply the quotient rule:

$\text{If } f(t) = \frac{u(t)}{v(t)}, \quad f'(t) = \frac{u'(t)v(t) - u(t)v'(t)}{v(t)^2}.$

Here:

• $u(t) = t^4$ and $u'(t) = 4t^3$,
• $v(t) = 6 \sin(t)$ and $v'(t) = 6 \cos(t)$.

Substitute into the formula: $f'(t) = \frac{(4t^3)(6 \sin(t)) - (t^4)(6 \cos(t))}{(6 \sin(t))^2}.$

Simplify the numerator: $f'(t) = \frac{24t^3 \sin(t) - 6t^4 \cos(t)}{36 \sin^2(t)}.$

Factorize: $f'(t) = \frac{6t^3 (4 \sin(t) - t \cos(t))}{36 \sin^2(t)}.$

Simplify further: $f'(t) = \frac{t^3 (4 \sin(t) - t \cos(t))}{6 \sin^2(t)}.$

(f) $f(y) = \frac{xy^6}{x^4 + 1}$

Here, $f(y)$ is a function of $y$, and $x$ is treated as a constant. Again, use the quotient rule:

$f'(y) = \frac{u'(y)v(y) - u(y)v'(y)}{v(y)^2},$

where:

• $u(y) = xy^6$ and $u'(y) = 6xy^5$,
• $v(y) = x^4 + 1$ (note that $v'(y) = 0$ since $x^4 + 1$ is constant with respect to $y$).

Substitute into the formula: $f'(y) = \frac{(6xy^5)(x^4 + 1) - (xy^6)(0)}{(x^4 + 1)^2}.$

Simplify: $f'(y) = \frac{6xy^5 (x^4 + 1)}{(x^4 + 1)^2}.$

Cancel common terms if possible: $f'(y) = \frac{6xy^5}{x^4 + 1}.$

Would you like further explanation or a detailed walkthrough for any of these steps?

Here are five related questions:

1. What are the conditions under which the quotient rule applies?
2. How does the chain rule integrate with the quotient rule for complex functions?
3. What are common errors to avoid when applying the quotient rule?
4. How can we simplify derivatives involving trigonometric functions?
5. How do constants like $x^4 + 1$ affect differentiation?

Tip: Always identify constants and variables clearly when applying differentiation rules to avoid unnecessary errors!