Let’s solve the requested problems from the image:

1. Differentiate $f(x) = 3x^2 - 2\cos x$

• The derivative of $3x^2$ is $6x$ (using the power rule).
• The derivative of $-2\cos x$ is $2\sin x$ (derivative of $\cos x$ is $-\sin x$).

So, $f'(x) = 6x + 2\sin x$

8. Differentiate $f(x) = \frac{\cot t}{e^t}$

We use the quotient rule $\left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2}$:

• $u = \cot t$, $u' = -\csc^2 t$
• $v = e^t$, $v' = e^t$

Substitute into the quotient rule: $f'(t) = \frac{(-\csc^2 t)(e^t) - (\cot t)(e^t)}{(e^t)^2}$

Simplify: $f'(t) = \frac{-e^t \csc^2 t - e^t \cot t}{e^{2t}}$

Factorize: $f'(t) = \frac{-e^t (\csc^2 t + \cot t)}{e^{2t}} = \frac{-(\csc^2 t + \cot t)}{e^t}$

13. Differentiate $y = \frac{t \sin t}{1 + t}$

We use the quotient rule:

• $u = t \sin t$, $u' = \sin t + t \cos t$ (product rule)
• $v = 1 + t$, $v' = 1$

Substitute into the quotient rule: $y' = \frac{(u'v - uv')}{v^2}$ $y' = \frac{(\sin t + t \cos t)(1 + t) - (t \sin t)(1)}{(1 + t)^2}$

Simplify: $y' = \frac{\sin t + t \cos t + t \sin t + t^2 \cos t - t \sin t}{(1 + t)^2}$ $y' = \frac{\sin t + t \cos t + t^2 \cos t}{(1 + t)^2}$

Final Answers:

1. $f'(x) = 6x + 2\sin x$
2. $f'(t) = \frac{-(\csc^2 t + \cot t)}{e^t}$
3. $y' = \frac{\sin t + t \cos t + t^2 \cos t}{(1 + t)^2}$

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

Expanding Your Knowledge:

1. What are common differentiation rules for trigonometric functions?
2. How does the quotient rule simplify when $v' = 0$?
3. Why is the product rule used in part of problem 13?
4. What happens if you differentiate $\cot t$ multiple times?
5. How can these techniques apply to integration?

Tip: Always identify if a problem involves product or quotient rules early to avoid errors.