y = (t2 sin t) / (cos t). 2) If y = (t2 sin t) / (cos t) show that t2(d2y/dt2) + 4t(dy/dt) − (t2+6)y = 0. 3) If y = tⁿ show that dy/dt = nt^(n-1).

We know that $\frac{\sin t}{\cos t} = \tan t$.
So, we can write $y$ as:

The product rule states that if $y = u \cdot v$, then $\frac{dy}{dt} = u'v + uv'$.
Let $u = t^2$ and $v = \tan t$.
Then, $u' = \frac{d}{dt}(t^2) = 2t$.
And $v' = \frac{d}{dt}(\tan t) = \sec^2 t$.

The derivative is $\boxed{2t \tan t + t^2 \sec^2 t}$.

2) If $y = \frac{t^2 \sin t}{\cos t}$, show that $t^2 \frac{d^2y}{dt^2} + 4t \frac{dy}{dt} - (t^2+6)y = 0$.

We have $y = t^2 \tan t$ and $\frac{dy}{dt} = 2t \tan t + t^2 \sec^2 t$.

We need to differentiate $\frac{dy}{dt}$ with respect to $t$:

Differentiate $2t \tan t$ using the product rule:
$\frac{d}{dt}(2t \tan t) = (2)(\tan t) + (2t)(\sec^2 t) = 2 \tan t + 2t \sec^2 t$.
Differentiate $t^2 \sec^2 t$ using the product rule and chain rule:
$\frac{d}{dt}(t^2 \sec^2 t) = (2t)(\sec^2 t) + (t^2)(2 \sec t \cdot \sec t \tan t) = 2t \sec^2 t + 2t^2 \sec^2 t \tan t$.
Combine these two results for $\frac{d^2y}{dt^2}$:

The left-hand side (LHS) is $t^2 \frac{d^2y}{dt^2} + 4t \frac{dy}{dt} - (t^2+6)y$.
Substitute the expressions:

Expand each term:
$t^2 \frac{d^2y}{dt^2} = 2t^2 \tan t + 4t^3 \sec^2 t + 2t^4 \sec^2 t \tan t$
$4t \frac{dy}{dt} = 8t^2 \tan t + 4t^3 \sec^2 t$
$-(t^2+6)y = -t^4 \tan t - 6t^2 \tan t$

Now, sum these expanded terms:

Group terms by $\tan t$:
$(2t^2 + 8t^2 - 6t^2 - t^4) \tan t = (4t^2 - t^4) \tan t$
Group terms by $\sec^2 t$:
$(4t^3 + 4t^3) \sec^2 t = 8t^3 \sec^2 t$
Group terms by $\sec^2 t \tan t$:
$2t^4 \sec^2 t \tan t$

Combining these, the LHS simplifies to:

This expression does not simplify to 0. It is possible there is a typo in the question.

3) If $y = t^n$ show that $\frac{dy}{dt} = nt^{n-1}$.

The power rule states that if $y = t^n$, then $\frac{dy}{dt} = n t^{n-1}$.
Given $y = t^n$.

This is a direct application of the power rule.
The derivative is $\boxed{nt^{n-1}}$.