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An antiderivative of $\sin x$ is $-\cos x$. The fundamental theorem of calculus gives

The product rule and chain rule give $y^{\prime }=\sin (x^2)+2x^2\cos (x^2)$. At zero the slope is zero and the curve passes through $(0,1)$. Consequently the tangent line is

For independent fair dice, the probability of avoiding one on each die is $5/6$. ThereforeThe complement of an event gives

For this separable differential equation, away from $x=1$, separation gives $\ln |y+1|=\ln |x-1|+C$, or $y+1=K(x-1)$ on a connected interval. The initial condition gives $2=-K$, henceThe maximal interval containing $x=0$ on which this solves the printed differential equation is $\boxed{-\infty <x<1}$. Although the linear formula extends through one, the right-hand side there is $0/0$ and the original equation is undefined. The equilibrium solution $y=-1$, lost if one divides by $y+1$, does not satisfy the initial condition.

Put $h=x-\pi /3$. The values $f(\pi /3)=1/2$, $f^{\prime }(\pi /3)=-\sqrt{3}/2$, and $f^{\prime \prime }(\pi /3)=-1/2$ give the Taylor expansionThe factorial in the quadratic term accounts for its coefficient $-1/4$.