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The finite geometric series has eleven terms, from $2^1$ through $2^{11}$. With common ratio $2$,Counting the powers explicitly avoids omitting the final $2048$ term.

Pair the first and last terms of the arithmetic series. There are $500$ pairs, each summing to $1001$, so

The repeated factor requires both a simple-pole and a double-pole term in the partial fraction decomposition. WriteMultiplying through gives $1=A(x-1{)}^2+B(x^2-1)+C(x+1)$. Substituting $x=-1$ and $x=1$ gives $A=1/4$ and $C=1/2$; the quadratic coefficient then gives $B=-1/4$. Therefore

Group the polynomial into two pairs, then use the difference of two squares:Thus $1$ is a double polynomial root and $-1$ is a simple polynomial root.

On the horizontal segment $y=0$, while $x$ decreases from $1$ to $-1$. Keeping this orientation givesThe minus sign follows from traversing opposite to the positive $x$ direction.

Only the $x$ component of the vector field is nonzero, so $\mathbf{F}\cdot d\mathbf{r}=(x^2+y^2)dx$. On the vertical segment, $dx=0$ throughout. Hence

The parameter derivative of the integrand is $\cos (a\log x)$, whose absolute value is bounded by one on $0<x<1$. This integrable bound justifies differentiation under the integral sign, givingThe original integral also exists: $|\sin (a\log x)/\log x|\le |a|$, and its apparent singularity at $x=1$ is removable. Set $x=e^{-t}$ to evaluate the derivative:ConsequentlySince $I(0)=0$, integration in the parameter fixes the constant and gives the logarithmic sine integral identity

The integrand and both limits depend on $x$, so use the Leibniz integral rule. The upper and lower endpoint contributions and the interior derivative giveFor $x\ne 0$, an antiderivative in $t$ is $t\sin (xt)/x+\cos (xt)/x^2$. Evaluating it and collecting terms givesAt $x=0$ the original Leibniz expression has zero endpoint terms and an integral over a zero-length interval, giving $\boxed{F^{\prime }(0)=0}$. The displayed quotient has a removable singularity and the same limiting value.

Assume real square-integrable functions on the same integration domain. Put $A=\int f^2$, $B=\int g^2$, and $C=\int fg$. If $B=0$, then $g=0$ almost everywhere, so $C=0$ and the claim follows. If $B>0$, nonnegativity of a square givesMultiplying by $B$ proves the integral form of the Cauchy-Schwarz inequality:For nonzero functions, equality holds exactly when they are proportional almost everywhere. The zero-function cases also give equality.

Substitution first makes the square-root branch transparent:On the given interval this is exactly $\sin x$, so direct differentiation givesin agreement with the chain-rule calculation. Outside this interval one must retain the absolute value; there are cusps at integer multiples of $\pi$.

The second derivative chain rule isPut $s=\sin x>0$ and $c_0=\cos x$. Substituting the preceding results and $y^{\prime }=2ys{c}_0$ givesTherefore

On $0<x<\pi$, $\sqrt{\log y}=\sqrt{{\sin }^{2}x}=\sin x>0$. The chain rule givesUsing $\sin 2x=2\sin x\cos x$ yields

For the composite square root, $y>1$ throughout the restricted $x$ interval. Differentiate with respect to $y$:The second expression comes from differentiating both $y^{-1}$ and $(\log y{)}^{-1/2}$. The endpoints where $y=1$ are excluded here.

The graph is even and periodic with period $\pi$, bounded between $1$ and $e$. It rises from each minimum at $k\pi$ to a maximum halfway to the next minimum, then falls back. It has horizontal tangents at all these extrema and repeats smoothly.

Differentiating givesThe exponential never vanishes, so all stationary points have $x=k\pi /2$. At $x=k\pi$, $y=1$ and $y^{\prime \prime }=2>0$; at $x=\pi /2+k\pi$, $y=e$ and $y^{\prime \prime }=-2e<0$. ThusThese are global minima and maxima because $0\le {\sin }^{2}x\le 1$. None of the stationary points is an inflection point.

For $r=\sqrt{x^2+y^2+z^2}>0$, the multivariable chain rule gives ${\partial }_{x}g(r)=g^{\prime }(r)x/r$, with the corresponding expressions for $y,z$. Thus the gradient of a radial function isThe origin needs a separate check because $\hat{\mathbf{r}}$ is undefined there. If the one-variable function has a finite right derivative at zero, the radial composite is differentiable at the origin exactly when $g^{\prime }(0^{+})=0$, in which case its gradient is zero. A nonzero radial slope gives incompatible opposite-direction derivatives.

Treat $R$ and $a$ as constants and use the total differential in the independent variables $n,V,T$:Each coefficient differentiates with respect to its displayed variable while holding the other two fixed; the formula applies where $V\ne 0$.