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Write $c=\cos \varphi$ and $s=\sin \varphi$. The needed coefficient derivatives are $c_x=s^2/r$, $s_x=-sc/r$, $c_y=-sc/r$ and $s_y=c^2/r$. Apply the product rule to both terms of $f_x=c{g}_r-(s/r)g_{\varphi }$ and use the chain rule for the derivatives of $g$. Collecting terms givesSimilarly, differentiating $f_y=s{g}_r+(c/r)g_{\varphi }$ givesOn addition, both the mixed partial derivative terms and the first angular derivative terms cancel. Using $s^2+c^2=1$, the Laplacian in polar coordinates is

The characteristic polynomial factors as $(r+1)(r+2)$, so the homogeneous solution is $A{e}^{-x}+B{e}^{-2x}$. The forcing $e^{-x}$ already occurs in that solution, so a constant multiple of $e^{-x}$ cannot be a particular solution. Use $y_p=Cx{e}^{-x}$ instead. Its derivatives are $C(1-x)e^{-x}$ and $C(x-2)e^{-x}$, so substitution gives $C{e}^{-x}$ and fixes $C=1$. ThusThe extra factor of $x$ accounts for the repeated forcing exponent in the method of undetermined coefficients.

On a smooth implicit branch, put $Q=x+z^2$ and require $Q\ne 0$. This is exactly the implicit function theorem condition, since the defining equation has vertical partial derivative $3Q$. Differentiating the cubic relation in $x$ and $y$ givesFor a second derivative, the dependence of $Q$ on $z$ must also be included: $Q_x=1+2z{z}_x$ and $Q_y=2z{z}_y$. ThusThe required cancellation follows immediately:The verification is local to smooth branches with $x+z^2\ne 0$; the implicit relation does not guarantee a differentiable function at its singular points.

The finite form of Taylor's theorem iswhere $\xi$ lies between $a$ and $x$ when $x\ne a$. Sufficient hypotheses are continuous derivatives up to order $n-1$ on the interval joining the endpoints and an $n$th derivative in its interior. The first $n$ terms end at degree $n-1$; the Taylor remainder uses the $n$th derivative. At $x=a$ the remainder is zero. This finite Taylor polynomial identity does not require convergence of an infinite Taylor series.

Write the sine argument as $\pi /2+\delta$, with $\delta =\frac{\pi }{2}(e^x-1)=\frac{\pi }{2}x+\frac{\pi }{4}{x}^2+O(x^3)$. The sine addition formula gives $\sin (\pi /2+\delta )=\cos \delta$. Its Maclaurin series is $1-{\delta }^2/2+O({\delta }^4)$, soThere is no linear term because the outer sine has zero derivative at $\pi /2$.

Apply the Leibniz integral rule, including both moving boundaries and the explicit parameter in the integrand. The upper bound has derivative $-\sin x$, the lower bound has derivative $\cos x$, and the integrand has partial derivative $-t^4{e}^{-x{t}^4}$ with respect to $x$. ThereforeThe formula remains valid when the bounds coincide or occur in reversed order, by the oriented definite integral convention. The integrand and its partial derivative are continuous on every finite integration interval.

At a point of tangency, the derivative of the curve equals the slope one of the straight line. The chain rule gives $2\cos 2x=1$. Since $0\le 2x\le \pi$, the unique solution is $x=\pi /6$. The corresponding height is $\sin (\pi /3)=\sqrt{3}/2$. Substitution in the tangent line gives

The probability density is an even function. Its derivative simplifies toIt increases for $z<0$ and decreases for $z>0$, with maximum height $h(0)=2\arctan (1/2)/\pi$. Thus the most probable value, or mode, is $\boxed{z=0}$. The original sketch shows the symmetric, slightly flattened peak and long tails.

There is a genuine qualification to the request for a mean: the mean does not exist. The integral representation shows that $h(z)\sim 1/(\pi z^2)$ as $|z|\to \infty$, soThe two divergent integrals cannot be subtracted to define an expected value. Symmetry does give the Cauchy principal value ${\lim }_{A\to \infty }{\int }_{-A}^{A}zh(z)dz=0$, but that is not the mean. Equivalently, adding the bounded variable $Y$ cannot make the first absolute moment of a Cauchy distribution finite.

Here $I_0=x$ up to a constant. Substitution of $n=0$ in the integration by parts recurrence givesIndeed, its derivative is $\ln x+1-1=\ln x$.

At height $z>0$, the elliptic paraboloid has a circular cross-section $0\le r\le \sqrt{z}$, of area $\pi z$. The cylindrical coordinates volume element therefore reduces the volume integral toFor the specified $f$, the derivative is $e^{-z}(z^{-1/2}+\frac{1}{2}{z}^{-3/2})$. Thus $z{f}^{\prime }(z)$ is integrable at zero and decays exponentially at infinity. Apply integration by parts first on a truncated interval and then take the endpoint limits:Both boundary terms vanish because $zf(z)=-\sqrt{z}{e}^{-z}$. With $z=u^2$, integration by substitution gives the Gaussian integral:The integrand is singular at the apex but the resulting improper integral is convergent.

Multiplication by the integrating factor and the product rule turn the left side into a single derivative:Using integration by substitution $u=\cos x$ gives $y\sin x=-{\cos }^{3}x/3+C$. Therefore the general solution isThere is one arbitrary constant of integration, as expected for a first-order ordinary differential equation.

The finite form of Taylor's theorem with the first $n$ terms iswhere $\xi$ lies strictly between $a$ and $x$ for $x\ne a$. A sufficient hypothesis is that $f$ has continuous derivatives through order $n-1$ on the interval joining the endpoints and an $n$th derivative in its interior. The displayed polynomial has degree at most $n-1$; its Taylor remainder uses the $n$th derivative, not the $(n+1)$st. At $x=a$ the remainder is zero. This is a finite Taylor polynomial identity and does not require an infinite Taylor series to converge.

For $f(x)=x^{1/5}$, the first two derivatives are $f^{\prime }(x)=x^{-4/5}/5$ and $f^{\prime \prime }(x)=-4x^{-9/5}/25$. Since $32=2^5$,The first three terms of the Taylor expansion, with $x-32=1$, therefore giveFor a check on accuracy, $f^{(3)}(x)=36x^{-14/5}/125$ is positive and decreasing on $[32,33]$. The Taylor remainder is consequently positive and less than $f^{(3)}(32)/6=3/1024000<3\times 10^{-6}$. The approximation is a slight underestimate.