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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.

With $\beta =0$, integrating ${\Phi }_x=2\alpha x+{\alpha }^{2}y+\alpha z$ givesThe other two partial derivatives require $K_y=K_z=0$. The origin condition fixes the remaining constant to zero. Hence the scalar potential isIts value at $(1,1,1)$ gives ${\alpha }^2+2\alpha =1$, so

A stationary point has zero gradient. The two partial derivatives giveSolving this system of linear equations gives

Put $s=x^2+y^2$. The gradient components areA stationary point requires both components to vanish. If $x\ne 0$, then $s=1$; the second component subsequently forces $y=0$, giving $(\pm 1,0)$. If $x=0$, either $y=0$ or, writing $u=y^2$,Only $u=1/2$ is nonnegative. Thus there are five stationary points, with valuesThe cases $x=0$ and $s=1$ exhaust the first partial derivative equation, so no other stationary points have been lost by division.

In the Cartesian coordinate system, the Laplacian of a scalar field is the sum of its three unmixed second partial derivatives. Each squared coordinate here contributes one, soEquivalently $\nabla \psi =(x,y,z)$ and its divergence is three.

Apply the Leibniz integral rule: differentiate both moving endpoints and the integrand's explicit $x$ dependence. The partial derivative is ${\partial }_x\sin (x{t}^2)=t^2\cos (x{t}^2)$. HenceThe signs distinguish upper and lower endpoint contributions. The formula also holds when the limits occur in reversed order, by the oriented-integral convention. If $\alpha =\beta$, the integral defining $f$ is identically zero and all displayed terms cancel, so

Set $D=1+x^2+y^2>0$. The quotient rule gives the partial derivativesSince $1+y+y^2=(y+1/2{)}^2+3/4>0$, the first equation $f_x=0$ forces $x=0$. The second then reduces to $1-y^2=0$. Thus the only stationary points and their field values areNo division by a potentially vanishing factor of $y$ was used, so no stationary points have been discarded.

The second partial derivatives are $f_{xx}=4$, $f_{xy}=f_{yx}=3+2y$ and $f_{yy}=2+2x+30y$. Evaluating at the origin gives the Hessian matrix

For $a=1$, integrate $Q=2xy\sin x$ with respect to $y$ to obtainIts other partial derivative is $f_x=y^2\sin x+x{y}^2\cos x+C^{\prime }(x)$. Comparing with $P$ gives $C^{\prime }(x)=0$. Thus the most general potential on the plane is

For $f(u,v)=\ln u+\ln v$, the needed partial derivatives areSubstituting these into part (ii) givesSince $v-x^2{y}^2=1$, the last two terms combine. Thus

Apply the product rule and the chain rule once more to each expression in part (i). The second derivative chain rule givesandFor example, the coefficient of $f_{uv}$ in the mixed partial derivative is $u_x{v}_y+v_x{u}_y=4x^{3}y+4x{y}^3$, while the final term is $v_{xy}{f}_v=4xy{f}_v$. The symmetry $f_{uv}=f_{vu}$ is used in combining the mixed terms under the usual twice-continuously-differentiable interpretation.

The chain rule gives $f_x=f_u{u}_x+f_v{v}_x$ and $f_y=f_u{u}_y+f_v{v}_y$. Since $u_x=2x$, $u_y=2y$, $v_x=2x{y}^2$ and $v_y=2x^{2}y$, the answer isEvery partial derivative of $f$ on the right is evaluated at $u=1+x^2+y^2$, $v=1+x^2{y}^2$.

First express the function in the original variables:The chain rule givesTaking its partial derivative with respect to $y$ yieldsThe numerator in the second fraction reduces to $4xy$, reproducingBoth logarithms are defined for every real $x,y$ because their arguments are at least one.

First $f_x=2x{y}^3$ and $f_y=3x^2{y}^2$. Differentiating again gives the four second-order partial derivatives:The two mixed partial derivatives agree, as expected for this smooth polynomial.

For an exact differential form, equality of the mixed partial derivatives requires $P_y=Q_x$. HereEquality for all $x>0$ and real $y$ forces $1/a=1/2$, so $\boxed{a=2}$. To construct the potential, integrate $P$ with respect to $x$:Differentiation with respect to $y$ gives $v_y=Q+C^{\prime }(y)$, so $C^{\prime }(y)=0$. Thus the most general potential on the connected domain $x>0$ isIts explicit partial derivatives verify sufficiency as well as necessity.

Let $t=y/x$. The product rule and the chain rule give the partial derivativesHence the terms containing ${\varphi }^{\prime }$ cancel:This also follows from the Euler theorem for homogeneous functions, because $u(\lambda x,\lambda y)=\lambda u(x,y)$ for $\lambda >0$.

The chain rule expresses the coordinate differential operators asThe coefficients are constant, so a second application givesandAdding cancels the mixed partial derivatives, and $a^2+b^2=1$ givesThe transformation has rotation matrix $\left(\begin{array}{cc}a& -b\\ b& a\end{array}\right)$, so this identity expresses the rotational invariance of the Laplacian.

For $u=XY$, the separated equations areOn a region containing the prescribed points, take $0<x<\pi$ and $0<y<\pi /2$. Integration givesThe two specified values, verified in the original PDF, giveDividing these relations eliminates $C$:The right-hand side is the square of the positive base on the left, which differs from one. Thus $\lambda =2$, and either original condition gives $C=1$. ThereforeIts partial derivatives satisfy $u_x/\sin x=2u$ and $\tan y{u}_y=2u$, verifying the equation wherever its coefficients are defined. Substitution at the two assigned points gives $1/(4e^{\sqrt{3}})$ and $3/(4e)$ exactly. The formula extends to any region where $\sin x\ne 0$ and $\cos y\ne 0$.

Let $P=\mu (x)(xy-16x-x^3)$ and $Q=\mu (x)(16+x^2)$. An exact differential form must satisfy $P_y=Q_x$. These partial derivatives giveTaking antiderivatives produces $\ln |\mu |=-\frac{1}{2}\ln (16+x^2)+C$. A convenient nonzero integrating factor for a differential one-form is thereforeAn explicit potential, which also proves global exactness, isIndeed $F_x=P$ and $F_y=Q$ for the chosen integrating factor.

The scaling relation says that $z$ is a homogeneous function of degree $n$. Fix $(x,y)$ and differentiate the identity with respect to the positive parameter $\lambda$. The chain rule givesSetting $\lambda =1$ proves the Euler theorem for homogeneous functions identityDifferentiate once more with respect to $\lambda$, using equality of the mixed partial derivatives for the assumed twice continuously differentiable function:At $\lambda =1$ this is the second-order Euler identity for homogeneous functions:The repeated parameter differentiation accounts for the factor $n(n-1)$, rather than $n^2$.