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For $f(x)=x{e}^{-2x^2}$, the product rule and chain rule giveThe positive exponential factor makes $f$ increasing for $0<x<1/2$ and decreasing for $1/2<x<1$. Its only interior stationary point is therefore a maximum at $x=1/2$. Checking it together with both endpoints,ConsequentlyChecking the endpoints is necessary for extrema on a closed interval.

Use integration by substitution with $u=x^4$, so $du=4x^{3}dx$. ThenDifferentiating the answer by the chain rule gives $x^3{e}^{-x^4}$, confirming the antiderivative. The real constant of integration $C$ is arbitrary.

The chain rule givesFor this to vanish for every differentiable $f$, its coefficient must be zero:The resulting $f(x-t)$ is a profile transported to the right at unit speed by the transport equation.

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.

Write $t_j=t^{(j)}(0)$ and $s=1+t_0>0$. The chain rule givesThe condition $t_3{t}_1<0$ ensures $t_1\ne 0$. Since $t_0\ne 0$, also $s\ne 1$ and $\ln s\ne 0$. Hence the first two nonzero terms are the constant and linear terms:The inequality $s>0$, together with continuity of $t$, makes the logarithm defined on a neighbourhood of zero. The remainder is written in Big O notation.

The curl of $\mathbf{G}=\mathbf{a}\times \mathbf{r}$ is $2\mathbf{a}$. Thus for $\mathbf{a}\ne 0$ it cannot be a gradient field, while for $\mathbf{a}=0$ it is zero and all its vector-field potentials are constants. This zero-vector case is allowed by the question.

For the radial vector field $\mathbf{H}$, write $\rho =|\mathbf{r}|$. Choose a radial primitiveFor $\rho >0$, the chain rule givesHence all vector-field potentials on a connected domain where the field is defined areThe assumed existence of the needed derivatives and integrals justifies this construction. If the domain includes the origin and the field is regular there, choose ${\rho }_0=0$ and extend the primitive there. Any two vector-field potentials differ by a constant on each connected component.

The chain rule gives $u_{xx}=f^{\prime \prime }(x+\lambda y)$ and $u_{yy}={\lambda }^2{f}^{\prime \prime }(x+\lambda y)$. The partial differential equation becomesFor this to hold for every twice differentiable $f$, the coefficient must vanish:For an affine $f$ alone any $\lambda$ would work, but the requirement is for arbitrary $f$.

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.

The original PDF specifies $x\to a$. The quotient is the difference quotient for the function $f(x)=x^x=e^{x\ln x}$ at the positive point $a$. The chain rule gives $f^{\prime }(x)=x^x(\ln x+1)$, so

The chain rule and the quotient rule giveThe exponential function and denominator are positive, so the stationary points occur exactly at $x=\pm 1$. Their values areThe derivative is negative for $|x|>1$ and positive for $|x|<1$, so $x=-1$ is a local minimum and $x=1$ a local maximum. They are also global extrema, since $-1/2\le x/(1+x^2)\le 1/2$.

Put $q={\phi }^{\prime }(y-\beta z)$. Apply implicit differentiation with respect to $x$ and $y$, using the chain rule:ConsequentlyAt a point where the assumed differentiable function $z(x,y)$ exists, the first equation forces $\alpha -\beta q\ne 0$. Combining the equations therefore givesThe assertion is understood on such differentiable branches of the implicit function; arbitrary parameter choices need not produce one everywhere.

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 $n^2$.

Holding $x$ fixed when taking the antiderivative with respect to $t$ givesUse the product rule and the chain rule to differentiate this explicit expression:This agrees term by term with part (i).

Write $I(x)=\ln g(x)$. For $x>-1$, differentiation under the integral sign givesIndeed, on any compact parameter interval $c\le x\le d$ with $c>-1$, the differentiated integrand is bounded by the integrable function $t^c$. The dominated convergence theorem therefore justifies the operation despite the improper lower endpoint. The apparent singularity at $t=1$ is removable.

Since $g=e^I$, the chain rule gives the first-order linear differential equation

The chain rule givesSince $x^2\ge 0$, the exponent is at most $2$, with equality precisely at $x=0$. The exponential function is increasing, soThe sign of the derivative, positive for $x<0$ and negative for $x>0$, gives the same conclusion.

Put $q=x^{2}yz$. The chain rule and product rule giveSumming these partial derivatives gives the Laplacian

The chain rule gives $d(e^{xy})=e^{xy}(ydx+xdy)$, so the given total differential is exact. Thus $f=e^{xy}+C$; two such functions differ by a constant because their gradient difference vanishes on the connected plane. The assigned value at the origin forces $C=-1$, giving