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The chain rule identifies an antiderivative directly:The upper limit in the PDF is $\sqrt{\pi /2}$. By the fundamental theorem of calculus, the required integral is

For a classical solution take $F,G$ twice differentiable, so that the second derivatives exist. Direct use of the chain rule giveswith the two profile arguments as specified. Thus $4y_{tt}=25y_{xx}$.

To prove generality, introduce characteristic coordinates $\xi =2x+5t$, $\eta =2x-5t$, whose Jacobian determinant is $-20\ne 0$. For $y=U(\xi ,\eta )$ the differential operator isThe equation becomes $U_{\xi \eta }=0$. Integrating once in each variable yieldsThis is the countertravelling decomposition of the one-dimensional wave equation, with wave speed $5/2$. Once differentiability alone is not sufficient for a classical second-order equation; the displayed argument uses the required second derivatives.

For $a>0$, write $a^x=e^{x\ln a}$ and apply the chain rule:This also covers $a=1$, for which both sides are zero.

Differentiate the outer sine, then the inner cosine. The chain rule gives

First take $F_y=x{e}^{\sin (xy)}\cos (xy)$, then differentiate with respect to $x$. The product rule and chain rule giveThe function is smooth, so the two mixed partial derivatives agree.

Write the rational expression as $(x^2+4{)}^{-1}$ and apply the chain rule:The denominator is positive for every real $x$, so the derivative is defined throughout the real line.

The outer exponential function differentiates to itself, and the inner sine has derivative cosine. The chain rule gives

For positive $a$, write $a^{-x}=\exp (-x\ln a)$. Differentiating by the chain rule givesThe natural logarithm of the constant base supplies the factor; it must not be omitted when $a\ne e$.

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

Set $\xi =x-ct$. The chain rule gives $u_t=-c{f}^{\prime }(\xi )$ and $u_x=f^{\prime }(\xi )$, so $u_t+c{u}_x=0$ for any differentiable $f$. The initial profile fixes $f(\xi )=\cos \xi$, givingThis is the translated initial profile, as described by the transport equation.

Differentiate the polar form of a complex number using the product rule and chain rule:A second differentiation givesThe two contributions to $2\dot{r}\dot{\theta }$ arise separately from differentiating $r\dot{\theta }$ and from differentiating the moving exponential factor.

Let $f(x)=\cosh \sqrt{x}$ near $x=2$. The chain rule givesSubstitute $x=2$ into the first three terms of the Taylor expansion:The quadratic coefficient is $f^{\prime \prime }(2)/2$, not $f^{\prime \prime }(2)$. The apparent square root creates no local difficulty at the positive expansion point.

On any interval where $\sin x>0$, the chain rule gives $y^{\prime }=\cos x/\sin x=\cot x$. Differentiating again,The domain condition is necessary because the real natural logarithm requires a positive argument.

For this parametric curve, the chain rule givesThe quotient is a finite slope when $\sin t\ne 0$. At $t=0$ and $t=\pi$, the horizontal component of the tangent vanishes while the vertical component does not; these have vertical tangent lines, rather than missing finite values of the displayed quotient.

Taking each partial derivative of the scalar potential givesThe factor $5y^4$ in the middle component comes from the chain rule. The potential is smooth, so the curl of a gradient identity gives $\boxed{\nabla \times \mathbf{F}=0}$.

Assume the needed derivatives exist continuously and $df/dg\ne 0$ on a neighborhood, so the inverse function theorem provides a local inverse. Put $p=df/dg$, $q=d^{2}f/d{g}^2$, and $r=d^{3}f/d{g}^3$. The chain rule gives $d/df=p^{-1}d/dg$, soDifferentiate once more:Thus the first three derivatives of an inverse function areAll forward derivatives on the right are evaluated at the corresponding value $g(f)$.

For an independent inversion, use the coordinate relationsThe positive-octant domain fixes their angle branches unambiguously. Let $s=\sqrt{x^2+y^2}=W\tan \theta$. Differentiating at fixed Cartesian coordinates givesand ${\varphi }_x=-y/s^2=-\sin \varphi /(W\tan \theta )$, ${\varphi }_y=x/s^2=\cos \varphi /(W\tan \theta )$, ${\varphi }_z=0$. Hence the inverse chain rule givesandMultiplying these coefficients by the forward coefficient matrix gives the identity. The original PDF correctly holds $x,y$ fixed in the final derivative; the transcription's $x,z$ subscript there is a typo.

The derivative of $\cot u$ is $-{\csc }^{2}u$. The chain rule therefore givesThe antiderivative applies on any interval containing no zero of $\sin (3x+1)$.

The base $2e$ is constant. Write the exponential function as $\exp [\cos x\ln (2e)]$ and apply the chain rule:

With $v=x+y$, the chain rule givesFor $v\ne 2$, this separable differential equation givesRestoring $v$ and absorbing constants, the nonconstant branches satisfyThe constant branch lost in division is $v=2$, giving the additional solution $\boxed{y=2-x}$. All branches must avoid $v=4/3$, where the original differential equation is undefined.