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Holding $y$ fixed, the chain rule giveswhere the local change of variable from $x$ to $u$ is valid.

For the internal energy $U(S,V)$, its exact differential identifies $T=U_S$ and $p=-U_V$. The Maxwell relation from equality of mixed partial derivatives is $T_V{|}_S=-p_S{|}_V$. At fixed volume, the constant-volume heat capacity satisfiesso $T_S{|}_V=T/C_V$, assuming nonzero $C_V$ and a locally invertible relation. Applying the chain rule to $p(T,V)$ gives the isentropic temperature-volume derivativeFor the stated one-mole ideal gas, the equation of state gives $p_T{|}_V=R/V$ and $C_V=3R/2$, so the right side is $-2T/(3V)$. Along an isentropic process, $p\propto V^{-5/3}$ and $T=pV/R\propto V^{-2/3}$, giving the same left side:

The chain rule gives

For $k\ne 0$, the chain rule gives $g_t=-f_{t}g/(2k)$ and $g_{xx}=(-f_{xx}/(2k)+f_x^2/(4k^2))g$. ConsequentlyThusThis is the potential form of the Cole-Hopf transformation, with $y=-2k{g}_x/g$. For $k>0$ it is a forward diffusion equation; at $k=0$ the exponential transformation is undefined, while negative $k$ gives a backward heat equation.

For a parametric curve with $dx/dt\ne 0$, the chain rule givesThe specified point has $\cos t=\sin t=1/\sqrt{2}$, hence $t=\pi /4$ in the given traversal. Therefore

First take $a>0$ and $|x|<a$, and put $u=\arcsin (x/a)$, $q=\sqrt{a^2-x^2}$. Then $u^{\prime }=1/q$. The chain rule givesConsequentlywhich proves the identity on the open domain of differentiability. The squared inverse sine is unchanged when $a$ is replaced by $-a$, so the identity also holds for negative $a$ with $|x|<|a|$. The value $a=0$ is excluded.

Put $u=x+y$. Then $u^{\prime }=1+\cos u=2{\cos }^2(u/2)$. Along the branch through $u(0)=0$, the chain rule yields $[\tan (u/2){]}^{\prime }=1$, hence $\tan (u/2)=x$. Choose the continuous branch through zero to obtainIndeed $y^{\prime }=(1-x^2)/(1+x^2)=\cos (2\arctan x)=\cos (x+y)$, and $y(0)=0$. The right-hand side has a bounded derivative with respect to $y$, so the usual uniqueness theorem for ordinary differential equations rules out another solution through the same point.

Assume $c\ne 0$, so $(\xi ,\eta )=(x-ct,x+ct)$ is an invertible coordinate change. The chain rule givesFor a twice continuously differentiable solution the mixed partial derivatives commute. Squaring the two operators therefore givesThus the one-dimensional wave equation becomes

For the real natural logarithm the domain is $x>0$. The chain rule gives

The product rule and chain rule give

The vector $m$ is constant, so $\nabla (m\cdot r)=m$. Also $\nabla (r^{-3})=-3r/r^5$, by the same chain rule. The product rule therefore yieldsThe three components follow by differentiating $(\alpha x+\beta y+\gamma z)(x^2+y^2+z^2{)}^{-3/2}$.

For $r=\sqrt{x^2+y^2+z^2}>0$, the chain rule gives ${\partial }_x(r^{-1})=-x/r^3$, with analogous formulas for the other components. HenceOn the right the numerator denotes the position vector, while its denominator is the cube of its magnitude.

Put $s=b^a{t}^{a}x$ and write $y^{\prime }$ for differentiation with respect to $s$. Work on a real branch where the powers are defined and differentiable; $b>0$, $t>0$ suffices for arbitrary real $a$, and the parameter choice below has $b>0$. At fixed $x$, $\partial s/\partial t=as/t$. The product rule and chain rule therefore give

Apply the chain rule to the exponential, differentiating its exponent $-x^2$:

The derivative of the hyperbolic sine is the hyperbolic cosine. Hence the chain rule gives

The product rule and chain rule give $y^{\prime }=\sin (x^2)+2x^2\cos (x^2)$. At zero the slope is zero and the curve passes through $(0,1)$. Consequently the tangent line is

Differentiating the hyperbolic cosine with the chain rule gives $g^{\prime }(x)=3{\cosh }^{2}x\sinh x$. For real $x$, $\cosh x>0$, so the only zero of the derivative is $\sinh x=0$, namely $x=0$. ThusSince $\cosh x\ge 1$, this stationary point is the global minimum.

For a real-valued expression, $\ln x>0$ requires $x>1$. Applying the chain rule twice gives

The numerator is the derivative of the denominator. Using the chain rule in reverse gives the antiderivativeThe absolute value makes this real on every interval where the denominator is nonzero. The integration constant may be chosen independently on the intervals separated by $x=(-3\pm \sqrt{17})/2$.

As $x\to 0$ through nonzero values, $1/x^2\to +\infty$ and $e^{-1/x^2}\to 0=f(0)$. Thus $f$ is continuous at zero.

For the derivative at zero use its definition, rather than substituting into a formula obtained away from zero:Indeed, putting $t=1/|h|$ makes the absolute value $t{e}^{-t^2}$, which tends to zero since Gaussian decay dominates every power. Therefore $f$ is continuous and differentiable at zero, with $f^{\prime }(0)=0$. Away from zero, the chain rule gives $f^{\prime }(x)=2x^{-3}{e}^{-1/x^2}$.

Apply the product rule and the chain rule to the exponential function: