On an interval where $\tan x$ is defined, use the substitution rule $t=\tan x$. Then $dx=dt/(1+t^2)$ and ${\cos }^{2}x=1/(1+t^2)$, givingThe inverse tangent expression is valid on each interval $(j\pi -\pi /2,j\pi +\pi /2)$ separately. The original integrand is continuous for all real $x$, since its denominator is at least $1$. To obtain one continuous antiderivative on the whole real axis, add $j\pi /2$ on the $j$th such interval and use the continuous limits at their endpoints.

Take $t=e^x>0$ in the substitution rule, so $dx=dt/t$. The transformed integral isThe two terms follow from the derivatives of the natural logarithm and the inverse tangent. Returning to $x$ gives the antiderivative

Use the substitution rule $t=\tan x$, with $dt=(1+{\tan }^{2}x)dx$. The endpoints become $1$ and $\sqrt{3}$, so

The original PDF places the $-1$ outside the square root. Use the substitution rule with $t=\sqrt{x}$, so $dx=2tdt$ and the endpoints become $2$ and $3$. Polynomial division givesHence the definite integral is

For the natural logarithm, $f(1)=0$, $f^{\prime }(1)=1$, $f^{\prime \prime }(1)=-1$, and $f^{(3)}(x)=2/x^3$. The quadratic Taylor polynomial therefore givesThe Taylor remainder is $R_3=2(0.1{)}^3/(6{\xi }^3)$ with $1<\xi <1.1$. It is positive, so the approximation is an underestimate. Moreover,Thus $1/3000\approx 0.0003334$ is a valid upper bound on the absolute error.

Taylor's theorem gives a finite Taylor polynomial of degree $n$, consisting of the first $n+1$ terms, and a remainder:In the Taylor theorem with Lagrange remainder form, assuming the required derivatives exist on an interval containing $a$ and $x$,for some $\xi$ strictly between $a$ and $x$ when $x\ne a$; at $x=a$ the remainder is zero. In particular, if $|f^{(n+1)}|\le M$ on that segment, then $|R_{n+1}(x)|\le M|x-a{|}^{n+1}/(n+1)!$. This is a finite expansion; it does not assume equality with an infinite Taylor series.

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 $n^2$.

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 Pythagorean theorem gives $c^2=a^2+b^2$. Taking a total differential gives $2cdc=2ada+2bdb$, soThe area of the right triangle is $A=ab/2$, so the product rule gives $dA=(bda+adb)/2$ andThe prescribed small changes therefore give $dA/A\approx 0.01-0.02=-0.01$: the area decreases by approximately $1\%$. For comparison, retaining the product of the two finite percentage changes gives $(1.01)(0.98)-1=-0.0102$, an exact decrease of $1.02\%$; the total differential retains only the first-order terms.