Derivatives and its Applications

Expand $f(x)=\sinh x$ in powers of $x$ up to three non-zero terms in the expansion.

Answer

Maclaurin’s theorem expands $\sinh x$ about $x=0$. Its even derivatives vanish there, and its odd derivatives equal $1$, so the first three nonzero terms come from powers $x$, $x^3$, and $x^5$.

$$\sinh x=x+\frac{x^3}{3!}+\frac{x^5}{5!}+\cdots$$

Thus the required expansion stops after the $x^5$ term.

$$\sinh x=x+\frac{x^3}{6}+\frac{x^5}{120}$$

Hence, the first three nonzero terms are $x+\frac{x^3}{6}+\frac{x^5}{120}$.

Evaluate

$$\lim_{x \to 0}\left(\frac{\sin x}{x}\right)^{1/x}$$

using L’Hospitals rule.

Answer

Set $L=(\sin x/x)^{1/x}$ and take logarithms. L’Hospital’s rule evaluates the exponent; the logarithm is valid near $x=0$ because $\sin x/x>0$ there.

$$\begin{aligned} \ln L&=\lim_{x\to0}\frac{\ln(\sin x/x)}{x}\\ &=\lim_{x\to0}\left(\cot x-\frac1x\right) \end{aligned}$$

Writing the last difference as a single quotient gives another $0/0$ form. Applying L’Hospital’s rule twice yields $0$.

$$\cot x-\frac1x=\frac{x\cos x-\sin x}{x\sin x} \longrightarrow 0$$

Therefore $\ln L=0$, and hence $L=1$ from either side.

$$\lim_{x\to0}\left(\frac{\sin x}{x}\right)^{1/x}=1$$

Hence, the limit is $1$.

Find the pedal equation of the polar curve

$$r^{2}=a^{2}\cos 2\theta.$$

Answer

For a polar curve, the perpendicular distance $p$ from the pole to its tangent satisfies $p=r^2/\sqrt{r^2+(dr/d\theta)^2}$. Differentiating the given equation supplies the derivative needed in this formula.

$$2r\frac{dr}{d\theta}=-2a^2\sin 2\theta$$

Using $r^2=a^2\cos 2\theta$ gives the denominator in the distance formula.

$$r^2+\left(\frac{dr}{d\theta}\right)^2 =\frac{a^4}{r^2}$$

Substitution gives the pedal relation, with $p\ge 0$.

$$p=\frac{r^3}{a^2}$$

Hence, the pedal equation is $r^3=a^2p$.

Find the asymptotes of the curve

$$\begin{aligned} &x^{3}+y^{3}-xy^{2}-x^{2}y\\ &\quad +x^{2}-y^{2}-1=0. \end{aligned}$$

Answer

For an oblique asymptote, set $y=mx+c$ and first equate the highest-degree terms to zero. Their factorization gives the possible slopes; the next-degree terms determine the intercepts, including both intercepts for the repeated root.

$$\begin{aligned} &x^3+y^3-xy^2-x^2y\\ &\quad=(x+y)(x-y)^2 \end{aligned}$$

The slope equation is $(1+m)(1-m)^2=0$, so $m=-1$ or $m=1$. For $m=-1$, cancellation gives $c=0$. For the repeated root $m=1$, the next coefficient gives $c(c-1)=0$.

$$y=-x,\qquad y=x,\qquad y=x+1$$

Hence, the three asymptotes are $y=-x$, $y=x$, and $y=x+1$.

Find the radius of curvature of

$$2x^{2}-y^{2}=25$$

at the point where it intersects the line $y=x$ where $x>0,y>0$.

Answer

The positive intersection with $y=x$ is $(5,5)$. Use the Cartesian radius-of-curvature formula $R=(1+(y')^2)^{3/2}/|y''|$ after implicit differentiation of the curve.

$$4x-2yy'=0$$

At $(5,5)$ this gives $y'=2$. Differentiating $y'=2x/y$ gives $y''=2/y-2xy'/y^2=-2/5$ there. Substitution into the formula yields the radius.

$$R=\frac{(1+2^2)^{3/2}}{2/5} =\frac{25\sqrt5}{2}$$

Hence, the radius of curvature is $\frac{25\sqrt5}{2}$ units.

Expand $y=\tan x$ in powers of $x$ up to three non-zero terms.

Answer

The Maclaurin expansion of $\tan x$ follows by dividing the series for $\sin x$ by that for $\cos x$. Keeping the first three non-zero powers gives the required expansion about $x=0$.

$$\tan x=x+\frac{x^3}{3}+\frac{2x^5}{15}+\cdots$$

The next omitted terms have degree at least seven, so they do not affect these three coefficients. The series is valid in a neighborhood of the origin, before the nearest pole of $\tan x$.

Hence, $\tan x=x+\frac{x^3}{3}+\frac{2x^5}{15}+\cdots$.

State L’Hospitals’s rule and hence evaluate the limit

$$\lim_{x\to 0}\frac{\log_{e} x}{\cot x}.$$

Answer

L’Hospital’s rule applies directly to quotients of type $0/0$ and $\infty/\infty$, provided the derivative quotient has a limit and the usual differentiability conditions hold. Here $x\to0^+$ is required because $\log_e x$ is real only for $x>0$; both numerator and denominator are unbounded in magnitude.

$$\lim_{x\to0^+}\frac{\log_e x}{\cot x}=\lim_{x\to0^+}\frac{1/x}{-\csc^2x}$$

Since $\sin x\sim x$ at zero, the derivative quotient tends to zero. Thus the original quotient also tends to zero by L’Hospital’s rule.

$$\lim_{x\to0^+}\frac{\log_e x}{\cot x}=\lim_{x\to0^+}\left(-\frac{\sin^2x}{x}\right)=0$$

Hence, the limit is $0$ (with $x\to0^+$).

Find the asymptotes of the curve

$$\begin{aligned} &x^{3}+2x^{2}y-xy^{2}-2y^{3}\\ &\quad +x^{2}-y^{2}+x+y+1=0. \end{aligned}$$

Answer

Factor the homogeneous cubic part to find the directions of the oblique asymptotes, then retain the next-degree terms to determine the constant shift for each direction. The cubic part factors as $(x-y)(x+y)(x+2y)$.

$$\begin{aligned} &x^3+2x^2y-xy^2-2y^3\\ &\quad=(x-y)(x+y)(x+2y) \end{aligned}$$

The first two directions require no shift. For the third, set $y=-x/2+c$ and make the coefficient of $x^2$ vanish; this gives $c=-1/2$.

$$y=x$$ $$y=-x$$ $$y=-\frac{x}{2}-\frac12$$

Hence, the asymptotes are $y=x$, $y=-x$, and $y=-x/2-1/2$.

Find the radius of curvature at origin of the curve

$$x^{3}+y^{3}=3axy.$$

Answer

The origin is a node because the lowest-degree terms of the implicit equation are $-3axy=0$, which give the coordinate axes as tangents. On either branch, the curve has quadratic contact with its tangent, and the second derivative determines the radius of curvature.

$$y\sim\frac{x^2}{3a}\quad\text{near the }x\text{-axis}$$ $$x\sim\frac{y^2}{3a}\quad\text{near the }y\text{-axis}$$

Each branch therefore has curvature magnitude $2/(3a)$ at the origin when $a>0$. The radius is its reciprocal, and both tangent branches give the same value.

$$\rho=\frac{3a}{2}$$

Hence, the radius of curvature at the origin is $3a/2$ on either branch ($a>0$).

Assuming the validity of the expansion, show that

$$e^{x}=1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\cdots.$$

Answer

Use the Maclaurin expansion of the exponential function, whose $n$th derivative at zero is $1$. Retaining the constant through cubic terms gives the requested expansion; the validity of the series is assumed as stated.

$$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots$$

Each coefficient is $f^{(n)}(0)/n!=1/n!$. The next term is $x^4/4!$, so the displayed terms give the polynomial through degree three.

Hence, $e^x=1+x+x^2/2!+x^3/3!+\cdots$.

Find the pedal equation of the polar curve

$$r=a\theta.$$

Answer

For a polar curve, the pedal formula is $p^2=r^4/(r^2+(dr/d\theta)^2)$. Here $r=a\theta$, so its radial derivative is the constant $a$; substituting it directly gives the relation between $p$ and $r$.

$$\frac{dr}{d\theta}=a$$

The denominator is $r^2+a^2$, and no elimination of $\theta$ is needed. The result is valid on the curve wherever the polar tangent and pedal length are defined.

$$p^2=\frac{r^4}{r^2+a^2}$$

Hence, the pedal equation is $p^2=r^4/(r^2+a^2)$.

Evaluate the limit

$$\lim_{x\to\infty}\frac{e^{px}}{x^{100}};\ p>0$$

and interpret the limit.

Answer

For $p>0$, the exponential grows faster than every fixed power as $x\to\infty$. Repeated application of L’Hospital’s rule also gives the limit: after 100 differentiations, the denominator is constant while the numerator remains a positive multiple of $e^{px}$.

$$\lim_{x\to\infty}\frac{e^{px}}{x^{100}}=\lim_{x\to\infty}\frac{p^{100}e^{px}}{100!}$$

The last expression diverges to positive infinity. The condition $p>0$ is essential; if $p=0$, the quotient instead tends to zero.

Hence, the limit is $+\infty$ for $p>0$.

Find the asymptotes of the curve

$$y^{3}-xy^{2}-x^{2}y+x^{3}+x^{2}-y^{2}=1.$$

Answer

Set $u=y-x$ and $v=y+x$. The cubic equation becomes a product $uv(u-1)$ equal to one, so its three large-coordinate branches approach the three linear factors’ zero sets.

$$(y-x)^2(y+x)+(x^2-y^2)=1$$

Equivalently, $uv(u-1)=1$. As $|v|$ grows, this requires $u\to0$ or $u\to1$; as $|u|$ grows, it requires $v\to0$. These limits give all three asymptotes.

$$y=x$$ $$y=x+1$$ $$y=-x$$

Hence, the asymptotes are $y=x$, $y=x+1$, and $y=-x$.

Find the radius of curvature of the curve

$$y=x^{2}(x-3)$$

at the point where the tangent is parallel to $x$-axis.

Answer

The tangent is horizontal where $y'=0$. Since $y=x^3-3x^2$, there are two such points, at $x=0$ and $x=2$; the radius formula gives the same result at both.

$$y'=3x(x-2)$$ $$y''=6x-6$$

At each point, $|y''|=6$ and $y'=0$. The Cartesian radius of curvature $\rho=(1+y'^2)^{3/2}/|y''|$ is therefore $1/6$.

$$\rho=\frac16\quad\text{at }(0,0)\text{ and }(2,-4)$$

Hence, the radius of curvature is $1/6$ at either horizontal tangent.

Expand $y=\cosh x$ in powers of $x$ up to three non-zero terms in the expansion.

Answer

Use Maclaurin’s theorem for $\cosh x$. Its odd derivatives vanish at zero and its even derivatives equal $1$ there, so only even powers occur in the expansion.

$$\cosh x=1+\frac{x^2}{2!}+\frac{x^4}{4!}+\frac{x^6}{6!}+\cdots$$

The first three non-zero terms are the constant and the terms in $x^2$ and $x^4$. The series converges for every real $x$.

Hence, through three non-zero terms, $\cosh x=1+x^2/2!+x^4/4!+\cdots$.

Find the pedal equation of the polar curve

$$r=a(1+\cos\theta).$$

Answer

For a polar curve, $p^2=r^4/(r^2+(dr/d\theta)^2)$. Here $dr/d\theta=-a\sin\theta$; the denominator simplifies using $r=a(1+\cos\theta)$ and $\sin^2\theta+\cos^2\theta=1$.

$$r^2+\left(\frac{dr}{d\theta}\right)^2=2ar$$

Substituting this expression in the pedal formula eliminates $\theta$ and gives the pedal equation. The relation applies to the cardioid $r=a(1+\cos\theta)$.

$$p^2=\frac{r^4}{2ar}=\frac{r^3}{2a}$$

Hence, the pedal equation is $p^2=r^3/(2a)$.

Evaluate

$$\lim_{x\to 0}x^{x}.$$

Answer

Take logarithms to resolve the indeterminate power. If $L=\lim_{x\to0^+}x^x$, then $\log_e L=\lim x\log_e x$; write the latter as a quotient and apply L’Hospital’s rule.

$$\log_e L=\lim_{x\to0^+}\frac{\log_e x}{1/x}$$

The quotient is of type $-\infty/\infty$. Differentiation gives $-x$, which tends to zero, and exponentiation returns the original limit.

$$L=e^0=1$$

Hence, $\lim_{x\to0^+}x^x=1$.

Find the radius of the curvature of the curve

$$\sqrt{x}+\sqrt{y}=\sqrt{a}$$

at the point where it cuts the line $y=x$.

Answer

The curve meets $y=x$ where $2\sqrt{x}=\sqrt a$, so the point is $(a/4,a/4)$ for $a>0$. Implicit differentiation gives slope $-1$ there; differentiating again gives the second derivative.

$$\left.\frac{dy}{dx}\right|_{(a/4,a/4)}=-1$$

At that point, $y''=4/a$. Using $\rho=(1+y'^2)^{3/2}/|y''|$ gives the radius of curvature.

$$\rho=\frac{(1+1)^{3/2}}{4/a}=\frac{a}{\sqrt2}$$

Hence, the radius of curvature is $a/\sqrt2$ for $a>0$.

List the indeterminate forms for which L’Hospitals rule can be applied. Evaluate

$$\lim_{x\to 0}x^{\tan x}$$

using L’Hospitals rule.

Answer

L’Hospital’s rule applies directly to quotient forms $0/0$ and $\infty/\infty$. The other common indeterminate forms are $0\cdot\infty$, $\infty-\infty$, $0^0$, $1^\infty$, and $\infty^0$; rewrite them as quotients first. Here $x\to0^+$ is required for a real power with base $x$. Take logarithms, then apply the rule to $\log_e x/\cot x$.

$$\log_e L=\lim_{x\to0^+}\frac{\log_e x}{\cot x}$$

Both numerator and denominator diverge in magnitude, and differentiating gives $-\sin^2x/x$, which tends to zero. Exponentiation returns the limit of the original expression.

$$L=e^0=1$$

Hence, $\lim_{x\to0^+}x^{\tan x}=1$.

Find the radius of curvature of

$$xy=c^{2}$$

at $(c,c)$.

Answer

Write the curve as $y=c^2/x$. At $(c,c)$, with $c\ne0$, its slope is $-1$ and its second derivative is $2/c$. The Cartesian radius formula uses the absolute value of the second derivative.

$$y'=-\frac{c^2}{x^2}$$ $$y''=\frac{2c^2}{x^3}$$

At $x=c$, $|y''|=2/|c|$. Substitution into $\rho=(1+y'^2)^{3/2}/|y''|$ gives a positive radius for either sign of $c$.

$$\rho=\sqrt2\,|c|$$

Hence, the radius of curvature at $(c,c)$ is $\sqrt2|c|$ ($c\ne0$).