Learning goals and preparation

We assume basic derivative rules, Chapter 6's limits and continuity theorems, and elementary integration. This chapter develops differential-calculus methods beyond the first chapter: implicit differentiation, the higher-order Leibniz rule, convexity and derivative inequalities, qualified l'Hôpital limits, and Taylor polynomials/series with error control.

7.1 Implicit relations and their local restrictions

Learning goals and setting

Prove local invertibility rather than only differentiating a formally assumed inverse. Use the Euclidean norm and its induced operator norm throughout. Let $U\subseteq\mathbb R^n$ be open, $F:U\to\mathbb R^n$ be $C^1$, $a\in U$, $b=F(a)$, and $A=DF(a)$ invertible. We will prove that some open neighborhoods $W$ of $a$ and $V$ of $b$ make $F:W\to V$ a $C^1$ bijection with derivative of its inverse equal to $[DF]^{-1}$.

Lemma: contraction on a closed ball

If a map $T$ takes a nonempty closed Euclidean ball $B$ into itself and $\|T(x)-T(y)\|\le q\|x-y\|$ for a fixed $0\le q<1$, then it has a unique fixed point in $B$.

Proof. Starting from any $x_0\in B$, define $x_{k+1}=T(x_k)$. If $q=0$, the inequality makes $T$ constant, and its value is the unique fixed point reached at the first iteration. For $0<q<1$, repeated contraction gives $\|x_{k+1}-x_k\|\le q^k\|x_1-x_0\|$. For $m>k$, summing consecutive differences bounds $\|x_m-x_k\|$ by $q^k\|x_1-x_0\|/(1-q)$, so the sequence is Cauchy. Coordinatewise real completeness gives a limit in $\mathbb R^n$, and closedness keeps it in $B$. Contraction implies continuity, so passing to the limit in $x_{k+1}=T(x_k)$ makes the limit fixed. Two fixed points would have distance at most $q$ times that same distance, forcing equality. This proof also gives the stated geometric error bound to the limit.

Existence and uniqueness of the local inverse

Continuity of $DF$ lets us choose $r>0$ with the closed ball $B=\overline B(a,r)\subset U$ and

$$\|I-A^{-1}DF(y)\|\le q=\frac12\quad\text{for all }y\in B.$$

For a target $x$ near $b$, set $T_x(y)=y-A^{-1}(F(y)-x)$. Its derivative has norm at most $q$ on the convex ball. Applying the one-variable fundamental theorem along a segment gives $\|T_x(y)-T_x(z)\|\le q\|y-z\|$. Also

$$\|T_x(y)-a\|\le q\|y-a\|+\|A^{-1}(x-b)\|.$$

Choose $V=B(b,\rho)$ with $\rho=(1-q)r/(2\|A^{-1}\|)$. For $x\in V$, the second term is less than $(1-q)r/2$, so $T_x$ maps $B$ into itself. The lemma supplies a unique $y=H(x)\in B$ satisfying $F(y)=x$. The fixed-point estimate at $a$ gives

$$\|H(x)-a\|\le\frac{\|A^{-1}(x-b)\|}{1-q}<r/2,$$

so all these roots are interior. Let $W=B(a,r)\cap F^{-1}(V)$. This is open, contains $a$, and $F:W\to V$ is bijective, because uniqueness holds even on the larger closed ball.

Continuity and differentiability of the inverse

For $x,z\in V$, compare the fixed-point equations:

$$\|H(x)-H(z)\|\le q\|H(x)-H(z)\|+\|A^{-1}\|\|x-z\|.$$

Thus $H$ is Lipschitz with constant $\|A^{-1}\|/(1-q)$. For $y\in B$, the matrix $A^{-1}DF(y)=I-E$ has $\|E\|\le q<1$. The convergent matrix series $\sum_{k=0}^{\infty}E^k$ is its inverse: multiplying finite partial sums gives $I-E^{N+1}$, which tends to $I$. Hence $DF(y)$ is invertible throughout $B$.

Fix $x\in V$, let $y=H(x)$, and take a small increment $k$ with $x+k\in V$. Set $h=H(x+k)-H(x)$. Differentiability of $F$ gives $k=DF(y)h+o(\|h\|)$. The Lipschitz estimate gives $\|h\|=O(\|k\|)$, so solving yields

$$h=[DF(y)]^{-1}k+o(\|k\|).$$

Therefore $DH(x)=[DF(H(x))]^{-1}$. The inverse matrices vary continuously: the adjugate-over-determinant formula is continuous where determinant is nonzero. Together with continuity of $DF$ and $H$, this proves $H$ is $C^1$. The theorem is now proved, including existence, uniqueness, neighborhoods and regularity.

Implicit functions as a corollary

Let $G:\mathbb R^{p+m}\supset U\to\mathbb R^m$ be $C^1$, $G(a,c)=0$, and $D_yG(a,c)$ invertible. Define $\Phi(x,y)=(x,G(x,y))$. Its block derivative is

$$D\Phi=\begin{pmatrix}I&0\\D_xG&D_yG\end{pmatrix},$$

which is invertible at $(a,c)$. The proved inverse theorem gives a local $C^1$ inverse. Shrink to product neighborhoods so that solving $\Phi(x,y)=(x,0)$ defines a unique $y=\varphi(x)$ and includes every zero in that product neighborhood. Differentiating the identity after existence is established gives

$$D\varphi(x)=-[D_yG(x,\varphi(x))]^{-1}D_xG(x,\varphi(x)).$$

The factor order is forced by solving the matrix equation and cannot be reversed.

If a differentiable branch $y=y(x)$ exists and satisfies $F(x,y(x))=0$, the chain rule gives $F_x+F_yy'=0$. Where $F_y\ne0$,

$$y'=-F_x/F_y.$$

This is not a global assertion that every curve is one function of $x$.

Worked example 1: a constraint with a vertical obstruction. For $x^2+y^2=25$, differentiation gives $2x+2yy'=0$, so $y'=-x/y$ when $y\ne0$. At $(3,4)$ this is $-3/4$. Differentiating again gives $2+2(y')^2+2yy''=0$, hence $y''=-[1+(y')^2]/y=-25/y^3$ on that branch. At $(5,0)$ the formula cannot divide by zero; the curve has a vertical tangent and cannot be represented by a differentiable $y(x)$ there.

7.2 Higher derivatives and the Leibniz formula

For sufficiently differentiable $f,g$, the $n$th derivative of a product is

$$\frac{d^n}{dx^n}(fg)=\sum_{k=0}^n\binom nkf^{(k)}g^{(n-k)}.$$

For $n=1$ this is the product rule. If true at order $n$, differentiate every term. Coefficients of $f^{(k)}g^{(n+1-k)}$ combine as $\binom n{k-1}+\binom nk=\binom{n+1}k$, proving the next order by induction. The endpoint coefficients are one. This argument explains the binomial pattern, not just the first few cases.

Worked example 2: a polynomial times an exponential. For $f=x^2$, $g=e^{ax}$ with constant $a$, derivatives of $f$ vanish above order two. Thus for $n\ge2$,

$$(x^2e^{ax})^{(n)}=e^{ax}\left(a^nx^2+2na^{n-1}x+n(n-1)a^{n-2}\right).$$

For $a=0$ and $n=2$, the term $a^{n-2}$ would involve the convention for $0^0$. State the displayed formula for $a\ne0$ and handle $a=0$ directly: the $n$th derivative of $x^2$ is identically $2$ for $n=2$ and identically $0$ for $n>2$.

Tangent bounds on intervals and at endpoints

Let $I$ be a nondegenerate interval. Assume $f$ is twice differentiable on its interior and $f''\ge0$ there. If an endpoint belongs to $I$, assume $f$ is continuous at that endpoint. A tangent at an included endpoint is used only when the corresponding one-sided derivative exists as a finite real number.

For interior points, the mean value theorem applied to $f'$ shows that $f'$ is nondecreasing. The following open-interval convexity and support-line proof gives the non-strict tangent bound for two interior points.

Convexity and support lines

A function on an interval is convex if $f((1-t)x+ty)\le(1-t)f(x)+tf(y)$ for $0\le t\le1$. If $f$ is differentiable on an open interval, it is convex exactly when $f'$ is nondecreasing.

Proof. A convex function has ordered secant slopes: for $x<z<y$, rearranging the convex inequality gives $(f(z)-f(x))/(z-x)\le(f(y)-f(x))/(y-x)\le(f(y)-f(z))/(y-z)$. Letting $z$ approach endpoints shows $f'(x)\le(f(y)-f(x))/(y-x)\le f'(y)$ and hence derivative monotonicity. Conversely if $f'$ is nondecreasing, mean value points in $(x,z)$ and $(z,y)$ give the first secant slope no greater than the second. Rearranging yields the convex inequality. In particular $f''\ge0$ suffices when the second derivative exists. The secant estimate also gives the support-line inequality $f(y)\ge f(x)+f'(x)(y-x)$.

If $f''>0$ throughout the interior, the mean value theorem makes $f'$ strictly increasing there. For an interior point $a$ and any $x\in I$ with $x\ne a$, the mean value theorem on the segment from $a$ to $x$ gives $f(x)>f(a)+f'(a)(x-a)$, including when $x$ is an included endpoint. With $f''\ge0$, the same argument gives the non-strict inequality.

For endpoint tangents, first compare secant slopes. If $x<y<z$ lie in $I$, the mean value theorem on $[x,y]$ and $[y,z]$ gives

$$\frac{f(y)-f(x)}{y-x}\le\frac{f(z)-f(y)}{z-y}.$$

The full chord slope is a weighted average of these two slopes, so it lies between them. If $f''>0$ in the interior, the two slopes are strictly ordered and the full chord slope lies strictly between them.

Let $a$ be an included left endpoint and let $x>a$. For $a<t<x$, the secant comparison gives

$$\frac{f(t)-f(a)}{t-a}\le\frac{f(x)-f(a)}{x-a}.$$

Let $t\downarrow a$. The left side tends to the finite right derivative $f'_+(a)$, proving $f(x)\ge f(a)+f'_+(a)(x-a)$. For strict positivity of $f''$, fix any $y\in(a,x)$. The same comparison first gives $f'_+(a)\le[f(y)-f(a)]/(y-a)$; the strict secant comparison then gives $[f(y)-f(a)]/(y-a)<[f(x)-f(a)]/(x-a)$. Thus the endpoint tangent inequality is strict. The strict inequality comes from the secant comparison at a fixed interior point.

At an included right endpoint $b$, take $x<t<b$. The corresponding comparison gives

$$\frac{f(b)-f(x)}{b-x}\le\frac{f(b)-f(t)}{b-t}\le f'_-(b),$$

where the second inequality follows by comparing secants and then taking the endpoint limit in the definition of the finite left derivative. This gives $f(x)\ge f(b)+f'_-(b)(x-b)$. If $f''>0$, the first inequality is strict for every fixed interior $t$, so the tangent bound is strict for $x<b$.

Worked example 3: two useful inequalities. Since $f=e^x$ has positive second derivative, its tangent at zero gives $e^x\ge1+x$ for all real $x$, with equality only at zero. Alternatively define $g=e^x-1-x$ and use its derivative's sign on either side of zero.

For $x>0$, define $h(x)=x-1-\ln x$. Its derivative is $1-1/x$, negative below one and positive above. Thus $h$ has minimum zero at one, proving $\ln x\le x-1$. Domain $x>0$ is essential; this is not an inequality for a real logarithm at negative inputs.

First/second derivative tests and global endpoint checks from Chapter 1 remain distinct. A zero second derivative at a stationary input is inconclusive: $x^4$ has a minimum, $-x^4$ a maximum, and $x^3$ neither at zero.

7.4 L'Hôpital limits with verified hypotheses

Theorem: Cauchy's mean value theorem

If $f,g$ are continuous on $[a,b]$ and differentiable on $(a,b)$ with $a<b$, some $c\in(a,b)$ satisfies

$$[f(b)-f(a)]g'(c)=[g(b)-g(a)]f'(c).$$

Proof. The auxiliary function $H(t)=[f(b)-f(a)]g(t)-[g(b)-g(a)]f(t)$ has equal endpoint values. Rolle's theorem gives $H'(c)=0$, precisely the identity. It is valid even when a denominator in a ratio formulation would be zero. Dividing to write derivative ratios requires separately checking those denominators; this is essential in proofs of L'Hopital's rule.

Learning goals and a precise one-sided setting

Identify genuine indeterminate forms, prove the rule rather than treating derivative division as algebra, and distinguish sufficient hypotheses from necessary ones. Let $f,g$ be differentiable on an open interval approaching an endpoint $b$, which may be finite or $+\infty$, from the left. Assume $g'$ is nonzero throughout a sufficiently late approach interval. The analogous right-hand approach and negative-infinity cases follow by reversing coordinates.

Cauchy's mean value theorem says that for distinct interior $x,y$, some $c$ between them satisfies

$$\frac{f(x)-f(y)}{g(x)-g(y)}=\frac{f'(c)}{g'(c)}.$$

The denominator $g(x)-g(y)$ cannot vanish, for otherwise Rolle's theorem on $g$ would give a zero derivative between them. This also shows $g$ takes any value, including zero, at most once there. By shortening the approach interval we may therefore ensure $g\ne0$, so the desired quotient is defined.

Theorem: finite derivative-ratio limit

If $f'/g'\to L\in\mathbb R$ as the endpoint is approached, then $f/g\to L$ provided either (i) both $f,g\to0$, or (ii) $|g|\to\infty$. In case (ii), no separate assumption that $f$ is unbounded is necessary.

Proof of (i). Let $h=f-Lg$. Given $\varepsilon>0$, throughout a sufficiently late interval $|h'/g'|<\varepsilon$. Cauchy's theorem gives $|h(x)-h(y)|\le\varepsilon|g(x)-g(y)|$ for any two points there. Fix $x$ and let $y$ approach the endpoint. Since $h(y),g(y)\to0$, this yields $|h(x)|\le\varepsilon|g(x)|$, so $|f(x)/g(x)-L|\le\varepsilon$. This argument works even when the endpoint is infinity, without pretending infinity is an endpoint value of a continuous function.

Proof of (ii). Given $\eta>0$, set $\varepsilon=\eta/4$. Again take $h=f-Lg$ and a late interval where the derivative ratio has magnitude below $\varepsilon$. Fix one point $x_0$ in it. Cauchy's theorem between $x_0$ and a later $x$ gives $|h(x)-h(x_0)|\le\varepsilon|g(x)-g(x_0)|$. Consequently

$$\left|\frac{f(x)}{g(x)}-L\right| \le\frac{|h(x_0)|}{|g(x)|}+\varepsilon\left|1-\frac{g(x_0)}{g(x)}\right|.$$

For sufficiently late $x$, the first term is less than $\eta/2$ and the absolute factor in the second is less than $2$. The total is therefore less than $\eta/2+2(\eta/4)=\eta$. Since $\eta>0$ was arbitrary, the quotient tends to $L$.

Theorem: infinite derivative-ratio limits

Under the same denominator assumptions, if $f'/g'\to+\infty$, then $f/g\to+\infty$ in either case (i) or (ii); replace signs for a limit of $-\infty$.

Proof. For any real threshold $M$, on a sufficiently late interval $f'/g'>M$. Cauchy's theorem makes every incremental ratio greater than $M$. In case (i), fix $x$ and pass the other point to the endpoint to obtain $f(x)/g(x)\ge M$. Since $M$ can be chosen above any requested threshold, this proves divergence to $+\infty$. In case (ii), fix $x_0$ and write the incremental ratio as $R(x)>M$:

$$\frac{f(x)}{g(x)}=R(x)\left(1-\frac{g(x_0)}{g(x)}\right)+\frac{f(x_0)}{g(x)}.$$

For $M>0$ and sufficiently late $x$, the multiplying factor exceeds $1/2$ and $f(x_0)/g(x)>-1$. Therefore $f(x)/g(x)>M/2-1$. Taking $M$ arbitrarily large proves divergence to $+\infty$. Applying the result to $-f$ covers the negative-infinity limit.

Worked example 4: turn a product into a quotient. As $x\to0^+$, $x\ln x=\ln x/(1/x)$. The numerator tends to negative infinity and denominator to positive infinity; derivative ratio is $-x\to0$. Thus $x\ln x\to0$ and $x^x=\exp(x\ln x)\to1$. Positivity of $x$ is retained through the logarithm.

The rule is not repeated differentiation of every quotient. For $(1+x)/x$ at zero from the right, numerator does not tend to zero, so the quotient is not of the required form. It tends to positive infinity, whereas an invalid derivative-ratio calculation would give one.

7.5 Taylor polynomials and a derived remainder

The order-$n$ Taylor polynomial at $a$ is $T_n(x)=\sum_{k=0}^nf^{(k)}(a)(x-a)^k/k!$, whose degree may be smaller than $n$ if leading derivatives vanish.

Theorem: Taylor's integral remainder

Let $n\ge0$ and let $f$ have continuous derivatives through order $n+1$ on the closed interval between $a$ and $x$. Then

$$f(x)=T_n(x)+\frac1{n!}\int_a^x f^{(n+1)}(t)(x-t)^n\,dt.$$

Consequently, if $|f^{(n+1)}(t)|\le M$ there, then $|f(x)-T_n(x)|\le M|x-a|^{n+1}/(n+1)!$.

Proof. At $n=0$, the identity is the fundamental theorem $f(x)-f(a)=\int_a^xf'(t)dt$. For $n\ge1$, integrate the remainder by parts, taking $u=(x-t)^n/n!$ and $dv=f^{(n+1)}(t)dt$. The boundary at $x$ is zero and at $a$ contributes $-f^{(n)}(a)(x-a)^n/n!$. The remaining integral is the order-$(n-1)$ remainder. Induction therefore yields the identity. This uses oriented integrals, so it also holds when $x<a$. Taking absolute values, reversing integration limits if needed, and integrating $|x-t|^n$ gives the stated bound. Continuity of the highest derivative makes every use of parts and the bound valid.

This theorem provides a sufficient regularity setting, not the weakest possible hypotheses. For a Taylor series to represent $f$ at a fixed $x$, the remainders must tend to zero as $n\to\infty$. A finite-order error bound alone does not prove that without control of its dependence on $n$.

With $h=x-a$, the substitution $t=a+sh$ gives

$$R_n(x)=\frac{h^{n+1}}{n!}\int_0^1(1-s)^nf^{(n+1)}(a+sh)ds.$$

This form controls either sign of $h$. If the derivative magnitude is at most $M$, then $|R_n|\le M|h|^{n+1}/(n+1)!$. Let $n$ be a nonnegative integer, let $f^{(n+1)}$ be continuous on the closed segment between $a$ and $x$, and suppose the integral remainder above holds. If $h\ne0$, put $w(s)=f^{(n+1)}(a+sh)$. The weight $(1-s)^n$ is positive on $[0,1)$ and has integral $1/(n+1)$. When $n=0$, it is the constant polynomial $1$ on all of $[0,1]$. If $w$ is nonconstant, continuity makes $w-\min w$ and $\max w-w$ strictly positive on some interior subintervals. The weight is positive there, so the corresponding integrals are both strictly positive. Thus the weighted average lies strictly between the minimum and maximum; the intermediate value theorem gives some $s_*\in(0,1)$ attaining it. If $w$ is constant, choose any interior $s_*$. Therefore

$$R_n(x)=\frac{f^{(n+1)}(a+s_*h)}{(n+1)!}h^{n+1}.$$

This also holds for $h<0$, with the sign retained in $h^{n+1}$. If $h=0$, directly $R_n(a)=0$ and no intermediate point is needed.

Worked example 5: a logarithmic approximation. At $a=0$, $\ln(1+x)$ has cubic Taylor polynomial $x-x^2/2+x^3/3$. On $0\le x\le1/5$, the fourth derivative magnitude is $6/(1+x)^4\le6$, so the error is at most $x^4/4$, and at $x=1/5$ at most $1/2500$. The polynomial is a controlled approximation, not an exact equality.

7.6 Series representation requires vanishing remainders

An infinite Taylor series represents a function at an input only if the remainder tends to zero as degree increases. For $e^x$, the remainder bound is at most $e^{|x|}|x|^{n+1}/(n+1)!$, which tends to zero for every fixed real $x$. The same factorial control with derivative bound one proves the sine and cosine series:

$$e^x=\sum_{k\ge0}\frac{x^k}{k!},\quad \sin x=\sum_{k\ge0}\frac{(-1)^kx^{2k+1}}{(2k+1)!},\quad \cos x=\sum_{k\ge0}\frac{(-1)^kx^{2k}}{(2k)!}.$$

The geometric identity gives $1/(1-x)=\sum_{k\ge0}x^k$ for $|x|<1$. Integrating its finite form with a retained remainder gives $\ln(1+x)=\sum_{k\ge1}(-1)^{k-1}x^k/k$ for $-1<x\le1$, with the endpoint one checked by an alternating/integral remainder and negative one excluded.

Smoothness alone is insufficient. The function equal to $e^{-1/x^2}$ for $x\ne0$ and zero at zero has all derivatives zero at zero, because exponential decay dominates each reciprocal power arising after differentiation. Its Taylor series is zero yet the function is positive away from zero. Thus derivative matching alone does not prove series equality.

Finite logarithm remainders including the endpoint

For every integer $m\ge1$ and $t\ne-1$, finite division gives

$$\frac1{1+t}=\sum_{k=0}^{m-1}(-t)^k+\frac{(-t)^m}{1+t}.$$

For fixed $-1<x\le1$, integrate over the oriented segment from zero to $x$. The fundamental theorem and $\ln1=0$ give

$$\ln(1+x)=\sum_{k=1}^m\frac{(-1)^{k-1}x^k}{k}+\int_0^x\frac{(-t)^m}{1+t}\,dt.$$

If $0\le x\le1$, the absolute remainder is at most $x^{m+1}/(m+1)\le1/(m+1)$. This includes $x=1$, proving convergence to $\ln2$ there. If $-1<x<0$, write $r=|x|<1$. The denominator is at least $1-r$ on the integration segment, so the absolute remainder is at most $r^{m+1}/[(m+1)(1-r)]$, which tends to zero. At $x=-1$, each series term is $-1/k$, giving divergence by the harmonic-series proof, and the real logarithm is undefined there. For $|x|>1$, choose $1<q<|x|$. The ratio of successive term magnitudes $|x|^k/k$ is $|x|k/(k+1)>q$ eventually. Iteration bounds those magnitudes below by one fixed positive magnitude times growing powers of $q$, so the terms do not tend to zero and the series diverges. Thus its real convergence set is exactly $-1<x\le1$, where its sum equals $\ln(1+x)$.

Full flat-function differentiation at zero

Use the previously proved real exponential series, positivity, reciprocal law and derivative identity $\exp'=\exp$. Define $F(0)=0$ and $F(x)=\exp(-1/x^2)$ for $x\ne0$. For every real $r\ge0$, set $u=1/x^2$. Choose an integer $k>r/2$. The nonnegative exponential series gives $\exp u\ge u^k/k!$ for $u>0$, hence

$$|x|^{-r}\exp(-1/x^2)=u^{r/2}\exp(-u)\le k!u^{r/2-k}\longrightarrow0.$$

At nonzero inputs, induction gives $F^{(n)}(x)=P_n(1/x)\exp(-1/x^2)$ for a polynomial $P_n$, starting with $P_0=1$ and using $P_{n+1}(z)=-z^2P_n'(z)+2z^3P_n(z)$. Suppose the order-$n$ derivative has this formula off zero and value zero at zero. Every polynomial has finite degree, so the estimate above first shows $F^{(n)}(x)\to0$ and then $F^{(n)}(h)/h\to0$. Thus $F^{(n)}$ is continuous at zero, its derivative at zero exists and equals zero, and the recurrence gives the next off-zero formula. Induction proves $F$ is smooth with every derivative zero at zero. Its Taylor series there is identically zero, whereas $F(x)>0$ for each $x\ne0$. This completes the claimed counterexample without treating smoothness as analyticity.

Worked example 6: retain the first surviving term. Taylor's theorem gives $e^x=1+x+x^2/2+O(x^3)$ near zero. Therefore $(e^x-1-x)/x^2\to1/2$. Replacing $e^x$ only by $1+x$ would erase the term that determines the limit.

7.7 Graded exercises

Foundation

1. Find $y'$ for $xy+y^2=6$ at $(1,2)$ and check that the point lies on the curve.
2. Find the third derivative of $x^2e^x$ using Leibniz's formula.
3. Prove $\ln(1+t)\le t$ for $t>-1$ using the logarithmic inequality above.

Core

4. Find $\lim_{x\to0}(e^{2x}-1)/x$ with hypotheses or a Taylor remainder.
5. Evaluate $\lim_{x\to\infty}(\ln x)/x^2$.
6. Find the second-order Taylor polynomial of $f(x)=1/x$ at $a=1$.
7. Approximate $\sin(0.2)$ by $x-x^3/6$ and bound its error by using its degree-four Taylor polynomial and the fifth derivative.
8. Find the infinite geometric expansion of $1/(1+2x)$ and its open convergence interval.

Synthesis

9. Use differentiation to prove $x^2\ge2x-1$ for all real $x$, and then supply an independent algebraic proof.
10. Derive the nth product derivative by induction, explicitly identifying why the two interior coefficients combine into $\binom{n+1}k$.

7.8 Full solutions

1. The point satisfies $1(2)+4=6$. Differentiation gives $y+xy'+2yy'=0$, so $y'=-y/(x+2y)=-2/5$ there; denominator five is nonzero.
2. Put $n=3,a=1$ in the formula to obtain $e^x(x^2+6x+6)$.
3. Set $x=1+t>0$ in $\ln x\le x-1$, giving the required inequality, equality only at $t=0$.
4. The quotient has $0/0$ form, denominator and derivative nonzero near but away from zero, and derivative ratio $2e^{2x}\to2$. Hence the limit is two. Equivalently $e^{2x}=1+2x+O(x^2)$.
5. Numerator and denominator both tend to infinity, denominator derivative $2x$ is nonzero for positive large $x$, and derivative ratio $(1/x)/(2x)=1/(2x^2)\to0$. The limit is zero.
6. Values are $f(1)=1,f'(1)=-1,f''(1)=2$, giving $T_2=1-(x-1)+(x-1)^2$.
7. The fourth-order polynomial is still $x-x^3/6$ because the fourth coefficient is zero. The fifth derivative magnitude is at most one, so error is at most $0.2^5/120=1/375000$.
8. Substitute $-2x$ into the geometric identity: $\sum_{k\ge0}(-2x)^k$ for $|x|<1/2$. Endpoint convergence is not supplied by the open-interval formula.
9. Define $g=x^2-2x+1$. Its derivative $2x-2$ is negative before one and positive after, so its global minimum is $g(1)=0$. Algebraically $g=(x-1)^2\ge0$. Both prove the same inequality, with equality only at one.
10. Assume $(fg)^{(n)}=\sum_k\binom nkf^{(k)}g^{(n-k)}$. Differentiating a term contributes once to $f^{(k+1)}g^{(n-k)}$ and once to $f^{(k)}g^{(n+1-k)}$. For an interior index $k$, these come from coefficients $\binom n{k-1}$ and $\binom nk$, whose sum is $\binom{n+1}k$ by Pascal's identity. The two endpoints each have coefficient one. This yields the $(n+1)$st formula; the ordinary product rule supplies the base case.

Summary and onward map

Implicit differentiation is local and requires a nonzero relevant coefficient. Leibniz's formula follows from product-rule induction. Derivatives prove inequalities by monotonicity or convex tangents. L'Hôpital limits need specific indeterminate-limit hypotheses, not mere unboundedness. Taylor remainders control approximation and determine when an infinite expansion is genuinely equal to the function.