Corrected practice exam 2 in differential calculus (201-SN2-RE)

$f(1)=\sqrt{4}=2$. The difference quotient is $\frac{\sqrt{3(1+h)+1}-2}{h}=\frac{\sqrt{4+3h}-2}{h}$, of the form $\frac{0}{0}$. A square root in the numerator calls for the conjugate: multiply top and bottom by $\sqrt{4+3h}+2$, which gives $\frac{(4+3h)-4}{h\left(\sqrt{4+3h}+2\right)}=\frac{3h}{h\left(\sqrt{4+3h}+2\right)}=\frac{3}{\sqrt{4+3h}+2}$. The factor $h$ cancels, which resolves the indeterminate form, and letting $h\to 0$: $f'(1)=\frac{3}{2+2}=\frac{3}{4}$. Remember the general mechanism: in any derivative from the definition, a factor $h$ MUST appear in the numerator; otherwise the limit would be infinite and the function would not be differentiable.

With the rules: $f(x)=(3x+1)^{1/2}$, so $f'(x)=\frac{1}{2}(3x+1)^{-1/2}\cdot 3=\frac{3}{2\sqrt{3x+1}}$. At $x=1$: $\frac{3}{2\times 2}=\frac{3}{4}$. The two methods agree, as expected: the rules are nothing but this limit computed once and for all, in general, so that it never has to be done again.

$g(3)=\frac{4\times 9}{8}=\frac{36}{8}=\frac{9}{2}$. Difference quotient: $\frac{1}{h}\left[\frac{4(3+h)^{2}}{8+h}-\frac{9}{2}\right]$. Put the two fractions over the common denominator $2(8+h)$: $\frac{8(3+h)^{2}-9(8+h)}{2(8+h)}$. Expand the numerator: $8\left(9+6h+h^{2}\right)-72-9h=72+48h+8h^{2}-72-9h=39h+8h^{2}=h(39+8h)$. The quotient becomes $\frac{h(39+8h)}{2(8+h)\,h}=\frac{39+8h}{2(8+h)}$, and as $h\to 0$: $g'(3)=\frac{39}{16}=2.4375$. Check with the quotient rule: $g'(x)=\frac{8x(x+5)-4x^{2}}{(x+5)^{2}}=\frac{4x^{2}+40x}{(x+5)^{2}}$, which equals $\frac{36+120}{64}=\frac{156}{64}=\frac{39}{16}$ at $x=3$.

$\frac{u(x+h)-u(x)}{h}=\frac{1}{h}\left[\frac{1}{2x+2h-3}-\frac{1}{2x-3}\right]=\frac{1}{h}\cdot\frac{(2x-3)-(2x+2h-3)}{(2x+2h-3)(2x-3)}=\frac{1}{h}\cdot\frac{-2h}{(2x+2h-3)(2x-3)}=\frac{-2}{(2x+2h-3)(2x-3)}$. As $h\to 0$: $u'(x)=\frac{-2}{(2x-3)^{2}}$. The sign is negative everywhere, which can be read on the graph: $u$ is decreasing on each of the two branches of its domain. Careful: decreasing ON EACH BRANCH, not on $\mathbb{R}$; a negative derivative says nothing across an asymptote.

For $v(x)=|x-2|$ at $a=2$: $v(2)=0$ and the quotient is $\frac{|2+h-2|-0}{h}=\frac{|h|}{h}$. From the left, $h<0$ so $|h|=-h$ and the quotient equals $-1$; from the right it equals $+1$. Both one-sided limits exist but they DIFFER: the limit does not exist, and $v$ is not differentiable at 2. Geometrically, this is a CORNER POINT: the curve has two half-tangents with different slopes. For $w(x)=\sqrt[3]{x}$ at $a=0$: $w(0)=0$ and the quotient is $\frac{h^{1/3}}{h}=h^{-2/3}=\frac{1}{h^{2/3}}$, which tends to $+\infty$ from BOTH sides, since $h^{2/3}>0$ whatever the sign of $h$. The limit is infinite, so it is not a number, so $w$ is not differentiable at 0. Geometrically, this is a VERTICAL TANGENT: the curve does have a tangent line, but that line has no slope. The difference is clear: for $v$, the limit does not exist because the two sides contradict each other; for $w$, the two sides agree, but on an infinite value. In both cases, $v$ and $w$ are nevertheless CONTINUOUS at the point studied: continuous does not imply differentiable.

A sum, so term by term, but two terms are traps. $\sqrt{2}$ is a NUMBER, not a variable: $\sqrt{2}\,x^{3}$ differentiates to $3\sqrt{2}\,x^{2}$, exactly as $7x^{3}$ would give $21x^{2}$. Likewise, $\pi^{2}$ is a constant: its derivative is $0$, not $2\pi$. The radical is rewritten first: $\sqrt[3]{8x^{4}}=\left(8x^{4}\right)^{1/3}=2x^{4/3}$, whose derivative is $2\cdot\frac{4}{3}x^{1/3}=\frac{8}{3}x^{1/3}=\frac{8}{3}\sqrt[3]{x}$. Altogether, $f'(x)=20x^{3}+3\sqrt{2}\,x^{2}+\frac{8}{3}\sqrt[3]{x}-2$. Check at $x=1$: $20+3\times 1.4142+2.6667-2\approx 24.91$.

A composition, and nothing else: rewrite the radical as a power, $g(x)=\left(3x^{4}-5x\right)^{5/3}$. Chain rule: $g'(x)=\frac{5}{3}\left(3x^{4}-5x\right)^{2/3}\left(12x^{3}-5\right)$. The derivative of the inside, $12x^{3}-5$, must not be forgotten: it is what distinguishes the chain rule from the plain power rule. You can also write $g'(x)=\frac{5\left(12x^{3}-5\right)}{3}\sqrt[3]{\left(3x^{4}-5x\right)^{2}}$. Check at $x=2$: $3(16)-10=38$ and $12(8)-5=91$, so $g'(2)=\frac{5}{3}\times 38^{2/3}\times 91\approx 1714.27$.

A PRODUCT of two compositions: the product rule governs the whole expression, and the chain rule applies inside each factor. $f'(t)=6\left(t^{3}+5\right)^{5}\left(3t^{2}\right)(4t-3)^{10}+\left(t^{3}+5\right)^{6}\cdot 10(4t-3)^{9}\cdot 4$. Simplify by factoring out the smallest common powers, $\left(t^{3}+5\right)^{5}$ and $(4t-3)^{9}$: $f'(t)=\left(t^{3}+5\right)^{5}(4t-3)^{9}\left[18t^{2}(4t-3)+40\left(t^{3}+5\right)\right]$. The bracket equals $72t^{3}-54t^{2}+40t^{3}+200=112t^{3}-54t^{2}+200$. Hence $f'(t)=\left(t^{3}+5\right)^{5}(4t-3)^{9}\left(112t^{3}-54t^{2}+200\right)$. Check at $t=1$: the unfactored form gives $6(6)^{5}(3)+6^{6}(40)=139\,968+1\,866\,240=2\,006\,208$, and the factored form $6^{5}\times 1\times 258=7776\times 258=2\,006\,208$. Simplifying is not optional: it is the only way to check your own computation.

The factor $2$ comes out of the derivative; only the quotient is differentiated. $\left(\frac{u}{v}\right)'=\frac{u'v-uv'}{v^{2}}$ with $u=3x^{2}+4$, $u'=6x$, $v=1-x$, $v'=-1$. Numerator: $6x(1-x)-\left(3x^{2}+4\right)(-1)=6x-6x^{2}+3x^{2}+4=-3x^{2}+6x+4$. Hence $h'(x)=\frac{2\left(-3x^{2}+6x+4\right)}{(1-x)^{2}}=\frac{-6x^{2}+12x+8}{(1-x)^{2}}$. The double negative in $-u v'$ is the most common error of the whole chapter: writing $u'v+uv'$ in the numerator of a quotient gives $6x-6x^{2}-3x^{2}-4$ here, an entirely wrong result. Check at $x=0$: $h'(0)=\frac{8}{1}=8$.

A quotient with a root in the denominator. $u=2t^{2}-1$, $u'=4t$; $v=\left(t^{2}+2\right)^{1/2}$, $v'=\frac{1}{2}\left(t^{2}+2\right)^{-1/2}\cdot 2t=\frac{t}{\sqrt{t^{2}+2}}$. Then $g'(t)=\frac{4t\sqrt{t^{2}+2}-\left(2t^{2}-1\right)\dfrac{t}{\sqrt{t^{2}+2}}}{t^{2}+2}$. Multiply top and bottom by $\sqrt{t^{2}+2}$ to clear the compound fraction: $g'(t)=\frac{4t\left(t^{2}+2\right)-t\left(2t^{2}-1\right)}{\left(t^{2}+2\right)^{3/2}}=\frac{4t^{3}+8t-2t^{3}+t}{\left(t^{2}+2\right)^{3/2}}=\frac{2t^{3}+9t}{\left(t^{2}+2\right)^{3/2}}=\frac{t\left(2t^{2}+9\right)}{\left(t^{2}+2\right)^{3/2}}$. Check at $t=1$: $\frac{1\times 11}{3^{3/2}}=\frac{11}{3\sqrt{3}}\approx 2.117$. Note the denominator raised to the power $\frac{3}{2}$: it is the trace of $\left(\sqrt{v}\right)^{2}$ multiplied by one more $\sqrt{v}$, and writing $\left(t^{2}+2\right)^{2}$ out of carelessness ruins everything.

A product in which each factor is a composition. $u=e^{3x}$, $u'=3e^{3x}$; $v=\ln\left(x^{2}+1\right)$, $v'=\frac{2x}{x^{2}+1}$. Hence $f'(x)=3e^{3x}\ln\left(x^{2}+1\right)+e^{3x}\cdot\frac{2x}{x^{2}+1}=e^{3x}\left[3\ln\left(x^{2}+1\right)+\frac{2x}{x^{2}+1}\right]$. Factoring out $e^{3x}$ is not cosmetic: the factored form shows at a glance that $f'$ has the sign of the bracket, since the exponential is always positive.

Do NOT differentiate this quotient as it stands. The properties of logarithms turn it into a sum: $g(x)=\ln\left(x^{2}+1\right)-\ln\sqrt{2x+3}=\ln\left(x^{2}+1\right)-\frac{1}{2}\ln(2x+3)$. Differentiation becomes immediate: $g'(x)=\frac{2x}{x^{2}+1}-\frac{1}{2}\cdot\frac{2}{2x+3}=\frac{2x}{x^{2}+1}-\frac{1}{2x+3}$. Differentiating the original form with the chain rule and then the quotient rule gives the same result after five more lines and two more chances to make a mistake. The lesson holds for the whole exam: simplifying first is part of the method, not an optional touch of elegance.

A composition of compositions: the power is the outer layer, the sine the middle one, and the linear function the inner one. Differentiate from the outside in, one layer at a time. $h'(x)=3\left(\sin(2x+1)\right)^{2}\cdot\cos(2x+1)\cdot 2=6\sin^{2}(2x+1)\cos(2x+1)$. Three layers, three factors: if your answer does not have as many factors as the function has layers, one is missing.

Neither the power rule nor the exponential rule applies: in $x^{x}$, the base AND the exponent vary. Use logarithmic differentiation. Set $y=x^{x}$ and take the logarithm of both sides: $\ln y=x\ln x$. Differentiate each side with respect to $x$, remembering that $y$ depends on $x$, which gives $\frac{y'}{y}$ on the left by the chain rule: $\frac{y'}{y}=\ln x+x\cdot\frac{1}{x}=\ln x+1$. Hence $y'=y(\ln x+1)$, that is, $k'(x)=x^{x}\left(\ln x+1\right)$. Check at $x=1$: $k'(1)=1\times(0+1)=1$. Writing $x\cdot x^{x-1}$ is the classic error: it treats the exponent as a constant.

Quotient rule with $u=e^{2x}$, $u'=2e^{2x}$, $v=1+e^{2x}$, $v'=2e^{2x}$. Numerator: $2e^{2x}\left(1+e^{2x}\right)-e^{2x}\cdot 2e^{2x}=2e^{2x}+2e^{4x}-2e^{4x}=2e^{2x}$. Hence $m'(x)=\frac{2e^{2x}}{\left(1+e^{2x}\right)^{2}}$. The $e^{4x}$ terms cancel, a sign that the computation was set up correctly. A useful remark: $m'>0$ everywhere, so $m$ is strictly increasing, and since $m(x)=\frac{1}{1+e^{-2x}}$ it goes from 0 to 1. This is the logistic function, the one that models saturation, and its derivative also equals $2m(1-m)$.

Differentiate term by term. $\left(2x^{3}\right)'=6x^{2}$. For $x^{2}y$, it is a PRODUCT of two functions of $x$: $\left(x^{2}y\right)'=2xy+x^{2}y'$. On the right, $\left(y^{3}\right)'=3y^{2}y'$ by the chain rule, and $(-5)'=0$. The differentiated equation is therefore $6x^{2}+2xy+x^{2}y'=3y^{2}y'$. Collect: $x^{2}y'-3y^{2}y'=-6x^{2}-2xy$, that is, $y'\left(x^{2}-3y^{2}\right)=-6x^{2}-2xy$, hence $\frac{dy}{dx}=\frac{-6x^{2}-2xy}{x^{2}-3y^{2}}=\frac{6x^{2}+2xy}{3y^{2}-x^{2}}$. The two forms are equivalent: top and bottom were simply multiplied by $-1$. The classic pitfall is forgetting the product in $x^{2}y$ and writing just $2xy$.

With $x=0$, the relation becomes $0+0=y^{3}-5$, so $y^{3}=5$ and $y=\sqrt[3]{5}\approx 1.710$. The point is $\left(0,\sqrt[3]{5}\right)$. The slope is $\frac{6(0)^{2}+2(0)y}{3y^{2}-0}=\frac{0}{3\sqrt[3]{25}}=0$: the tangent line there is HORIZONTAL, with equation $y=\sqrt[3]{5}$. Note that we never needed to solve for $y$ in terms of $x$, which would be impossible by hand here: that is exactly the point of implicit differentiation.

Check: $1^{2}+1\times 2+2^{2}=1+2+4=7$, so the point does lie on the curve. Implicit differentiation: $2x+\left(y+xy'\right)+2yy'=0$, where the term $xy$ calls for the product rule. Collect: $y'(x+2y)=-2x-y$, so $\frac{dy}{dx}=\frac{-2x-y}{x+2y}$. At $(1,2)$: $\frac{-2-2}{1+4}=-\frac{4}{5}$. Equation of the tangent line: $y-2=-\frac{4}{5}(x-1)$, that is, $y=-\frac{4}{5}x+\frac{14}{5}$. The curve $x^{2}+xy+y^{2}=7$ is a tilted ellipse, which no single function $y=f(x)$ describes in full: once again, only the implicit route works.

A horizontal tangent means $y'=0$, so a zero numerator and a nonzero denominator: $-2x-y=0$, that is, $y=-2x$. Substitute into the equation of the curve: $x^{2}+x(-2x)+(-2x)^{2}=x^{2}-2x^{2}+4x^{2}=3x^{2}=7$, so $x=\pm\sqrt{\frac{7}{3}}\approx\pm 1.528$. The two points are $\left(\sqrt{\tfrac{7}{3}},-2\sqrt{\tfrac{7}{3}}\right)\approx(1.53, -3.06)$ and its reflection through the origin $\approx(-1.53, 3.06)$. Check that the denominator $x+2y=x-4x=-3x$ is not zero there, which holds since $x\ne 0$: without this check, you could declare horizontal a tangent that is in fact indeterminate.

A vertical tangent means that the slope does not exist, so a zero denominator and a nonzero numerator: $x+2y=0$, that is, $y=-\frac{x}{2}$. Substituting: $x^{2}-\frac{x^{2}}{2}+\frac{x^{2}}{4}=\frac{3x^{2}}{4}=7$, so $x=\pm\sqrt{\frac{28}{3}}\approx\pm 3.055$, with $y=\mp 1.528$. What this means for the formula of c): it is valid only away from these two points. An implicit differentiation formula always carries an implicit condition of existence, namely that its denominator is nonzero, and forgetting it amounts to claiming that a curve has a slope where it is vertical. Notice the symmetry of the result: the horizontal and the vertical tangents of this ellipse correspond to each other when $x$ and $y$ are swapped, which reflects the symmetry of the equation itself.

First derivative, chain rule: $f'(x)=5\left(2x^{3}+1\right)^{4}\cdot 6x^{2}=30x^{2}\left(2x^{3}+1\right)^{4}$. Second derivative, product rule on $30x^{2}$ and $\left(2x^{3}+1\right)^{4}$: $f''(x)=60x\left(2x^{3}+1\right)^{4}+30x^{2}\cdot 4\left(2x^{3}+1\right)^{3}\cdot 6x^{2}=60x\left(2x^{3}+1\right)^{4}+720x^{4}\left(2x^{3}+1\right)^{3}$. FACTOR NOW, not later: $f''(x)=60x\left(2x^{3}+1\right)^{3}\left[\left(2x^{3}+1\right)+12x^{3}\right]=60x\left(2x^{3}+1\right)^{3}\left(14x^{3}+1\right)$. Third derivative: three factors, so three terms. $f'''(x)=60\left(2x^{3}+1\right)^{3}\left(14x^{3}+1\right)+60x\cdot 18x^{2}\left(2x^{3}+1\right)^{2}\left(14x^{3}+1\right)+60x\left(2x^{3}+1\right)^{3}\cdot 42x^{2}$. Factor out $60\left(2x^{3}+1\right)^{2}$: the bracket becomes $\left(2x^{3}+1\right)\left(14x^{3}+1\right)+18x^{3}\left(14x^{3}+1\right)+42x^{3}\left(2x^{3}+1\right)$, that is, $\left(28x^{6}+16x^{3}+1\right)+\left(252x^{6}+18x^{3}\right)+\left(84x^{6}+42x^{3}\right)=364x^{6}+76x^{3}+1$. Hence $f'''(x)=60\left(2x^{3}+1\right)^{2}\left(364x^{6}+76x^{3}+1\right)$. At $x=1$: $60\times 9\times 441=540\times 441=238\,140$.

$g(x)=x^{-1}$. Compute the first few: $g'=-x^{-2}$, $g''=2x^{-3}$, $g'''=-6x^{-4}$, $g^{(4)}=24x^{-5}$. Two patterns appear: the sign alternates, hence a factor $(-1)^{n}$, and the coefficient is $1, 2, 6, 24$, that is, $n!$. The exponent is $-(n+1)$. Hence $g^{(n)}(x)=\frac{(-1)^{n}\,n!}{x^{n+1}}$. Proof by induction: the formula holds for $n=1$; if it holds for $n$, differentiating $(-1)^{n}n!\,x^{-(n+1)}$ gives $(-1)^{n}n!\cdot\left(-(n+1)\right)x^{-(n+2)}=(-1)^{n+1}(n+1)!\,x^{-(n+2)}$, which is indeed the formula for $n+1$.