Exercises — Derivatives of inverse trig functions — all six

$\frac{d}{dx}\arccos x=-\frac{1}{\sqrt{1-x^2}}.$ Why the minus: as the ratio $x$ grows, the angle $\arccos x$ shrinks (arccos runs from $\pi$ down to $0$). A function that decreases has a negative slope. Same $\sqrt{1-x^2}$ shape as arcsin, opposite sign.

$\frac{d}{dx}\arctan x=\frac{1}{1+x^2},\frac{d}{dx}\mathrm{arccot}x=-\frac{1}{1+x^2}.$ No square root at all — that cleanliness comes from the identity ${\sec }^{2}y=1+{\tan }^{2}y$ (see Pythagorean Identities), which needs no square root when we solve. The co-partner just flips the sign.

Only $\mathrm{arcsec}$ and $\mathrm{arccsc}$ carry the $|x|$: $\frac{d}{dx}\mathrm{arcsec}x=\frac{1}{|x|\sqrt{x^2-1}}.$ At $x=2$: $|2|=2$ and $\sqrt{4-1}=\sqrt{3}$, so the value is $\frac{1}{2\sqrt{3}}=\frac{\sqrt{3}}{6}\approx \mathrm{0.2887.}$

Inside $u=5x$, so $u^{\prime }=5$. $f^{\prime }(x)=\frac{1}{1+(5x{)}^2}\cdot 5=\frac{5}{1+25x^2}.$ The lone extra factor is the derivative of the inside — miss it and you are off by a factor of $5$.

Inside $u=\sqrt{x}=x^{1/2}$, so $u^{\prime }=\frac{1}{2}{x}^{-1/2}=\frac{1}{2\sqrt{x}}$. Also $u^2=x$. $g^{\prime }(x)=\frac{1}{\sqrt{1-(\sqrt{x}{)}^2}}\cdot \frac{1}{2\sqrt{x}}=\frac{1}{\sqrt{1-x}}\cdot \frac{1}{2\sqrt{x}}=\frac{1}{2\sqrt{x(1-x)}}.$ Note how $(\sqrt{x}{)}^2$ collapses to $x$ — always simplify the inside-squared before writing the answer.

Inside $u=e^x$, so $u^{\prime }=e^x$. Arccos brings a minus sign. $h^{\prime }(x)=-\frac{1}{\sqrt{1-(e^x{)}^2}}\cdot e^x=\frac{-e^x}{\sqrt{1-e^{2x}}}.$ Valid only where $e^x<1$, i.e. $x<0$ — outside that the ratio exceeds $1$ and $\arccos$ is undefined.

Inside $u=x^3$, $u^{\prime }=3x^2$, and $u^2=x^6$. $p^{\prime }(x)=\frac{3x^2}{1+x^6},p^{\prime }(1)=\frac{3}{1+1}=\frac{3}{2}.$

Product rule: $(uv{)}^{\prime }=u^{\prime }v+u{v}^{\prime }$ with $u=x,v=\arctan x$. $f^{\prime }(x)=1\cdot \arctan x+x\cdot \frac{1}{1+x^2}=\arctan x+\frac{x}{1+x^2}.$ No further collapse — the two pieces are genuinely different kinds of term (an angle plus a rational function).

Inside $u=1/x=x^{-1}$, so $u^{\prime }=-x^{-2}=-\frac{1}{x^2}$, and $u^2=\frac{1}{x^2}$. $g^{\prime }(x)=\frac{1}{1+\frac{1}{x^2}}\cdot \left(-\frac{1}{x^2}\right)=\frac{x^2}{x^2+1}\cdot \left(-\frac{1}{x^2}\right)=-\frac{1}{1+x^2}.$ This equals $\frac{d}{dx}\mathrm{arccot}x$ exactly — reassuring, since $\arctan (1/x)=\mathrm{arccot}x$ for $x>0$.

$\frac{1}{\sqrt{1-x^2}}-\left(-\frac{1}{\sqrt{1-x^2}}\right)=\frac{2}{\sqrt{1-x^2}}.$ The two derivatives add (because subtracting the negative flips it), giving twice the arcsin derivative.

Recognise $\frac{2x}{1-x^2}$ as the double-angle for tangent: if $x=\tan \theta$ then $\frac{2x}{1-x^2}=\tan (2\theta )$, so $y=2\arctan x$ on $|x|<1$. Then $\frac{dy}{dx}=2\cdot \frac{1}{1+x^2}=\frac{2}{1+x^2}.$ (Brute-force quotient-rule differentiation gives the same $\frac{2}{1+x^2}$ — the identity just saves the algebra.)

$\frac{d}{dx}(\arctan x+\mathrm{arccot}x)=\frac{1}{1+x^2}+\left(-\frac{1}{1+x^2}\right)=0.$ A zero derivative on an interval means the function is constant (a flat graph). Evaluate at $x=1$: $\arctan 1+\mathrm{arccot}1=\frac{\pi }{4}+\frac{\pi }{4}=\frac{\pi }{2}.$ So $\arctan x+\mathrm{arccot}x=\frac{\pi }{2}$ for all $x$.

Since $\frac{d}{dx}\arctan x=\frac{1}{1+x^2}$, the antiderivative is $\arctan x$. ${\int }_0^1\frac{dx}{1+x^2}=[\arctan x{]}_0^1=\arctan 1-\arctan 0=\frac{\pi }{4}-0=\frac{\pi }{4}.$

The integrand is $\frac{d}{dx}\arcsin x$, so ${\int }_0^{1/2}\frac{dx}{\sqrt{1-x^2}}=[\arcsin x{]}_0^{1/2}=\arcsin \frac{1}{2}-\arcsin 0=\frac{\pi }{6}-0=\frac{\pi }{6}.$

Inside $u=x/a$, $u^{\prime }=1/a$, $u^2=x^2/a^2$. $\frac{d}{dx}\arctan \left(\frac{x}{a}\right)=\frac{1}{1+\frac{x^2}{a^2}}\cdot \frac{1}{a}=\frac{a^2}{a^2+x^2}\cdot \frac{1}{a}=\frac{a}{a^2+x^2}.$ So the antiderivative of $\frac{1}{a^2+x^2}$ is $\frac{1}{a}\arctan \left(\frac{x}{a}\right)$. Then ${\int }_0^a\frac{dx}{a^2+x^2}=\frac{1}{a}[\arctan \frac{x}{a}{]}_0^a=\frac{1}{a}(\arctan 1-\arctan 0)=\frac{1}{a}\cdot \frac{\pi }{4}=\frac{\pi }{4a}.$

Look at the two red curve pieces. For $x>1$ the arcsec curve rises from $0$ toward $\frac{\pi }{2}$; for $x<-1$ it rises from $\frac{\pi }{2}$ toward $\pi$. Both pieces go uphill, so the slope is positive on each. The raw algebra gives $\frac{1}{x\sqrt{x^2-1}}$, which would be negative for $x<-1$ (since $x<0$ there). Wrapping $x$ in $|x|$ keeps the denominator positive on both sides, so the derivative stays positive everywhere arcsec is increasing: $\frac{d}{dx}\mathrm{arcsec}x=\frac{1}{|x|\sqrt{x^2-1}}.$ Check at $x=-2$: $\frac{1}{|-2|\sqrt{3}}=\frac{1}{2\sqrt{3}}>0$ ✓ (the bare-$x$ version would wrongly give $-\frac{1}{2\sqrt{3}}$).

$\frac{d}{dx}\arcsin x=\frac{1}{\sqrt{1-x^2}}.$ As $x\to 1^{-}$, the quantity $1-x^2\to 0^{+}$, so $\sqrt{1-x^2}\to 0^{+}$, and the whole fraction $\to +\infty$. In the figure the arcsin curve rears up vertically at $x=1$: the tangent line's slope blows up. Geometrically, near a ratio of $1$ the angle changes enormously for a tiny change in ratio — that steepness is the exploding derivative. The same thing happens (mirror image) as $x\to -1^{+}$.

Inside $u=x^2$, $u^{\prime }=2x$, and $u^2=x^4$. Since $u=x^2\ge 0$, we have $|u|=x^2$. $y^{\prime }=\frac{1}{|x^2|\sqrt{(x^2{)}^2-1}}\cdot 2x=\frac{2x}{x^2\sqrt{x^4-1}}=\frac{2}{x\sqrt{x^4-1}}.$ Domain: arcsec needs $|u|\ge 1$, i.e. $x^2\ge 1$, so $|x|\ge 1$. And the derivative needs $x^4>1$, i.e. $|x|>1$ strictly (equality gives a vertical tangent). Note $\frac{2}{x\sqrt{x^4-1}}$ genuinely keeps the sign of $x$: it is positive for $x>1$ and negative for $x<-1$, which is correct because $\mathrm{arcsec}(x^2)$ increases then decreases across the two branches.

Let $u=\frac{x-1}{x+1}$. Quotient rule: $u^{\prime }=\frac{(1)(x+1)-(x-1)(1)}{(x+1{)}^2}=\frac{(x+1)-(x-1)}{(x+1{)}^2}=\frac{2}{(x+1{)}^2}.$ Now the tricky simplification of $1-u^2$: $1-u^2=1-\frac{(x-1{)}^2}{(x+1{)}^2}=\frac{(x+1{)}^2-(x-1{)}^2}{(x+1{)}^2}=\frac{4x}{(x+1{)}^2}.$ (Difference of squares: $(x+1{)}^2-(x-1{)}^2=4x$.) So $\sqrt{1-u^2}=\frac{2\sqrt{x}}{|x+1|}$ (valid where $x>0$, giving $x+1>0$ so $|x+1|=x+1$). With arccos's minus sign: $y^{\prime }=-\frac{1}{\sqrt{1-u^2}}\cdot u^{\prime }=-\frac{x+1}{2\sqrt{x}}\cdot \frac{2}{(x+1{)}^2}=-\frac{1}{\sqrt{x}(x+1)}.$ Clean result: $y^{\prime }=-\frac{1}{\sqrt{x}(x+1)}$, valid for $x>0$.

Recognition sign rule ::: sin/tan/sec positive, their co-partners negative Chain-rule extra factor for $\arctan (5x)$ ::: multiply by $u^{\prime }=5$ giving $\frac{5}{1+25x^2}$ ${\int }_0^1\frac{dx}{1+x^2}$ ::: $\arctan 1-\arctan 0=\frac{\pi }{4}$ Why $|x|$ in arcsec derivative ::: keeps the slope positive on both branches Limit of $\frac{d}{dx}\arcsin x$ as $x\to 1^{-}$ ::: $+\infty$ (vertical tangent)