Inverse Functions and Inverse Trigonometric Derivatives

(a) Need $y\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ with $\sin y=-\frac12$: $y=-\frac{\pi}{6}$.

(b) Need $y\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ with $\tan y=-1$: $y=-\frac{\pi}{4}$, not $\frac{3\pi}{4}$.

(c) $\sec y=-2$ means $\cos y=-\frac12$; in $[0,\pi]$ that is $y=\frac{2\pi}{3}$.

(d) $\sin\frac{5\pi}{6}=\frac12$, and $\arcsin\frac12=\frac{\pi}{6}$ — cancellation fails because $\frac{5\pi}{6}$ is outside the branch.

(e) With $\theta=\arcsin\frac35$, $\cos^2\theta=1-\frac{9}{25}=\frac{16}{25}$, and $\cos\theta\ge0$ on the branch, so $\cos\theta=\frac45$.

(a) $u=3x$, $u'=3$: $y'=\dfrac{3}{\sqrt{1-9x^2}}$, needing $9x^2<1$.

(b) $u=x^2$, $u'=2x$: $y'=\dfrac{2x}{1+x^4}$, and $1+x^4\ge1$ is never zero.

(c) $u=\sqrt{x}$, $u'=\dfrac{1}{2\sqrt{x}}$, $u^2=x$, and $\arccos$ carries a minus sign:

$y'=-\frac{1}{2\sqrt{x}\,\sqrt{1-x}}=-\frac{1}{2\sqrt{x-x^2}}$

which needs $x>0$ (for $u'$) and $x<1$ (for the radical).

(a) $f'(x)=2-\sin x\ge2-1=1>0$ for every $x$, so $f$ is strictly increasing on the interval $\mathbb{R}$, hence one-to-one; it is continuous and differentiable everywhere.

(b) Find $a$ with $f(a)=1$: $f(0)=0+\cos 0=1$, so $g(1)=0$ and the point of tangency is $(1,0)$. Then $f'(0)=2$, so

$g'(1)=\frac{1}{f'(0)}=\frac12,\qquad y=0+\tfrac12(x-1)=\frac{x-1}{2}$

Check by reflection: the tangent to $f$ at $(0,1)$ has slope $2$, and its mirror image has slope $\frac12$.

Product rule with $u=\arcsin x$ and $v=\left(1-x^2\right)^{1/2}$, so $u'=\dfrac{1}{\sqrt{1-x^2}}$ and $v'=\dfrac{-x}{\sqrt{1-x^2}}$:

$f'(x)=\frac{1}{\sqrt{1-x^2}}\cdot\sqrt{1-x^2}+\arcsin x\cdot\frac{-x}{\sqrt{1-x^2}}=1-\frac{x\arcsin x}{\sqrt{1-x^2}}$

Both $u'$ and $v'$ need $1-x^2>0$, so the formula holds on $(-1,1)$; at $x=\pm1$ the quotient runs to $-\infty$ and the graph meets the axis vertically.

$f'(0)=1-0=1$ and $f(0)=0$.

With $u=3x$, $u'=3$:

$y'=\frac{3}{|3x|\sqrt{(3x)^2-1}}=\frac{3}{3|x|\sqrt{9x^2-1}}=\frac{1}{|x|\sqrt{9x^2-1}}$

using $|3x|=3|x|$. The formula needs $|3x|>1$, i.e. $|x|>\frac13$.

At $x=-1$: $y'(-1)=\dfrac{1}{1\cdot\sqrt{8}}=\dfrac{1}{2\sqrt2}=\dfrac{\sqrt2}{4}\approx0.354$.

Sign check: dropping the bars would give $-\frac{\sqrt2}{4}$, but $\operatorname{arcsec}(3x)=\arccos\left(\frac{1}{3x}\right)$ is increasing at $x=-1$, so the slope must be positive.

(a) With $u=\frac1x$, $u'=-\frac{1}{x^2}$, and multiplying the second fraction top and bottom by $x^2$:

$F'(x)=\frac{1}{1+x^2}+\frac{-1/x^2}{1+1/x^2}=\frac{1}{1+x^2}-\frac{1}{x^2+1}=0$

(b) The domain $(-\infty,0)\cup(0,\infty)$ is not an interval, so the zero-derivative corollary of the MVT gives one constant on each piece only. Test each: $F(1)=\frac{\pi}{4}+\frac{\pi}{4}=\frac{\pi}{2}$, and $F(-1)=-\frac{\pi}{4}-\frac{\pi}{4}=-\frac{\pi}{2}$. The constants genuinely differ.