Composite, Implicit, and Inverse Functions

2

-0.5

-2

0.5

$\frac{1}{e}$

$0$

$e$

$1$

No, it doesn't vanish given $q'(q^{-1}(x))$ equals inverse direct slope $q$ at $x$

Yes, disappearing occurs because $q'(q^{-1}(x))$ equates inverse reciprocal slope $q$ at $x$

Vanishing doesn't follow since the existence of $(\frac{dq}{dx})^{-1}$ relies upon direct relationship between $q$ and its slope

Disappearance isn't guaranteed unless additionally provided info on behavior of $q$ around specified point

Quotient rule because it deals with division between two functions which applies indirectly here.

Implicit differentiation because it allows for differentiation with respect to $y$ indirectly through $g^{-4}(y)$.

Power rule because it applies directly to functions raised to a power like $g^{-4}(y)$.

Chain rule because it can be used when one function is composed within another like $e^{g(x)}$.

$1$

$\frac{\sqrt{3}}{2}$

$\frac{1}{2}$

$\frac{1}{\sqrt{3}}$

It occurs wherever <math-inline>r^{-}</math-inline> attains its greatest or least value unrelated to $e^{\pi}$.

The extremum emerges at $({r^{-}}')(e^{\pi})$ when $r$ is non-differentiable.

It's located at $({r^{-}}')(e^{\pi})=0$.

No such extremum exists because `$r^{-}$'s derivative never zeroes out.

$\frac{1}{2}$

$\frac{2}{\sqrt{3}}$

$\frac{\sqrt{3}}{2}$

Undefined

$-3$

$3$

$\frac{1}{3}$

$-\frac{1}{3}$

$\frac{1}{7}$

$\frac{-1}{7}$

-7

7

$(h^{-1})' (b)=(h')^{-1}(b)$ where h(b)=a.

$(h^{-1})'(b)=\frac{-1}{h'(h^{-1}(b))}$ where h(b)=a.

$(h^{-1})'(b)=\frac{1}{h'(h^{-1}(b))}$ where h(b)=a.

$(h^{-1})'(b)=\frac{1}{(h')^{-1}(b)}$ where h(b)=a.