Composite, Implicit, and Inverse Functions

2

-0.5

-2

0.5

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 $e^{g(x)}$.

$-7$

$-49$

$\frac{1}{7}$

$7$

The original function, $f(x)$, cannot have any discontinuities or sharp corners on its domain.

The original function, $f(x)$, must be monotonically increasing or decreasing and have a non-zero derivative everywhere on its domain.

There are no specific conditions; all functions have continuous and differentiable inverses.

The original function, $f(x)$, can have zero as its derivative anywhere on its domain.

$1$

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

$\frac{1}{2}$

Undefined

$\left( \arcsin(u)' \right) = u' \cdot \sqrt{1-u^2}$

$\left( \arcsin(u)' \right) = \frac{u'}{\sqrt{1-u^2}}$

$\left( \arcsin(u)' \right) = -\dfrac{u'}{\sqrt{u^2-1}}$

$\left( \arcsin(u)' \right) = \sin(u')$

$0$

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

$-\frac{1}{2}$

$\sqrt{\frac{3}{4}}$

$\frac{1}{12}$

$\frac{1}{9}$

$\frac{1}{6}$

$\frac{1}{4}$

$\frac{1}{2}$

$1$

$\frac{\pi}{4}$

Undefined

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