Composite, Implicit, and Inverse Functions

$g(c)$ is undefined due to nonexistence or continuity of $\operatorname{arcsec}(x)$ in its respective domain.

$g'(c)$ is discontinuous at $c$ as $\operatorname{arcsec}(x)$ is known to have asymptotes making differentiation impossible.

$g'(c)=\frac{1}{|c|\sqrt{c^2-1}}$ because $\operatorname{arcsec}(x)$ is continuously differentiable wherever it exists.

$g'(c)=\frac{-1}{|c|\sqrt{c^2+1}}$ because secant slopes are negative when dealing with $\operatorname{arcsec}(x)$

$\frac{-x}{(1-x^2)^{\frac{3}{2}}}$

$\frac{x}{(1-x^2)\sqrt{1-x^2}}$

$\frac{0}{\sqrt{1-x^2}}$

$\frac{-1}{\sqrt{1-x^2}}$

Solve explicitly for $y$ in terms of $x$, differentiate normally, and then raise $y'$ to power -6.

Take logarithm on both sides, apply implicit differentiation, solve for $y'$ and only then take power -6.

Differentiate once implicitly for $y'$, substitute into original equation, and solve algebraically before raising it.

Apply implicit differentiation directly on both sides, solve for $y'$, then raise it to power -6.

$(\sin(f^{-1}(0.5)))^{-1}$

$-\left(\cos(f^{-1}(0.5))\right)^{-1}$

$(\cos(f^{-1}(0.5)))^{-2}$

$-\sin(f^{-1}(0.5))$

$-\frac{5}{4 - y^2 \cdot (15+y^2)}$

$-\frac{5}{4 + y^2 \cdot (25-y^2)}$

$-\frac{5}{8+ y^3 \cdot (20-y^3)}$

$-\frac{7}{3 + y \cdot (30-y)}$

$\frac{1}{|x|\sqrt{x^2-1}}$

$\frac{-1}{|x|\sqrt{x^2-1}}$

$\frac{-\sin(x)}{|\cos(x)|}$

$(\sec^{-10}(x))' = -10\sec^{-11}(x)\cdot \sec(x)\tan ( x )$

$1$

$0$

$-\frac{\pi}{4}$

$\frac{\pi}{4}$

$2e^x\cos^{-1}(e^x)$

$2e^x\sin(e^x)$

$-\frac{2e^x}{1-e^2x}$

$-\frac{2e^x}{1+e^2x}$

$-\frac{\sqrt{2}}{4}$

$-2$

$-\sqrt{2}$

$\sqrt{2}$

$-\frac{9}{54 \cdot \sqrt{8}}$

$-\frac{\sqrt{8}}{18}$

$-\frac{11}{54 \cdot \sqrt{8}}$

$\frac{20}{27}$