Composite, Implicit, and Inverse Functions

$\arctan(x) \cdot \ln(|x|)$

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

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

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

$g(c)$ 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 $\operatorname{arcsec}(x)$

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

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

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

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

$12\sin(3x)$

$-\frac{12}{\sqrt{1-(3x)^2}}$

$4 - \cos^{-1}(3x)$

$-4\sin(3x)$

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.

By applying Rolle’s Theorem

By using L'Hôpital's Rule

By using the Main Theorem for Inverse Functions

By applying the Squeeze Theorem

$-\ln(-)$

$z \cot(-)$

$\tanh()$

$\ln(|z| \pm \sqrt{z^2+1})$

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

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

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

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

$\dfrac{-7/8}{{(17)}^{5/4}(4+x)}$

$\dfrac{-15/(32+8x)}{{(17)}^{5/4}(4+x)}$

$\dfrac{-7/(64+16x)}{{(17)}^{5/4}}$

$\dfrac{-7/64}{{17}^{5/4}(4+x)}$

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

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

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

$\sqrt{1-x^2}$