§2.6 Derivatives of inverse functions (continued)

Inverse trigonometric functions and their derivatives

The link between the derivative of a function and the derivative of its inverse

Summary

Let $r(x) = \arctan(x)$. Rewriting as $\tan(r(x)) = x$ and differentiating both sides gives $\sec^2(r(x)) \cdot r'(x) = 1$. Solving for $r'(x)$ produces

With $\theta = \arctan(x)$, a right triangle has legs $1$ and $x$ and hypotenuse $\sqrt{1+x^2}$. Reading cosine off this triangle gives $\cos(\arctan(x)) =$

Which of the following is the derivative of $r(x) = \arctan(x)$?

For $f(x) = x^3\arctan(x) + e^x\ln(x)$, which structure does $f'(x)$ have?

For $p(t) = 2^{t\arcsin(t)}$, which structure does $p'(t)$ have?

For $h(z) = (\arcsin(5z) + \arctan(4-z))^{27}$, which structure does $h'(z)$ have?

Using the identity $\cot(\arctan(y)) = \frac{1}{y}$, compute $s'(2)$ for $s(y) = \cot(\arctan(y))$ as a decimal.

For $m(v) = \ln(\sin^2(v)+1)$, which structure does $m'(v)$ have?

For $g(w) = \arctan\left(\frac{\ln(w)}{1+w^2}\right)$, which combination of rules does the derivative require?

For $f(x) = \frac{1}{4}x^3 + 4$ with inverse $g$, we know $f(2) = 6$, $f'(2) = 3$, and $g'(6) = \frac{1}{3}$. The relationship between $f'(2)$ and $g'(6)$ is that they are

For $f(x) = \frac{1}{4}x^3 + 4$ with inverse $g(x) = \sqrt[3]{4x-16}$, compute $g'(6)$ as a fraction.

A graph shows $y = f(x)$ passing through $(1, -1)$ with tangent slope $f'(1) \approx 2$. For the inverse function, $(f^{-1})'(-1)$ is approximately

If $h$ is invertible and differentiable with $h(a) = b$, then $(h^{-1})'(b)$ equals

Let $h(x) = x + \sin(x)$, which passes through $(\frac{\pi}{2}, \frac{\pi}{2}+1)$ and satisfies $h'(x) = 1 + \cos(x)$. Compute $(h^{-1})'(\frac{\pi}{2}+1)$.

For $r(z) = \arctan(\ln(\arcsin(z)))$, how many chain rule layers fire in computing $r'(z)$?

The derivative of $\arcsin(x)$ is $\frac{1}{\sqrt{1-x^2}}$ on the interval $-1 \lt x \lt 1$.