역함수의 미분법

$y = -4(x-\sqrt{3}/4)+\pi/6$

(a) By the chain rule, $$f'(x) = \frac{1}{2\arctan(x) + 3\arcsin(x) + 5} \cdot \frac{d}{dx}\left( 2\arctan(x) + 3\arcsin(x) + 5\right)\,,$$ and by the sum and constant multiple rules, it follows that $$f'(x) = \frac{1}{2\arctan(x) + 3\arcsin(x) + 5} \cdot \left(\frac{2}{1+x^2} + \frac{3}{\sqrt{1-x^2}}\right)\,.$$ (b) By the chain rule applied to the outermost function, $$r'(z) = \frac{1}{1+\left(\ln(\arcsin(z))\right)^2} \cdot \frac{d}{dz} \left( \ln(\arcsin(z)) \right)\,.$$ Applying the chain rule again to evaluate $\frac{d}{dz} \left( \ln(\arcsin(z)) \right)$, $$r'(z) = \frac{1}{1+\left(\ln(\arcsin(z))\right)^2} \cdot \left( \frac{1}{\arcsin(z)} \right) \cdot \frac{d}{dz}\left(\arcsin(z)\right)\,.$$ Finally, using the basic rule for $\frac{d}{dz}\left(\arcsin(z)\right)$, $$r'(z) = \frac{1}{1+\left(\ln(\arcsin(z))\right)^2} \cdot \left( \frac{1}{\arcsin(z)} \right) \cdot \frac{1}{\sqrt{1-z^2}}\,.$$ (c) First, by the product rule, $$q'(t) = \arctan^2(3t) \cdot \frac{d}{dt}\left(\arcsin^4(7t)\right) + \arcsin^4(7t) \cdot \frac{d}{dt}\left(\arctan^2(3t)\right)\,.$$ Next, we recall that $\arcsin^4(7t) = \left(\arcsin(7t)\right)^4$ and $\arctan^2(3t) = \left( \arctan(3t) \right)^2$, and apply the chain rule to find the derivative of each of these functions to get $$q'(t) = \arctan^2(3t) \cdot \left[4\arcsin^3(7t) \frac{d}{dt}\left( \arcsin(7t) \right)\right] + \arcsin^4(7t) \cdot \left[2\arctan(3t) \frac{d}{dt}\left(\arctan(3t)\right) \right]\,.$$ Finally, applying the chain rule one more time to evaluate the remaining derivatives, $$q'(t) = \arctan^2(3t) \cdot \left[4\arcsin^3(7t) \left( \frac{7}{\sqrt{1-(7t)^2}} \right)\right] + \arcsin^4(7t) \cdot \left[2\arctan(3t) \left(\frac{3}{1+(3t)^2}\right) \right]\,.$$ (d) By the chain rule, $$g'(v) = \frac{1}{ \frac{\arctan(v)}{\arcsin(v) + v^2} } \cdot \frac{d}{dv}\left[ \frac{\arctan(v)}{\arcsin(v) + v^2} \right]$$ Thus, by the quotient rule, $$g'(v) = \frac{1}{\frac{\arctan(v)}{\arcsin(v) + v^2}} \cdot \frac{(\arcsin(v) + v^2) \cdot \frac{1}{1+v^2} - \arctan(v) \cdot \left(\frac{1}{\sqrt{1-v^2}} + 2v \right)}{(\arcsin(v) + v^2)^2}$$

(a) $f$ is an invertible function because its graph passes the horizontal line test. (b) To find a formula for $g = f^{-1}$, we take the equation $y = \frac{1}{4}x^3 + 4$ and solve for $x$ in terms of $y$. Doing so, we find that $y - 4 = \frac{1}{4}x^3$, so $x^3 = 4(y-4)$, and thus $$x = \sqrt[3]{4(y-4)} = f^{-1}(y)\,.$$ Writing the inverse as a function of $x$, we have found that $f^{-1}(x) = g(x) = \sqrt[3]{4x-16}$. (c) First, $f'(x) = \frac{3}{4}x^2$, and thus $f'(2) = 3$. Next, $g'(x) = \frac{1}{3}(4x-16)^{-2/3} \cdot 4$, so that $g'(6) = \frac{1}{3} \cdot 8^{-2/3} \cdot 4 = \frac{1}{3} \cdot \frac{1}{4} \cdot 4 = \frac{1}{3}$. What is special about these two derivative values is that they happen to be reciprocals. Noting that $f(2) = 3$ and $g(3) = 2$, we know that $g'(3) = \frac{1}{f'(2)}$ by the rule for the derivative of an inverse function, and this matches our finding above that $g'(3) = \frac{1}{3}$.

(a) The function $h$ is invertible because its graph passes the horizontal line test. (b) It does not seem possible to solve the equation $y = x + \sin(x)$ for $x$ in terms of $y$. (c) By the rule for the derivative of an invertible function, we have $(h^{-1})'(\frac{\pi}{2} + 1) = \frac{1}{h'\left(\frac{\pi}{2} \right)}$. Thus, we first compute $h'(x)$ and find that $h'(x) = 1 + \cos(x)$. Since $h'\left(\frac{\pi}{2} \right) = 1 + \cos\left(\frac{\pi}{2} \right) = 1$, it follows that $(h^{-1})'(\frac{\pi}{2} + 1) = \frac{1}{1} = 1$.