역삼각함수의 미분

First, let's interpret the given information: when the input of our function is $2x+1$ for some $x$, then its output is $\dfrac{5x-9}{3x+7}$, for that same $x$. We're asked to evaluate $f^{-1}(7)$, which is the number $y$ with the property that $f(y)=7$. If the output of our function is 7, that means

$$\begin{aligned} 7&=\frac{5x-9}{3x+7} \\ \text{and so} \\ 7(3x+7)&=5x-9\\ x&=-\frac{29}{8} \\ \text{So, when $x=-\dfrac{29}{8}$, our equation $f(2x+1)=\dfrac{5x-9}{3x+7}$ becomes:} \\ f\left(2\cdot\frac{-29}{8}+1\right)&=\dfrac{5\cdot\frac{-29}{8}-9}{3\cdot\frac{-29}{8}+7} \\ \text{Or, equivalently:} \\ f\left(-\frac{25}{4}\right)&=7 \end{aligned}$$

Therefore, $f^{-1}(7)=-\dfrac{25}{4}$.

If $f^{-1}(y)=0$, that means $f(0)=y$. So, we want to find out what we plug into $f^{-1}$ to get 0. Since we only know $f^{-1}$ in terms of a variable $x$, let's figure out what $x$ gives us an output of 0:

$$\begin{aligned} \frac{2x+3}{x+1}&=0\\ 2x+3&=0\\ x&=-\frac{3}{2} \\ \text{Now, the equation $f^{-1}(4x-1)=\dfrac{2x+3}{x+1}$ with $x=\dfrac{-3}{2}$ tells us:} \\ f^{-1}\left(4\cdot\frac{-3}{2}-1\right)&=\frac{2\cdot\frac{-3}{2}+3}{\frac{-3}{2}+1} \\ \text{Or, equivalently:} \\ f^{-1}(-7)&=0 \end{aligned}$$

Therefore, $f(0)=-7$.

• Solution 1: We begin by differentiating implicitly. Following the usual convention, we use $y'$ to mean $y'(x)$.

$$\begin{aligned} \arcsin(x+2y)&=x^2+y^2 \qquad\text{Using the chain rule:}\\ \frac{1+2y'}{\sqrt{1-(x+2y)^2}}&=2x+2yy'\\ \frac{1}{\sqrt{1-(x+2y)^2}}+\frac{2y'}{\sqrt{1-(x+2y)^2}}&=2x+2yy'\\ \frac{2y'}{\sqrt{1-(x+2y)^2}}-2yy'&=2x-\frac{1}{\sqrt{1-(x+2y)^2}}\\ y'\left(\frac{2}{\sqrt{1-(x+2y)^2}}-2y\right)&=2x-\frac{1}{\sqrt{1-(x+2y)^2}}\\ y'&=\frac{2x-\frac{1}{\sqrt{1-(x+2y)^2}}}{\frac{2}{\sqrt{1-(x+2y)^2}}-2y} \left(\frac{\sqrt{1-(x+2y)^2}}{\sqrt{1-(x+2y)^2}}\right)\\ y'&=\frac{2x\sqrt{1-(x+2y)^2}-1}{2-2y\sqrt{1-(x+2y)^2}} \end{aligned}$$

• Solution 2: We begin by taking the sine of both sides of the equation.

$$\begin{aligned} \arcsin(x+2y)&=x^2+y^2\\ x+2y&=\sin(x^2+y^2) \\ \text{Now, we differentiate implicitly.} \\ 1+2y'&=\cos(x^2+y^2)\cdot(2x+2yy')\\ 1+2y'&=2x\cos(x^2+y^2)+2yy'\cos(x^2+y^2)\\ 2y'-2yy'\cos(x^2+y^2)&=2x\cos(x^2+y^2)-1\\ y'\left(2-2y\cos(x^2+y^2)\right)&=2x\cos(x^2+y^2)-1\\ y'&=\frac{2x\cos(x^2+y^2)-1}{2-2y\cos(x^2+y^2)} \end{aligned}$$

• We used two different methods, and got two answers that look pretty different. However, the answers ought to be equivalent. To see this, we remember that for all values of $x$ and $y$ that we care about (those pairs $(x,y)$ in the domain of our curve), the equality $$\arcsin(x+2y)=x^2+y^2$$ holds. Drawing a triangle:

where the adjacent side (in red) come from the Pythagorean Theorem. Then, $\cos(x^2+y^2)=\sqrt{1-(x+2y)^2}$, so using our second solution:

$$\begin{aligned} y'&=\frac{2x\cos(x^2+y^2)-1}{2-2y\cos(x^2+y^2)}\\ &=\frac{2x\sqrt{1-(x+2y)^2}-1}{2-2y\sqrt{1-(x+2y)^2}} \end{aligned}$$

which is exactly the answer from our first solution.