Differentiating Inverse Functions - Calculus 1

1. Let $f(x) = x^5 + 3x - 2$. What is the equation of the tangent line to the graph of $y=f^{-1}(x)$ at the point where $x=2$?
   Answer: $y - 1 = \frac{1}{8}(x - 2)$ To find the equation of the tangent line, we need a point and a slope. The point is on the graph of $y=f^{-1}(x)$ and has an x-coordinate of 2. Let $g(x)=f^{-1}(x)$. 1. Find the point: We need $(2, g(2))$. To find $g(2)$, we must solve $f(y)=2$. So, $y^5 + 3y - 2 = 2$, which simplifies to $y^5 + 3y - 4 = 0$. By inspection, $y=1$ is the solution. Thus, $g(2)=1$. The point of tangency is $(2, 1)$. 2. Find the slope: The slope is $g'(2)$. Using the formula, $g'(2) = \frac{1}{f'(g(2))} = \frac{1}{f'(1)}$. 3. Calculate $f'(x)$. $f'(x) = 5x^4 + 3$. 4. Evaluate $f'(1) = 5(1)^4 + 3 = 8$. 5. The slope is $g'(2) = 1/8$. 6. Using the point-slope form $y - y_1 = m(x - x_1)$, we get $y - 1 = \frac{1}{8}(x - 2)$.
2. The functions $f$ and $g$ are inverses. The point $(2, 5)$ lies on the graph of $y=f(x)$, and the slope of the tangent line to the graph of $f$ at this point is $-\frac{3}{4}$. What is the equation of the tangent line to the graph of $y=g(x)$ at $x=5$?
   Answer: $y - 2 = -\frac{4}{3}(x - 5)$ We are given $f(2)=5$ and $f'(2) = -3/4$. Since $g=f^{-1}$, the point $(5, 2)$ lies on the graph of $g$. The slope of the tangent line to $g$ at $x=5$ is $g'(5)$. Using the inverse function rule, $g'(5) = \frac{1}{f'(g(5))}$. Since $f(2)=5$, we have $g(5)=2$. Therefore, $g'(5) = \frac{1}{f'(2)} = \frac{1}{-3/4} = -\frac{4}{3}$. The equation of the tangent line to $g$ at the point $(5, 2)$ is $y - 2 = -\frac{4}{3}(x - 5)$.
3. Let $g$ be the inverse of a twice-differentiable function $f$. Given $f(1)=2$, $f'(1)=3$, and $f''(1)=4$, find $g''(2)$.
   Answer: $-4/27$ The formula for the second derivative of an inverse function is $g''(x) = -\frac{f''(g(x))}{[f'(g(x))]^3}$. To evaluate $g''(2)$, we first need $g(2)$. Since $f(1)=2$, we have $g(2)=1$. Substituting $x=2$ and $g(2)=1$ into the formula, we get $g''(2) = -\frac{f''(1)}{[f'(1)]^3}$. We are given $f'(1)=3$ and $f''(1)=4$. Plugging these values in: $g''(2) = -\frac{4}{(3)^3} = -\frac{4}{27}$.
4. Let $f$ be an invertible function and let $g=f^{-1}$. The line $y=2x+5$ is tangent to the graph of $f$ at the point $(-1, 3)$. Which of the following is an equation for the line tangent to the graph of $g$ at the point $(3, -1)$?
   Answer: $y = \frac{1}{2}x - \frac{5}{2}$ When you encounter problems involving inverse functions and their tangent lines, remember that inverse functions "flip" both coordinates and slopes in specific ways. Since the line $y = 2x + 5$ is tangent to $f$ at $(-1, 3)$, we know that $f(-1) = 3$ and $f'(-1) = 2$. Because $g = f^{-1}$, the point $(-1, 3)$ on $f$ becomes the point $(3, -1)$ on $g$ - the coordinates are swapped. The key insight is the relationship between derivatives of inverse functions: $(f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}$. At the point $(3, -1)$ on $g$, we have $g'(3) = \frac{1}{f'(g(3))} = \frac{1}{f'(-1)} = \frac{1}{2}$. So the tangent line to $g$ at $(3, -1)$ has slope $\frac{1}{2}$ and passes through $(3, -1)$. Using point-slope form: $y - (-1) = \frac{1}{2}(x - 3)$, which simplifies to $y = \frac{1}{2}x - \frac{5}{2}$. Looking at the wrong answers: A) uses slope 2 instead of $\frac{1}{2}$, missing the reciprocal relationship. B) has the correct reciprocal slope but uses $-\frac{1}{2}$, incorrectly making it negative. D) has the right slope but wrong y-intercept, likely from an arithmetic error when solving for the line equation. Study tip: For inverse function tangent lines, always remember: swap the point coordinates and take the reciprocal of the slope. The derivative relationship $(f^{-1})'(a) = \frac{1}{f'(f^{-1}(a))}$ is your key formula.
5. The function $f(x) = \int_2^x \sqrt{1+t^3} \, dt$ is one-to-one for $x \ge -1$. What is the value of $(f^{-1})'(0)$?
   Answer: $\frac{1}{3}$ We need $(f^{-1})'(0) = \frac{1}{f'(f^{-1}(0))}$. First, find $b = f^{-1}(0)$ by solving $f(b)=0$. $\int_2^b \sqrt{1+t^3} \, dt = 0$ when the limits of integration are equal, so $b=2$. Next, we find $f'(x)$ using the Fundamental Theorem of Calculus, Part 1: $f'(x) = \sqrt{1+x^3}$. Now, evaluate $f'$ at $b=2$: $f'(2) = \sqrt{1+2^3} = \sqrt{9} = 3$. Finally, $(f^{-1})'(0) = \frac{1}{f'(2)} = \frac{1}{3}$.
6. Let $f(x) = x^3 + 4x - 1$. If $g(x) = f^{-1}(x)$, what is the value of $g'(4)$?
   Answer: $1/7$ The formula for the derivative of an inverse function is $g'(a) = \frac{1}{f'(g(a))}$. Here, $a=4$. First, we need to find $g(4) = f^{-1}(4)$. This is the value $x$ such that $f(x) = 4$. We solve $x^3 + 4x - 1 = 4$, which simplifies to $x^3 + 4x - 5 = 0$. By inspection, $x=1$ is the solution, since $1^3 + 4(1) - 5 = 0$. So, $g(4) = 1$. Next, we find the derivative of $f(x)$, which is $f'(x) = 3x^2 + 4$. We evaluate this derivative at $x=g(4)=1$: $f'(1) = 3(1)^2 + 4 = 7$. Finally, we apply the formula: $g'(4) = \frac{1}{f'(1)} = \frac{1}{7}$.
7. Let $f(x) = x^3+2x+1$ and let $g(x) = f^{-1}(x)$. If $k(x) = x \cdot g(x)$, what is the value of $k'(4)$?
   Answer: $9/5$ To find $k'(4)$, we first need to find the derivative of $k(x) = x \cdot g(x)$ using the product rule: $k'(x) = (1) \cdot g(x) + x \cdot g'(x)$. Now, we evaluate this at $x=4$: $k'(4) = g(4) + 4g'(4)$. We need to find two values: $g(4)$ and $g'(4)$. 1. Find $g(4) = f^{-1}(4)$. We need to solve $f(x)=4$. $x^3+2x+1=4$, which simplifies to $x^3+2x-3=0$. By inspection, $x=1$ is the solution ($1^3+2(1)-3=0$). So, $g(4)=1$. 2. Find $g'(4)$. We use the formula $g'(4) = \frac{1}{f'(g(4))} = \frac{1}{f'(1)}$. First, find $f'(x)$: $f'(x) = 3x^2+2$. Then evaluate at $x=1$: $f'(1) = 3(1)^2+2 = 5$. So, $g'(4) = 1/5$. 3. Substitute these values back into the expression for $k'(4)$: $k'(4) = g(4) + 4g'(4) = 1 + 4(1/5) = 1 + 4/5 = 9/5$.
8. Let $f(x) = \int_2^x \sqrt{t^3+1} \, dt$. For $x>0$, $f$ is invertible. Let $g=f^{-1}$. Find $g'(0)$.
   Answer: $1/3$ We need to find $g'(0) = (f^{-1})'(0) = \frac{1}{f'(f^{-1}(0))}$. First, we must find the value of $b = f^{-1}(0)$, which means we solve $f(b)=0$. So, $\int_2^b \sqrt{t^3+1} \, dt = 0$. An integral is zero if its upper and lower limits are equal, so $b=2$. This means $f^{-1}(0)=2$. Next, we need $f'(x)$. By the Fundamental Theorem of Calculus Part 1, $f'(x) = \frac{d}{dx} \int_2^x \sqrt{t^3+1} \, dt = \sqrt{x^3+1}$. We need to evaluate this at $x=f^{-1}(0)=2$. So, $f'(2) = \sqrt{2^3+1} = \sqrt{9} = 3$. Finally, $g'(0) = \frac{1}{f'(2)} = \frac{1}{3}$.
9. Let $f(x) = \tan(x)$ for $-\pi/2 < x < \pi/2$, and let $g(x)=f^{-1}(x)$. Consider the function $h(x) = x^2 g(x)$. Find $h'(1)$.
   Answer: $\frac{\pi+1}{2}$ This problem tests your understanding of inverse function derivatives and the product rule. When you see a function involving an inverse function that you need to differentiate, remember that you'll likely need the inverse function derivative formula. First, identify what $g(x)$ is: since $f(x) = \tan(x)$, we have $g(x) = f^{-1}(x) = \arctan(x)$. So $h(x) = x^2 \arctan(x)$. To find $h'(x)$, use the product rule: $h'(x) = 2x \arctan(x) + x^2 \cdot \frac{d}{dx}[\arctan(x)]$. Since $\frac{d}{dx}[\arctan(x)] = \frac{1}{1+x^2}$, we get: $h'(x) = 2x \arctan(x) + \frac{x^2}{1+x^2}$ At $x = 1$: $h'(1) = 2(1)\arctan(1) + \frac{1^2}{1+1^2} = 2 \cdot \frac{\pi}{4} + \frac{1}{2} = \frac{\pi}{2} + \frac{1}{2} = \frac{\pi+1}{2}$ Choice A, $\frac{\pi+4}{2}$, likely comes from incorrectly using $\arctan(1) = \pi$ instead of $\frac{\pi}{4}$. Choice B, $\frac{1}{2}$, results from forgetting the $2x\arctan(x)$ term entirely and only computing the second term. Choice C, $\frac{5}{2}$, might arise from computational errors or misremembering that $\arctan(1) = 1$ instead of $\frac{\pi}{4}$. The correct answer is D. Study tip: Always memorize that $\arctan(1) = \frac{\pi}{4}$ and $\frac{d}{dx}[\arctan(x)] = \frac{1}{1+x^2}$. These appear frequently in calculus problems involving inverse trig functions.
10. Let $f(x) = x^3 + 2x - 1$. If $g$ is the inverse function of $f$, what is the value of $g'(2)$?
    Answer: $1/5$ The formula for the derivative of an inverse function is $g'(a) = \frac{1}{f'(g(a))}$. Here, $a=2$. First, we need to find $g(2)$. If $g(2)=x$, then $f(x)=2$. So we solve $x^3 + 2x - 1 = 2$, which simplifies to $x^3 + 2x - 3 = 0$. By inspection, $x=1$ is a solution. Thus, $g(2)=1$. Next, we find the derivative of $f(x)$, which is $f'(x) = 3x^2 + 2$. Now we evaluate $f'(g(2)) = f'(1) = 3(1)^2 + 2 = 5$. Finally, $g'(2) = \frac{1}{f'(1)} = \frac{1}{5}$.
11. Let $g(x)$ be the inverse of $f(x) = x e^{x^2}$. Find $g'(e)$.
    Answer: $\frac{1}{3e}$ When you encounter a problem asking for the derivative of an inverse function, you need to use the inverse function derivative formula: if $g(x)$ is the inverse of $f(x)$, then $g'(a) = \frac{1}{f'(g(a))}$. First, find where $f(x) = e$. Since $f(x) = xe^{x^2}$, you need to solve $xe^{x^2} = e$. By inspection or careful analysis, $x = 1$ works: $1 \cdot e^{1^2} = e$. Therefore, $g(e) = 1$. Next, find $f'(x)$ using the product rule: $f'(x) = e^{x^2} + x \cdot e^{x^2} \cdot 2x = e^{x^2}(1 + 2x^2)$. At $x = 1$: $f'(1) = e^1(1 + 2 \cdot 1^2) = e(1 + 2) = 3e$. Therefore: $g'(e) = \frac{1}{f'(g(e))} = \frac{1}{f'(1)} = \frac{1}{3e}$ Choice D is correct. Choice A gives $\frac{1}{e^{e}(1+2e^2)}$, which incorrectly evaluates $f'$ at $x = e$ instead of $x = 1$. Choice B gives $3e$, which is actually $f'(1)$ rather than its reciprocal—a common error of forgetting the inverse relationship. Choice C gives $\frac{1}{e}$, which might result from incorrectly computing $f'(1) = e$ instead of $3e$. Study tip: Always remember that $g'(a) = \frac{1}{f'(g(a))}$—you evaluate $f'$ at the output of the inverse function, not at the input. Finding $g(a)$ first is crucial.
12. Let $g$ be the inverse of a differentiable function $f$. The tangent line to the graph of $f$ at the point where $x=3$ is given by the equation $y = 5x - 11$. Which of the following statements about the graph of $g$ must be true?
    Answer: The graph of $g$ has a tangent line at $x=4$ with a slope of $1/5$. When you encounter inverse functions and tangent lines, the key relationship to remember is that inverse functions "swap" x and y coordinates, and their derivatives are reciprocals at corresponding points. From the tangent line equation $y = 5x - 11$ at $x = 3$, you can extract crucial information: the slope is $f'(3) = 5$, and the point on $f$ is $(3, 4)$ since $y = 5(3) - 11 = 4$. This means $f(3) = 4$. Since $g$ is the inverse of $f$, the point $(3, 4)$ on $f$ corresponds to the point $(4, 3)$ on $g$. The derivative formula for inverse functions tells us that $g'(x) = \frac{1}{f'(g(x))}$. At $x = 4$: $g'(4) = \frac{1}{f'(g(4))} = \frac{1}{f'(3)} = \frac{1}{5}$. Therefore, $g$ has a tangent line at $x = 4$ with slope $\frac{1}{5}$, making D correct. A is wrong because it places the tangent line at $x = 3$ instead of $x = 4$. B incorrectly assumes the tangent line equation can be found by simply manipulating the original equation algebraically. C uses the wrong slope—it should be $\frac{1}{5}$, not $5$. Study tip: Always identify corresponding points first (swap coordinates), then apply the reciprocal derivative relationship. The x-value where you evaluate $g'$ is the y-value from the original function.
13. Let $f$ be a one-to-one differentiable function such that $f(1)=3$, $f(3)=5$, $f'(1)=2$, and $f'(3)=4$. Let $g=f^{-1}$. What is the value of $(f \circ g)'(5)$?
    Answer: $1$ This problem tests your understanding of composite functions and inverse function derivatives. When you see a composition like $(f \circ g)'$ where $g = f^{-1}$, recognize that you're dealing with the derivative of $f(f^{-1}(x))$. Let's find $(f \circ g)'(5) = (f(f^{-1}(x)))'$ evaluated at $x = 5$. Since $f(f^{-1}(x)) = x$ by the definition of inverse functions, we have $(f \circ g)'(x) = \frac{d}{dx}[x] = 1$ for all values in the domain. Therefore, $(f \circ g)'(5) = 1$, which is answer choice B. Now let's examine why the other answers are incorrect: A) $\frac{1}{4}$ likely comes from incorrectly applying the inverse function derivative formula $(f^{-1})'(y) = \frac{1}{f'(f^{-1}(y))}$. Since $f^{-1}(5) = 3$ and $f'(3) = 4$, you might calculate $(f^{-1})'(5) = \frac{1}{4}$. However, this gives you $g'(5)$, not $(f \circ g)'(5)$. C) 4 represents $f'(3)$, which you might get by incorrectly thinking $(f \circ g)'(5) = f'(f^{-1}(5)) = f'(3) = 4$. This misses the chain rule application. D) 8 might result from incorrectly multiplying $f'(3) = 4$ by something else, perhaps attempting an erroneous chain rule calculation. Study tip: Remember that $f \circ f^{-1} = x$ always, so its derivative is always 1. Don't overthink these problems—the composition of a function with its inverse always simplifies dramatically.
14. An implicit function is defined by the equation $y^3 + xy = 4$. This equation defines $y$ as a differentiable function of $x$, say $y=f(x)$, and also defines $x$ as a differentiable function of $y$, say $x=g(y)$. Since $g=f^{-1}$, what is the value of $g'(3)$?
    Answer: $-\frac{58}{9}$ We want to find $g'(3) = (f^{-1})'(3)$, which is $\frac{1}{f'(f^{-1}(3))}$. First, find $x = f^{-1}(3)$ by setting $y=3$ in the original equation: $3^3 + x(3) = 4 \implies 27 + 3x = 4 \implies 3x = -23 \implies x = -23/3$. So, we need $\frac{1}{f'(-23/3)}$. Now, find $f'(x) = dy/dx$ by implicit differentiation with respect to $x$: $3y^2 \frac{dy}{dx} + y + x \frac{dy}{dx} = 0 \implies \frac{dy}{dx}(3y^2+x) = -y \implies \frac{dy}{dx} = \frac{-y}{3y^2+x}$. Evaluate this at the point $(-23/3, 3)$: $f'(-23/3) = \frac{-3}{3(3^2) + (-23/3)} = \frac{-3}{27 - 23/3} = \frac{-3}{58/3} = -\frac{9}{58}$. The derivative of the inverse is the reciprocal: $g'(3) = \frac{1}{-9/58} = -\frac{58}{9}$.
15. Let $g(x)$ be the inverse of $f(x) = e^x + 2x$. What is the value of $g'(1)$?
    Answer: $\frac{1}{3}$ We need to compute $g'(1) = (f^{-1})'(1)$. The formula is $(f^{-1})'(1) = \frac{1}{f'(f^{-1}(1))}$. First, find $b = f^{-1}(1)$ by solving $f(b)=1$, which is $e^b + 2b = 1$. By inspection, $b=0$ is the solution. Next, compute the derivative $f'(x) = e^x + 2$. Evaluate the derivative at $b=0$: $f'(0) = e^0 + 2 = 1 + 2 = 3$. Finally, the derivative of the inverse at $x=1$ is the reciprocal of this value: $g'(1) = \frac{1}{f'(0)} = \frac{1}{3}$.
16. The function $f(x) = x^5 + x^3 + 2x$ is one-to-one. If $L$ is the tangent line to the graph of $y=f^{-1}(x)$ at the point where $y=1$, what is the slope of $L$?
    Answer: $1/10$ The slope of the tangent line to $y=f^{-1}(x)$ is $(f^{-1})'(x)$. We need to find the slope at the point on the curve where the y-coordinate is 1. Let this point be $(a,1)$. Then $f^{-1}(a) = 1$, which implies $f(1)=a$. Calculating $a$, we get $a = f(1) = 1^5 + 1^3 + 2(1) = 4$. So we need to find the slope at $x=4$, which is $(f^{-1})'(4)$. The formula is $(f^{-1})'(4) = \frac{1}{f'(f^{-1}(4))}$. Since $f^{-1}(4)=1$, this is $\frac{1}{f'(1)}$. First, we find the derivative: $f'(x) = 5x^4 + 3x^2 + 2$. Then, $f'(1) = 5(1)^4 + 3(1)^2 + 2 = 10$. The slope is $\frac{1}{10}$.
17. Let $f(x)$ be a differentiable function and let $g(x) = f^{-1}(x)$. Suppose the graph of $f$ passes through $(4, 1)$ and the tangent line to $f$ at $x=4$ is parallel to the line $y = 3x+5$. What is the value of $g'(1)$?
    Answer: $\frac{1}{3}$ This question tests your understanding of the derivative of inverse functions, which follows a specific relationship. When you see $g(x) = f^{-1}(x)$, remember that the derivative of an inverse function is the reciprocal of the original function's derivative at the corresponding point. The key formula is: $g'(a) = \frac{1}{f'(b)}$ where $f(b) = a$. Since the graph of $f$ passes through $(4, 1)$, we know $f(4) = 1$. This means the inverse function satisfies $g(1) = 4$, so we need to find $g'(1)$. The tangent line to $f$ at $x = 4$ is parallel to $y = 3x + 5$, which has slope 3. Therefore, $f'(4) = 3$. Using the inverse function derivative formula: $g'(1) = \frac{1}{f'(4)} = \frac{1}{3}$ Looking at the wrong answers: Choice A gives $-\frac{1}{3}$, which incorrectly adds a negative sign—there's no reason for the derivative to be negative here. Choice B gives $3$, which is $f'(4)$ itself; this represents forgetting to take the reciprocal. Choice D gives $1$, which might come from incorrectly using the point $(4,1)$ without applying the proper inverse derivative relationship. Study tip: For inverse function derivatives, always remember the reciprocal relationship and identify the correct corresponding points. If $f(a) = b$, then $g'(b) = \frac{1}{f'(a)}$. Draw a quick diagram if needed to keep track of which point maps to which.
18. Let $f(x) = x^5 + 3x^3 + 2x - 1$. What is the value of $(f^{-1})'(5)$?
    Answer: $\frac{1}{16}$ To find $(f^{-1})'(5)$, we use the formula $(f^{-1})'(a) = \frac{1}{f'(f^{-1}(a))}$ with $a=5$. First, we need to find $b = f^{-1}(5)$, which means we need to solve $f(b) = 5$. So, $b^5 + 3b^3 + 2b - 1 = 5$, which simplifies to $b^5 + 3b^3 + 2b = 6$. By inspection, $b=1$ is the solution. Next, we find the derivative of $f(x)$: $f'(x) = 5x^4 + 9x^2 + 2$. Now we evaluate $f'$ at $b=1$: $f'(1) = 5(1)^4 + 9(1)^2 + 2 = 16$. Finally, $(f^{-1})'(5) = \frac{1}{f'(1)} = \frac{1}{16}$.
19. Let $h(x) = [f(x)]^3$ and $g(x) = f^{-1}(x)$. The line $y-8 = 12(x-1)$ is tangent to the graph of $h(x)$ at $x=1$. Find the value of $g'(2)$.
    Answer: $1$ This problem tests your understanding of the chain rule, inverse function derivatives, and how to extract information from tangent lines. When you see a composition like $h(x) = [f(x)]^3$ with tangent line information, you need to work backwards to find the underlying function's properties. Since the line $y - 8 = 12(x - 1)$ is tangent to $h(x)$ at $x = 1$, you know two key facts: $h(1) = 8$ and $h'(1) = 12$. From $h(1) = [f(1)]^3 = 8$, you can deduce that $f(1) = 2$ (since $2^3 = 8$). Using the chain rule, $h'(x) = 3[f(x)]^2 \cdot f'(x)$. At $x = 1$: $h'(1) = 3[f(1)]^2 \cdot f'(1) = 3(2)^2 \cdot f'(1) = 12f'(1) = 12$. Therefore, $f'(1) = 1$. Now you can find $g'(2)$. Since $g(x) = f^{-1}(x)$, the derivative of an inverse function formula gives you: $g'(a) = \frac{1}{f'(g(a))}$. You need $g(2)$, which means solving $f^{-1}(2) = ?$, or equivalently, $f(?) = 2$. From earlier work, $f(1) = 2$, so $g(2) = 1$. Therefore: $g'(2) = \frac{1}{f'(g(2))} = \frac{1}{f'(1)} = \frac{1}{1} = 1$. Choice A ($\frac{1}{2}$) incorrectly uses $\frac{1}{f(1)}$. Choice B ($\frac{1}{12}$) mistakenly uses $\frac{1}{h'(1)}$. Choice D (2) incorrectly uses $f(1)$ instead of applying the inverse derivative formula. The answer is C. Remember: for inverse function derivatives, always use $[f^{-1}]'(a) = \frac{1}{f'(f^{-1}(a))}$ and work carefully to identify the correct input values.
20. Let $h(x) = f^{-1}(x)$ for the function $f(x) = \frac{x+1}{x-1}$ on the domain $x>1$. Which of the following is an expression for $h'(x)$?
    Answer: $\frac{-2}{(x-1)^2}$ This function is its own inverse. To find $f^{-1}(x)$, we set $y = \frac{x+1}{x-1}$ and solve for $x$: $y(x-1) = x+1 \implies yx-y = x+1 \implies yx-x = y+1 \implies x(y-1) = y+1 \implies x = \frac{y+1}{y-1}$. So, $f^{-1}(x) = \frac{x+1}{x-1}$. This means $h(x) = f(x)$. Therefore, $h'(x) = f'(x)$. Using the quotient rule on $f(x)$, we get $f'(x) = \frac{(1)(x-1) - (x+1)(1)}{(x-1)^2} = \frac{x-1-x-1}{(x-1)^2} = \frac{-2}{(x-1)^2}$.

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