Derivatives of sin & cos - Calculus 1

Let $f(x) = x^2 \cos(x)$ . What is the second derivative, $f''(x)$ ?
Answer: $(2-x^2)\cos(x) - 4x\sin(x)$ To find the second derivative, we must apply the product rule twice. First, find the first derivative $f'(x)$ : $f'(x) = (\frac{d}{dx}x^2)\cos(x) + x^2(\frac{d}{dx}\cos(x)) = 2x\cos(x) - x^2\sin(x)$ . Next, differentiate $f'(x)$ term by term, applying the product rule to each term: $f''(x) = [(\frac{d}{dx}2x)\cos(x) + 2x(\frac{d}{dx}\cos(x))] - [(\frac{d}{dx}x^2)\sin(x) + x^2(\frac{d}{dx}\sin(x))]$ $f''(x) = [2\cos(x) + 2x(-\sin(x))] - [2x\sin(x) + x^2(\cos(x))]$ $f''(x) = 2\cos(x) - 2x\sin(x) - 2x\sin(x) - x^2\cos(x)$ Combine like terms: $f''(x) = (2-x^2)\cos(x) - 4x\sin(x)$ .
Let $g(x) = \sin(\pi \cos(x))$ . What is the value of $g'(\frac{\pi}{2})$ ?
Answer: $-\pi$ This requires the chain rule. Let the outer function be $f(u) = \sin(u)$ and the inner function be $u(x) = \pi \cos(x)$ . The derivative is $g'(x) = f'(u(x)) \cdot u'(x)$ . $f'(u) = \cos(u)$ , so $f'(u(x)) = \cos(\pi \cos(x))$ . $u'(x) = \frac{d}{dx}(\pi \cos(x)) = -\pi \sin(x)$ . Combining these, we get: $g'(x) = \cos(\pi \cos(x)) \cdot (-\pi \sin(x))$ . Now, evaluate at $x = \frac{\pi}{2}$ : $g'(\frac{\pi}{2}) = \cos(\pi \cos(\frac{\pi}{2})) \cdot (-\pi \sin(\frac{\pi}{2}))$ . Since $\cos(\frac{\pi}{2}) = 0$ and $\sin(\frac{\pi}{2}) = 1$ , we have: $g'(\frac{\pi}{2}) = \cos(\pi \cdot 0) \cdot (-\pi \cdot 1) = \cos(0) \cdot (-\pi) = 1 \cdot (-\pi) = -\pi$ .
The position of a particle moving along the x-axis is given by $s(t) = 3\sin(t) - \cos(2t)$ for $t \ge 0$ . What is the acceleration of the particle at $t = \frac{\pi}{2}$ ?
Answer: $-7$ Acceleration is the second derivative of the position function. First, we find the velocity function, $v(t) = s'(t)$ : $v(t) = \frac{d}{dt}(3\sin(t) - \cos(2t)) = 3\cos(t) - (-\sin(2t) \cdot 2) = 3\cos(t) + 2\sin(2t)$ . Next, we find the acceleration function, $a(t) = v'(t)$ : $a(t) = \frac{d}{dt}(3\cos(t) + 2\sin(2t)) = -3\sin(t) + 2(\cos(2t) \cdot 2) = -3\sin(t) + 4\cos(2t)$ . Finally, we evaluate the acceleration at $t = \frac{\pi}{2}$ : $a(\frac{\pi}{2}) = -3\sin(\frac{\pi}{2}) + 4\cos(2 \cdot \frac{\pi}{2}) = -3\sin(\frac{\pi}{2}) + 4\cos(\pi) = -3(1) + 4(-1) = -3 - 4 = -7$ .
A particle's position is described by $p(t) = t - 2\sin(t)$ . At what time $t$ in $[0, \pi]$ is the particle momentarily at rest?
Answer: $\pi/3$ When you encounter a particle motion problem asking when the particle is "momentarily at rest," you need to find when the velocity equals zero. A particle is at rest when it's not moving, which means its instantaneous rate of change of position is zero. To find the velocity, you take the derivative of the position function: $v(t) = p'(t) = \frac{d}{dt}(t - 2\sin(t)) = 1 - 2\cos(t)$ Setting the velocity equal to zero: $1 - 2\cos(t) = 0$ Solving for $t$ : $2\cos(t) = 1$ , so $\cos(t) = \frac{1}{2}$ In the interval $[0, \pi]$ , cosine equals $\frac{1}{2}$ when $t = \frac{\pi}{3}$ . This makes D the correct answer. Let's check why the other options are wrong. For A) $t = 0$ : $\cos(0) = 1$ , so $v(0) = 1 - 2(1) = -1 \neq 0$ . For B) $t = \frac{\pi}{6}$ : $\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$ , so $v(\frac{\pi}{6}) = 1 - 2(\frac{\sqrt{3}}{2}) = 1 - \sqrt{3} \neq 0$ . For C) $t = \frac{\pi}{2}$ : $\cos(\frac{\pi}{2}) = 0$ , so $v(\frac{\pi}{2}) = 1 - 2(0) = 1 \neq 0$ . Remember: "at rest" always means velocity equals zero, not position equals zero. Always differentiate the position function to find velocity, then set that derivative equal to zero and solve.
What is the equation of the line tangent to the graph of $f(x) = \cos(2x) - \sin(x)$ at $x = \frac{\pi}{6}$ ?
Answer: $y = -\frac{3\sqrt{3}}{2}(x - \frac{\pi}{6})$ First, find the point of tangency. The y-coordinate is $f(\frac{\pi}{6}) = \cos(2\cdot\frac{\pi}{6}) - \sin(\frac{\pi}{6}) = \cos(\frac{\pi}{3}) - \sin(\frac{\pi}{6}) = \frac{1}{2} - \frac{1}{2} = 0$ . The point is $(\frac{\pi}{6}, 0)$ . Next, find the slope of the tangent line by finding the derivative, $f'(x)$ . Using the chain rule for $\cos(2x)$ , we get: $f'(x) = -\sin(2x) \cdot 2 - \cos(x) = -2\sin(2x) - \cos(x)$ . Now, evaluate the slope at $x = \frac{\pi}{6}$ : $m = f'(\frac{\pi}{6}) = -2\sin(2\cdot\frac{\pi}{6}) - \cos(\frac{\pi}{6}) = -2\sin(\frac{\pi}{3}) - \cos(\frac{\pi}{6}) = -2(\frac{\sqrt{3}}{2}) - \frac{\sqrt{3}}{2} = -\sqrt{3} - \frac{\sqrt{3}}{2} = -\frac{3\sqrt{3}}{2}$ . Using the point-slope form $y - y_1 = m(x - x_1)$ , the equation is $y - 0 = -\frac{3\sqrt{3}}{2}(x - \frac{\pi}{6})$ .
The function $f(x) = e^x \cos(x)$ is decreasing on which of the following intervals?
Answer: ( $\frac{\pi}{4}, \frac{5\pi}{4})$ A function is decreasing where its derivative is negative. First, find $f'(x)$ using the product rule: $f'(x) = (\frac{d}{dx}e^x)\cos(x) + e^x(\frac{d}{dx}\cos(x)) = e^x \cos(x) + e^x(-\sin(x)) = e^x(\cos(x) - \sin(x))$ . To find where $f(x)$ is decreasing, we need to solve $f'(x) < 0$ . Since $e^x$ is always positive, we need to solve $\cos(x) - \sin(x) < 0$ , which is equivalent to $\cos(x) < \sin(x)$ . On the unit circle, $\sin(x) = \cos(x)$ at $x = \frac{\pi}{4}$ and $x = \frac{5\pi}{4}$ . The inequality $\cos(x) < \sin(x)$ holds for $x$ values between these two points. Thus, the function is decreasing on the interval $(\frac{\pi}{4}, \frac{5\pi}{4})$ .
If the equation $y\sin(x) = \cos(y)$ defines $y$ as a differentiable function of $x$ , what is the value of $\frac{dy}{dx}$ at the point $(0, \frac{\pi}{2})$ ?
Answer: $-\frac{\pi}{2}$ We use implicit differentiation with respect to $x$ . Apply the product rule to the left side and the chain rule to the right side: $\frac{d}{dx}(y\sin(x)) = \frac{d}{dx}(\cos(y))$ $(\frac{dy}{dx})\sin(x) + y(\cos(x)) = -\sin(y)(\frac{dy}{dx})$ Now, we solve for $\frac{dy}{dx}$ : $(\frac{dy}{dx})\sin(x) + \sin(y)(\frac{dy}{dx}) = -y\cos(x)$ $\frac{dy}{dx}(\sin(x) + \sin(y)) = -y\cos(x)$ $\frac{dy}{dx} = \frac{-y\cos(x)}{\sin(x) + \sin(y)}$ Substitute the point $(x, y) = (0, \frac{\pi}{2})$ into the expression: $\frac{dy}{dx} = \frac{-(\frac{\pi}{2})\cos(0)}{\sin(0) + \sin(\frac{\pi}{2})} = \frac{-(\frac{\pi}{2})(1)}{0 + 1} = -\frac{\pi}{2}$ .
The equation $\sin(y) + \cos(x) = 1$ defines $y$ implicitly as a function of $x$ . Find $\frac{dy}{dx}$ at the point $(\pi/2, 0)$ .
Answer: 1 Differentiate both sides of the equation with respect to $x$ , remembering to use the chain rule for the term involving $y$ . $\frac{d}{dx}(\sin(y)) + \frac{d}{dx}(\cos(x)) = \frac{d}{dx}(1)$ $\cos(y) \cdot \frac{dy}{dx} - \sin(x) = 0$ Now, solve for $\frac{dy}{dx}$ : $\cos(y) \frac{dy}{dx} = \sin(x)$ $\frac{dy}{dx} = \frac{\sin(x)}{\cos(y)}$ Finally, substitute the coordinates of the point $(\pi/2, 0)$ into this expression: $\frac{dy}{dx} \bigg|_{(\pi/2, 0)} = \frac{\sin(\pi/2)}{\cos(0)} = \frac{1}{1} = 1$ .
What is the slope of the tangent line to the curve $y = \cos^3(x)$ at $x = \frac{\pi}{6}$ ?
Answer: $-\frac{9}{8}$ When you see a question asking for the slope of a tangent line, you're being tested on derivatives and the chain rule. The slope of the tangent line at any point is simply the derivative of the function evaluated at that point. To find the derivative of $y = \cos^3(x)$ , you need the chain rule since this is a composite function. Let $u = \cos(x)$ , so $y = u^3$ . Using the chain rule: $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = 3u^2 \cdot (-\sin(x)) = 3\cos^2(x) \cdot (-\sin(x)) = -3\cos^2(x)\sin(x)$ Now evaluate this derivative at $x = \frac{\pi}{6}$ : - $\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$ - $\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$ So: $\frac{dy}{dx}\bigg|_{x=\frac{\pi}{6}} = -3\left(\frac{\sqrt{3}}{2}\right)^2 \cdot \frac{1}{2} = -3 \cdot \frac{3}{4} \cdot \frac{1}{2} = -\frac{9}{8}$ This confirms answer D is correct. Looking at the wrong answers: A) $\frac{9}{4}$ likely comes from forgetting the negative sign and miscalculating the fraction. B) $-\frac{1}{8}$ suggests errors in both the chain rule application and trigonometric values. C) $\frac{9}{8}$ gets the magnitude right but misses the crucial negative sign from the derivative of cosine. Study tip: Always remember that $\frac{d}{dx}[\cos(x)] = -\sin(x)$ , and when using the chain rule, carefully track each step to avoid sign errors and computational mistakes.
Let $g(x) = x - \sin(x)$ . On the interval $[0, 2\pi]$ , where is the graph of $g(x)$ concave up?
Answer: $(0, \pi)$ To determine where a function is concave up, you need to find where its second derivative is positive. This is a fundamental connection in calculus: concavity is determined by the sign of the second derivative. For $g(x) = x - \sin(x)$ , let's find the second derivative. First, $g'(x) = 1 - \cos(x)$ . Then, $g''(x) = \sin(x)$ . The function is concave up where $g''(x) = \sin(x) > 0$ . On the interval $[0, 2\pi],\sin(x) > 0$ precisely when $x \in (0, \pi)$ . At $x = 0$ and $x = \pi,\sin(x) = 0$ , so these are inflection points where concavity changes. For $x \in (\pi, 2\pi),\sin(x) < 0$ , making the function concave down in this region. Looking at the wrong answers: Choice A, $(\pi/2, 3\pi/2)$ , includes the interval $(\pi, 3\pi/2)$ where $\sin(x) < 0$ , so the function is concave down there. Choice B, $(\pi, 2\pi)$ , is exactly where $\sin(x) < 0$ , making the function concave down throughout. Choice C, $(0, 2\pi)$ , incorrectly includes the second half of the interval where the function is concave down. The correct answer is D: $(0, \pi)$ . Study tip: Always remember that concave up means $f''(x) > 0$ and concave down means $f''(x) < 0$ . For trigonometric functions, sketch the basic sine and cosine curves to quickly identify where they're positive or negative.
For $0 \le x \le 2\pi$ , find the sum of all values of $x$ where the tangent lines to $f(x) = \sin(x)$ and $g(x) = \cos(x)$ are parallel.
Answer: $5\pi/2$ When you encounter a problem about parallel tangent lines, you're looking for points where two functions have equal derivatives, since parallel lines have identical slopes. To find where $f(x) = \sin(x)$ and $g(x) = \cos(x)$ have parallel tangent lines, you need to solve $f'(x) = g'(x)$ . Taking derivatives: $f'(x) = \cos(x)$ and $g'(x) = -\sin(x)$ . Setting them equal: $\cos(x) = -\sin(x)$ . Dividing both sides by $\cos(x)$ (being careful about when $\cos(x) = 0$ ): $1 = -\tan(x)$ , so $\tan(x) = -1$ . Over the interval $[0, 2\pi]$ , this occurs at $x = \frac{3\pi}{4}$ and $x = \frac{7\pi}{4}$ . You can verify that $\cos(x) \neq 0$ at these points, so our division was valid. The sum is $\frac{3\pi}{4} + \frac{7\pi}{4} = \frac{10\pi}{4} = \frac{5\pi}{2}$ , which is choice B. Choice A ( $\frac{3\pi}{2}$ ) only accounts for one solution rather than both. Choice C ( $2\pi$ ) might come from incorrectly thinking the solutions are at $\frac{\pi}{2}$ and $\frac{3\pi}{2}$ , but $\tan(x)$ is undefined at these points. Choice D ( $4\pi$ ) could result from doubling the correct answer or finding incorrect solutions. Remember: when finding parallel tangent lines between two curves, always set their derivatives equal and solve completely over the given interval. Don't forget to check that your algebra steps are valid at the solutions you find.
Let $f(x) = x^2 \sin(x)$ . What is the slope of the tangent line to the graph of $f$ at $x = \pi/2$ ?
Answer: $\pi$ To find the slope of the tangent line, we need to compute the derivative $f'(x)$ and evaluate it at $x = \pi/2$ . Using the product rule, $(uv)' = u'v + uv'$ , where $u = x^2$ and $v = \sin(x)$ . $f'(x) = \frac{d}{dx}(x^2) \cdot \sin(x) + x^2 \cdot \frac{d}{dx}(\sin(x))$ $f'(x) = 2x \sin(x) + x^2 \cos(x)$ Now, evaluate $f'(x)$ at $x = \pi/2$ : $f'(\pi/2) = 2(\pi/2) \sin(\pi/2) + (\pi/2)^2 \cos(\pi/2)$ $f'(\pi/2) = \pi \cdot 1 + (\pi^2/4) \cdot 0$ $f'(\pi/2) = \pi$ Thus, the slope of the tangent line at $x = \pi/2$ is $\pi$ .
Let $f(x) = \sin(x) + \cos(x)$ . Which of the following is equivalent to $f^{(4)}(x)$ , the fourth derivative of $f(x)$ ?
Answer: $f(x)$ When you encounter questions about higher-order derivatives of trigonometric functions, look for patterns that repeat every few derivatives. Trigonometric functions have cyclical derivative patterns that make problems like this much easier once you recognize them. Let's find the fourth derivative of $f(x) = \sin(x) + \cos(x)$ by computing each derivative step by step: $f(x) = \sin(x) + \cos(x)$ $f'(x) = \cos(x) - \sin(x)$ $f''(x) = -\sin(x) - \cos(x) = -f(x)$ $f'''(x) = -\cos(x) + \sin(x) = -f'(x)$ $f^{(4)}(x) = \sin(x) + \cos(x) = f(x)$ Notice the beautiful pattern: after four derivatives, we're back to the original function! This confirms that answer choice D is correct. Now let's see why the other options are wrong. Choice A suggests $f^{(4)}(x) = -f'(x)$ , but we found that $f'''(x) = -f'(x)$ , so this confuses the third and fourth derivatives. Choice B claims $f^{(4)}(x) = -f(x)$ , but this is actually $f''(x)$ —another derivative mix-up. Choice C suggests $f^{(4)}(x) = f'(x)$ , but there's no step in our calculation that gives us the first derivative again. Study tip: Memorize that both $\sin(x)$ and $\cos(x)$ return to themselves after four derivatives. This cyclical pattern (period 4) appears frequently in calculus problems, so recognizing it will save you time on exams.
For which of the following values of $x$ in the interval $[0, 2\pi]$ does the function $f(x) = x - 2\cos(x)$ have a horizontal tangent line?
Answer: $x = \frac{7\pi}{6}$ and $x = \frac{11\pi}{6}$ A horizontal tangent line occurs where the derivative is equal to zero. First, find the derivative of $f(x)$ : $f'(x) = \frac{d}{dx}(x - 2\cos(x)) = 1 - 2(-\sin(x)) = 1 + 2\sin(x)$ . Set the derivative equal to zero and solve for $x$ : $1 + 2\sin(x) = 0$ $2\sin(x) = -1$ $\sin(x) = -\frac{1}{2}$ In the interval $[0, 2\pi]$ , the sine function is negative in the third and fourth quadrants. The reference angle is $\frac{\pi}{6}$ . The solution in the third quadrant is $x = \pi + \frac{\pi}{6} = \frac{7\pi}{6}$ . The solution in the fourth quadrant is $x = 2\pi - \frac{\pi}{6} = \frac{11\pi}{6}$ .
The third derivative of $f(x) = \sin(x) + \cos(x)$ is equal to its first derivative for certain values of $x$ . Which of the following is such a value?
Answer: $\pi/4$ This problem tests your understanding of trigonometric derivatives and when you need to find where two different derivatives of the same function are equal. First, let's find the derivatives of $f(x) = \sin(x) + \cos(x)$ . The first derivative is $f'(x) = \cos(x) - \sin(x)$ . The second derivative is $f''(x) = -\sin(x) - \cos(x)$ . The third derivative is $f'''(x) = -\cos(x) + \sin(x)$ . We need $f'''(x) = f'(x)$ , so we set up the equation: $-\cos(x) + \sin(x) = \cos(x) - \sin(x)$ Rearranging: $2\sin(x) = 2\cos(x)$ , which simplifies to $\sin(x) = \cos(x)$ . This equation is satisfied when $\tan(x) = 1$ , which occurs at $x = \frac{\pi}{4} + n\pi$ for integer values of $n$ . Among our choices, this gives us $x = \frac{\pi}{4}$ , making D correct. Let's check why the other options fail: For A, at $x = \pi$ , we have $\sin(\pi) = 0$ and $\cos(\pi) = -1$ , so they're not equal. For B, at $x = \frac{\pi}{2}$ , we get $\sin(\frac{\pi}{2}) = 1$ and $\cos(\frac{\pi}{2}) = 0$ , which aren't equal. For C, at $x = 0$ , we have $\sin(0) = 0$ and $\cos(0) = 1$ , so they're not equal either. Study tip: When comparing derivatives of trigonometric functions, look for special angle values where sine and cosine have simple relationships, like $\frac{\pi}{4}$ where they're equal.
If $f(x) = 2\sin(x)\cos(x)$ , find $f'(\pi/6)$ .
Answer: 1 When you encounter a function like $f(x) = 2\sin(x)\cos(x)$ , you need to find its derivative using differentiation rules, then evaluate at the given point. To find $f'(x)$ , apply the product rule since you have two functions multiplied together: $\sin(x)$ and $\cos(x)$ . The product rule states that $(uv)' = u'v + uv'$ . For $f(x) = 2\sin(x)\cos(x)$ : - $u = \sin(x)$ , so $u' = \cos(x)$ - $v = \cos(x)$ , so $v' = -\sin(x)$ Therefore: $f'(x) = 2[\cos(x) \cdot \cos(x) + \sin(x) \cdot (-\sin(x))] = 2[\cos^2(x) - \sin^2(x)]$ This simplifies to $f'(x) = 2\cos(2x)$ using the double angle identity $\cos(2x) = \cos^2(x) - \sin^2(x)$ . Now evaluate at $x = \pi/6$ : $f'(\pi/6) = 2\cos(2 \cdot \pi/6) = 2\cos(\pi/3) = 2 \cdot \frac{1}{2} = 1$ Looking at the wrong answers: (A) $\sqrt{3}$ would result from confusing $\cos(\pi/3)$ with $\cos(\pi/6) = \sqrt{3}/2$ . (B) 2 comes from forgetting that $\cos(\pi/3) = 1/2$ , not 1. (D) $1/2$ results from calculating $\cos(\pi/3)$ correctly but forgetting to multiply by 2. Study tip: Recognize that $2\sin(x)\cos(x) = \sin(2x)$ , so you could also differentiate to get $2\cos(2x)$ directly. Memorizing common trigonometric identities saves time on derivative problems.
If $f(x) = e^x \cos(x)$ , find $f'(\pi)$ .
Answer: $-e^\pi$ Use the product rule, $(uv)' = u'v + uv'$ , with $u = e^x$ and $v = \cos(x)$ . Then $u' = e^x$ and $v' = -\sin(x)$ . $f'(x) = (e^x)(\cos(x)) + (e^x)(-\sin(x))$ $f'(x) = e^x(\cos(x) - \sin(x))$ Now, evaluate at $x = \pi$ : $f'(\pi) = e^\pi(\cos(\pi) - \sin(\pi))$ $f'(\pi) = e^\pi(-1 - 0) = -e^\pi$ .
If $y = \cos^3(2x)$ , what is $\frac{dy}{dx}$ at $x = \pi/12$ ?
Answer: $-9/4$ When you encounter a function like $y = \cos^3(2x)$ , you're dealing with a composition that requires the chain rule. The expression $\cos^3(2x)$ means $[\cos(2x)]^3$ , so you have three nested functions: the outer power of 3, the middle cosine function, and the inner function $2x$ . To find $\frac{dy}{dx}$ , apply the chain rule step by step. First, the derivative of $[\cos(2x)]^3$ is $3[\cos(2x)]^2 \cdot \frac{d}{dx}[\cos(2x)]$ . Next, $\frac{d}{dx}[\cos(2x)] = -\sin(2x) \cdot 2 = -2\sin(2x)$ . Combining these: $\frac{dy}{dx} = 3\cos^2(2x) \cdot (-2\sin(2x)) = -6\cos^2(2x)\sin(2x)$ . Now evaluate at $x = \frac{\pi}{12}$ . Since $2x = \frac{\pi}{6}$ , you need $\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$ and $\sin(\frac{\pi}{6}) = \frac{1}{2}$ . Substituting: $\frac{dy}{dx} = -6 \cdot \left(\frac{\sqrt{3}}{2}\right)^2 \cdot \frac{1}{2} = -6 \cdot \frac{3}{4} \cdot \frac{1}{2} = -\frac{9}{4}$ . This confirms answer B. Answer A ( $\frac{9}{4}$ ) has the correct magnitude but wrong sign—likely from forgetting the negative sign in the derivative of cosine. Answer C ( $-\frac{9}{8}$ ) suggests an error in computing $\cos^2(\frac{\pi}{6})$ , possibly using $\frac{\sqrt{3}}{4}$ instead of $\frac{3}{4}$ . Answer D ( $\frac{3\sqrt{3}}{4}$ ) appears to come from incorrectly leaving $\sqrt{3}$ unsquared. Remember: with nested functions, work from outside to inside with the chain rule, and always double-check your signs when differentiating trigonometric functions.
If $g(x) = \frac{\cos(x)}{1 - \sin(x)}$ , what is the value of $g'(0)$ ?
Answer: 1 To find $g'(0)$ , we use the quotient rule: $(\frac{u}{v})' = \frac{u'v - uv'}{v^2}$ , with $u = \cos(x)$ and $v = 1 - \sin(x)$ . $u' = -\sin(x)$ and $v' = -\cos(x)$ . $g'(x) = \frac{(-\sin(x))(1 - \sin(x)) - (\cos(x))(-\cos(x))}{(1 - \sin(x))^2}$ $g'(x) = \frac{-\sin(x) + \sin^2(x) + \cos^2(x)}{(1 - \sin(x))^2}$ Using the identity $\sin^2(x) + \cos^2(x) = 1$ : $g'(x) = \frac{1 - \sin(x)}{(1 - \sin(x))^2} = \frac{1}{1 - \sin(x)}$ Now, evaluate at $x = 0$ : $g'(0) = \frac{1}{1 - \sin(0)} = \frac{1}{1 - 0} = 1$ .
Evaluate the limit: $\lim_{h \to 0} \frac{\cos(\frac{\pi}{3} + h) - \frac{1}{2}}{h}$
Answer: $-\frac{\sqrt{3}}{2}$ This limit represents the definition of the derivative of a function $f(x)$ at a point $x=a$ , which is given by $\lim_{h \to 0} \frac{f(a+h) - f(a)}{h} = f'(a)$ . By comparing the given limit to this definition, we can identify $f(x) = \cos(x)$ and $a = \frac{\pi}{3}$ . So, the value of the limit is equal to $f'(\frac{\pi}{3})$ . First, find the derivative of $f(x)$ : $f'(x) = -\sin(x)$ . Then, evaluate the derivative at $x = \frac{\pi}{3}$ : $f'(\frac{\pi}{3}) = -\sin(\frac{\pi}{3}) = -\frac{\sqrt{3}}{2}$ .

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