Derivatives of cos, sin, e^x, ln x - Calculus 1

Let $f(x) = e^x + x$ and let $g(x)$ be the inverse function of $f(x)$ . What is the value of $g'(1)$ ?
Answer: $\frac{1}{2}$ 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 $g'(x) = \frac{1}{f'(g(x))}$ . To find $g'(1)$ , you first need to determine what value $g(1)$ equals. Since $g$ is the inverse of $f$ , we have $g(1) = a$ where $f(a) = 1$ . So you need to solve $e^x + x = 1$ . Testing $x = 0$ : $f(0) = e^0 + 0 = 1 + 0 = 1$ . Perfect! So $g(1) = 0$ . Now apply the inverse function formula: $g'(1) = \frac{1}{f'(g(1))} = \frac{1}{f'(0)}$ . Find $f'(x)$ : $f'(x) = e^x + 1$ , so $f'(0) = e^0 + 1 = 1 + 1 = 2$ . Therefore, $g'(1) = \frac{1}{2}$ , which is answer choice D. Answer choice A (2) would result from incorrectly thinking $g'(1) = f'(0)$ without taking the reciprocal. Answer choice B (1) might come from forgetting that the derivative of $e^x$ is $e^x$ , not 1, leading to $f'(0) = 1$ . Answer choice C ( $\frac{1}{e+1}$ ) could result from incorrectly computing $f'(0)$ as $e + 1$ instead of 2. Remember: for inverse function derivatives, always find where the original function equals your input value first, then apply the reciprocal formula $g'(x) = \frac{1}{f'(g(x))}$ .
At which value of $x$ does the graph of the function $f(x) = \frac{\ln x}{x}$ have a horizontal tangent line?
Answer: $x = e$ A horizontal tangent line occurs where the derivative $f'(x)$ is equal to zero. Using the quotient rule, $f'(x) = \frac{(\frac{1}{x}) \cdot x - (\ln x) \cdot 1}{x^2} = \frac{1 - \ln x}{x^2}$ . Setting $f'(x) = 0$ requires the numerator to be zero: $1 - \ln x = 0$ , which implies $\ln x = 1$ . The solution is $x = e$ .
What is the smallest positive value of $x$ for which the derivatives of $f(x) = \sin(2x)$ and $g(x) = 2\sin(x)$ are equal?
Answer: $2\pi/3$ First, find the derivatives: $f'(x) = 2\cos(2x)$ and $g'(x) = 2\cos(x)$ . Set them equal: $2\cos(2x) = 2\cos(x)$ , which simplifies to $\cos(2x) = \cos(x)$ . Using the double-angle identity $\cos(2x) = 2\cos^2(x) - 1$ , we get $2\cos^2(x) - 1 = \cos(x)$ . This is a quadratic equation in $\cos(x)$ : $2\cos^2(x) - \cos(x) - 1 = 0$ . Factoring gives $(2\cos(x) + 1)(\cos(x) - 1) = 0$ . The solutions are $\cos(x) = 1$ (which gives $x=2k\pi$ ) and $\cos(x) = -1/2$ (which gives $x = 2\pi/3 + 2k\pi$ and $x = 4\pi/3 + 2k\pi$ ). The smallest positive solution is $x = 2\pi/3$ .
Evaluate the limit: $\lim_{h \to 0} \frac{\ln(e^2 + h) - 2}{h}$
Answer: $e^{-2}$ This limit represents the definition of the derivative of a function $f(x)$ at a point $x=a$ , which is $\lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$ . Here, $f(x) = \ln(x)$ and $a = e^2$ . Note that $f(a) = \ln(e^2) = 2$ . The limit is therefore equal to $f'(e^2)$ . The derivative is $f'(x) = \frac{1}{x}$ . Evaluating at $x=e^2$ gives $f'(e^2) = \frac{1}{e^2} = e^{-2}$ .
Find the slope of the line normal to the graph of $f(x) = e^x \cos(x)$ at the point where $x=\pi$ .
Answer: $e^{-\pi}$ To find the slope of the normal line, we first need the slope of the tangent line, which is given by the derivative $f'(x)$ . Using the product rule, $f'(x) = \frac{d}{dx}(e^x) \cos(x) + e^x \frac{d}{dx}(\cos(x)) = e^x \cos(x) - e^x \sin(x)$ . Next, we evaluate the derivative at $x=\pi$ to find the slope of the tangent line at that point: $m_{tan} = f'(\pi) = e^\pi \cos(\pi) - e^\pi \sin(\pi) = e^\pi(-1) - e^\pi(0) = -e^\pi$ . The slope of the normal line is the negative reciprocal of the slope of the tangent line. So, $m_{normal} = -\frac{1}{m_{tan}} = -\frac{1}{-e^\pi} = \frac{1}{e^\pi} = e^{-\pi}$ .
Evaluate the limit: $\lim_{x \to \pi} \frac{\cos x + 1}{x - \pi}$
Answer: 0 This limit is in the form of the definition of the derivative. Let $f(x) = \cos x$ . The derivative of $f$ at $a=\pi$ is given by $f'(\pi) = \lim_{x \to \pi} \frac{f(x) - f(\pi)}{x - \pi}$ . Since $f(\pi) = \cos(\pi) = -1$ , the expression is $\lim_{x \to \pi} \frac{\cos x - (-1)}{x - \pi} = \lim_{x \to \pi} \frac{\cos x + 1}{x - \pi}$ . So the limit is equal to $f'(\pi)$ . The derivative is $f'(x) = -\sin x$ . Evaluating at $x=\pi$ gives $f'(\pi) = -\sin(\pi) = 0$ . Alternatively, using L'Hôpital's Rule gives $\lim_{x \to \pi} \frac{-\sin x}{1} = -\sin(\pi) = 0$ .
Let $f(x) = \ln(e^{2x} + 1)$ . Find $f'(0)$ .
Answer: 1 This problem tests your ability to find derivatives of composite functions, specifically when the natural logarithm contains a more complex expression. When you see $\ln$ of something other than just $x$ , immediately think chain rule. To find $f'(x)$ where $f(x) = \ln(e^{2x} + 1)$ , you need the chain rule. The derivative of $\ln(u)$ is $\frac{1}{u} \cdot u'$ , where $u = e^{2x} + 1$ . First, find $u' = \frac{d}{dx}(e^{2x} + 1) = 2e^{2x}$ (using the chain rule again since the exponent is $2x$ ). So $f'(x) = \frac{1}{e^{2x} + 1} \cdot 2e^{2x} = \frac{2e^{2x}}{e^{2x} + 1}$ . Now evaluate at $x = 0$ : $f'(0) = \frac{2e^{0}}{e^{0} + 1} = \frac{2 \cdot 1}{1 + 1} = \frac{2}{2} = 1$ . Looking at the wrong answers: Choice A (0) might result from incorrectly thinking the derivative of a logarithm is always zero or from computational errors. Choice B (1/2) could come from forgetting the factor of 2 in the derivative of $e^{2x}$ or from inverting the final fraction. Choice C (2) likely results from forgetting to substitute $x = 0$ and leaving the answer as $2e^{2x}$ , or from errors in applying the quotient structure. Remember: when differentiating $\ln$ of a composite function, always use $\frac{u'}{u}$ , and don't forget to find the derivative of the inner function completely before substituting your specific value.
Let $f(x) = x e^x$ . What is the value of $f''(2)$ , the second derivative of $f$ evaluated at $x=2$ ?
Answer: $4e^2$ To find the second derivative, we must apply the product rule twice. The first derivative is $f'(x) = (1)e^x + x(e^x) = e^x(1+x)$ . The second derivative is $f''(x) = (e^x)(1+x) + (e^x)(1) = e^x(1+x+1) = e^x(x+2)$ . Evaluating at $x=2$ , we get $f''(2) = e^2(2+2) = 4e^2$ .
The position of a particle moving along an axis is given by $p(t) = e^t \sin(t)$ . What is the acceleration of the particle at time $t=\pi$ ?
Answer: $-2e^\pi$ Acceleration is the second derivative of the position function. We need to find $p''(t)$ and evaluate it at $t=\pi$ . First, find the velocity, $v(t) = p'(t)$ . Using the product rule: $v(t) = (\frac{d}{dt}e^t)\sin(t) + e^t(\frac{d}{dt}\sin(t)) = e^t \sin(t) + e^t \cos(t) = e^t(\sin(t) + \cos(t))$ . Next, find the acceleration, $a(t) = v'(t)$ . Using the product rule again: $a(t) = (\frac{d}{dt}e^t)(\sin(t) + \cos(t)) + e^t(\frac{d}{dt}(\sin(t) + \cos(t)))$ $a(t) = e^t(\sin(t) + \cos(t)) + e^t(\cos(t) - \sin(t))$ $a(t) = e^t\sin(t) + e^t\cos(t) + e^t\cos(t) - e^t\sin(t) = 2e^t\cos(t)$ . Finally, evaluate the acceleration at $t=\pi$ : $a(\pi) = 2e^\pi\cos(\pi) = 2e^\pi(-1) = -2e^\pi$ .
Let $f(x) = e^x$ and $g(x) = x^2 + 1$ . If $h(x) = f(\ln(g(x)))$ , what is the value of $h'(1)$ ?
Answer: $2$ The function $h(x)$ can be simplified before differentiation. Since $e^x$ and $\ln x$ are inverse functions, $h(x) = f(\ln(g(x))) = e^{\ln(g(x))} = g(x)$ . So, $h(x) = x^2 + 1$ . The derivative is $h'(x) = 2x$ . Evaluating at $x=1$ gives $h'(1) = 2(1) = 2$ .
If $f(x) = \ln(\cos^2(x))$ , what is $f'(\pi/4)$ ?
Answer: $-2$ We can solve this using the chain rule. Let the outer function be $\ln(u)$ and the inner function be $u = \cos^2(x)$ . The derivative is $f'(x) = \frac{1}{\cos^2(x)} \cdot \frac{d}{dx}(\cos^2(x))$ . To find the derivative of $\cos^2(x)$ , we use the chain rule again: $\frac{d}{dx}(\cos(x))^2 = 2(\cos(x))(-\sin(x)) = -2\sin(x)\cos(x)$ . Combining these, $f'(x) = \frac{1}{\cos^2(x)} \cdot (-2\sin(x)\cos(x)) = -2\frac{\sin(x)}{\cos(x)} = -2\tan(x)$ . Finally, evaluate at $x=\pi/4$ : $f'(\pi/4) = -2\tan(\pi/4) = -2(1) = -2$ . Alternatively, one could first simplify $f(x) = \ln(\cos^2(x)) = 2\ln(\cos(x))$ for values where $\cos(x)>0$ , which is true at $x=\pi/4$ . Then $f'(x) = 2 \cdot \frac{1}{\cos(x)} \cdot (-\sin(x)) = -2\tan(x)$ , yielding the same result.
If $f(x) = \sin(\ln x)$ , what is $f'(e)$ ?
Answer: $\frac{\cos(1)}{e}$ Using the chain rule, $f'(x) = \cos(\ln x) \cdot \frac{d}{dx}(\ln x) = \cos(\ln x) \cdot \frac{1}{x}$ . Now, substitute $x = e$ : $f'(e) = \cos(\ln e) \cdot \frac{1}{e} = \cos(1) \cdot \frac{1}{e} = \frac{\cos(1)}{e}$ .
Let $f(x) = e^{2x} \cos(x)$ . Find the value of $x$ in $(0, \pi)$ where $f'(x) = 0$ .
Answer: $\arctan(2)$ Using the product rule, $f'(x) = (2e^{2x})\cos(x) + e^{2x}(-\sin(x)) = e^{2x}(2\cos(x) - \sin(x))$ . Setting $f'(x) = 0$ , since $e^{2x}$ is never zero, we must have $2\cos(x) - \sin(x) = 0$ . This implies $2\cos(x) = \sin(x)$ . Assuming $\cos(x) \neq 0$ , we can divide by $\cos(x)$ to get $\tan(x) = 2$ . The value of $x$ in $(0, \pi)$ that satisfies this is $x = \arctan(2)$ . Since the tangent is positive, this angle is in the first quadrant, so it is in the specified interval.
On which of the following open intervals is the function $f(x) = x - 2\sin(x)$ increasing?
Answer: $(\pi/3, 5\pi/3)$ A function is increasing where its derivative is positive. $f'(x) = 1 - 2\cos(x)$ . We need to solve $1 - 2\cos(x) > 0$ , which simplifies to $1 > 2\cos(x)$ , or $\cos(x) < 1/2$ . On the interval $[0, 2\pi]$ , $\cos(x) = 1/2$ at $x = \pi/3$ and $x = 5\pi/3$ . The inequality $\cos(x) < 1/2$ is true for $x$ values between these two points. Thus, the function is increasing on the interval $(\pi/3, 5\pi/3)$ .
Let $f(x) = A e^x - B \cos(x)$ . Given that $f'(0) = 3$ and $f''(0) = 5$ , what is the value of $A + B$ ?
Answer: 5 First, find the first and second derivatives of $f(x)$ . $f'(x) = A e^x + B \sin(x)$ . $f''(x) = A e^x + B \cos(x)$ . Now use the given conditions. From $f'(0) = 3$ , we have $A e^0 + B \sin(0) = 3 \Rightarrow A(1) + B(0) = 3 \Rightarrow A = 3$ . From $f''(0) = 5$ , we have $A e^0 + B \cos(0) = 5 \Rightarrow A(1) + B(1) = 5 \Rightarrow A + B = 5$ . The question asks for $A+B$ , which is directly given by the second condition.
For the function $g(t) = e^t (\sin t + \cos t)$ , find a value of $t$ in the interval $(0, 2\pi)$ for which the tangent line to the graph of $g$ is horizontal.
Answer: $\pi/2$ A horizontal tangent line occurs when the derivative is zero. Using the product rule: $g'(t) = (e^t)(\sin t + \cos t) + e^t(\cos t - \sin t)$ . Factoring out $e^t$ gives $g'(t) = e^t(\sin t + \cos t + \cos t - \sin t) = 2e^t \cos t$ . Since $e^t$ is never zero, we must have $\cos t = 0$ . In the interval $(0, 2\pi)$ , this occurs at $t = \pi/2$ and $t = 3\pi/2$ . Of the choices provided, $\pi/2$ is a correct answer.
Let $h(x) = e^{\cos(2x)}$ . What is the value of $h'(\pi/4)$ ?
Answer: -2 This requires the chain rule. The derivative of $e^u$ is $e^u \cdot u'$ . Here, $u = \cos(2x)$ , and $u' = -\sin(2x) \cdot 2$ . So, $h'(x) = e^{\cos(2x)} \cdot (-\sin(2x) \cdot 2) = -2\sin(2x)e^{\cos(2x)}$ . Now, evaluate at $x = \pi/4$ : $h'(\pi/4) = -2\sin(\pi/2)e^{\cos(\pi/2)} = -2(1)e^0 = -2(1)(1) = -2$ .
If $y = \cos^3(\ln x)$ , what is $\frac{dy}{dx}$ ?
Answer: $-\frac{3}{x} \cos^2(\ln x) \sin(\ln x)$ This requires applying the chain rule multiple times. Let $y = u^3$ , $u = \cos(v)$ , and $v = \ln x$ . Then $\frac{dy}{du} = 3u^2$ , $\frac{du}{dv} = -\sin(v)$ , and $\frac{dv}{dx} = \frac{1}{x}$ . By the chain rule, $\frac{dy}{dx} = \frac{dy}{du} \frac{du}{dv} \frac{dv}{dx} = 3u^2 \cdot (-\sin v) \cdot \frac{1}{x}$ . Substituting back gives $\frac{dy}{dx} = 3\cos^2(\ln x) \cdot (-\sin(\ln x)) \cdot \frac{1}{x} = -\frac{3}{x} \cos^2(\ln x) \sin(\ln x)$ .
Let $h(x) = \frac{\sin x}{e^x}$ . What is the $x$ -coordinate of the first positive critical point of $h(x)$ ?
Answer: $\pi/4$ Critical points occur where the derivative is zero or undefined. We find the derivative using the quotient rule: $h'(x) = \frac{(\cos x)(e^x) - (\sin x)(e^x)}{(e^x)^2} = \frac{e^x(\cos x - \sin x)}{e^{2x}} = \frac{\cos x - \sin x}{e^x}$ . To find critical points, we set $h'(x) = 0$ . Since $e^x$ is never zero, we must have $\cos x - \sin x = 0$ , which means $\cos x = \sin x$ . This is equivalent to $\tan x = 1$ . The first positive value of $x$ for which this is true is $x = \pi/4$ .
If $f(x) = \cos^2(x)$ , find the value of $f''(\pi/6)$ .
Answer: -1 First find the first derivative using the chain rule: $f'(x) = 2\cos(x)(-\sin(x)) = -2\sin(x)\cos(x)$ . This can be simplified using the double-angle identity to $f'(x) = -\sin(2x)$ . Now, find the second derivative: $f''(x) = -\cos(2x) \cdot 2 = -2\cos(2x)$ . Finally, evaluate at $x=\pi/6$ : $f''(\pi/6) = -2\cos(2 \cdot \pi/6) = -2\cos(\pi/3) = -2(1/2) = -1$ .

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