Derivatives of Trigonometry and Logarithmic Functions - AP Calculus BC

A velocity function is $v(x)=2\sin x-\ln x$ . What is $v'(x)$ for $x>0$ ?
Answer: $2\cos x-\dfrac{1}{x}$ This problem involves differentiating trigonometric and logarithmic functions. The derivative of sin(x) is cos(x), and the derivative of ln(x) is 1/x. For v(x) = 2sin(x) - ln(x), we apply these rules to get v'(x) = 2cos(x) - 1/x. A student might incorrectly think the derivative of sin(x) is -cos(x), leading to the wrong answer -2cos(x) - 1/x. Always remember that sine differentiates to positive cosine, while cosine differentiates to negative sine.
A decay model is $m(t)=9e^{-t}+\cos t$ . What is $m'(t)$ ?
Answer: $-9e^{-t}-\sin t$ This question tests the skill of differentiating combinations of exponential and trigonometric functions. The derivative of a $e^{-t}$ is a (-1) $e^{-t}$ = -a $e^{-t}$ , and the derivative of cos t is -sin t. For m(t) = 9 $e^{-t}$ + cos t, combine to get -9 $e^{-t}$ - sin t. Thus, m'(t) = -9 $e^{-t}$ - sin t. A tempting distractor is choice B, 9 $e^{-t}$ - sin t, which forgets the negative sign from the exponential's chain rule. Always apply the chain rule correctly for exponentials and include signs for trigonometric derivatives.
For $r(x)=e^{x}\cos x$ in a physics model, what is $r'(x)$ ?
Answer: $e^{x}(\cos x-\sin x)$ This question tests basic function differentiation using the product rule with exponential and trigonometric functions. To find r'(x), we apply the product rule to $e^x$ ·cos(x): r'(x) = ( $e^x$ )'·cos(x) + $e^x$ ·(cos(x))' = $e^x$ ·cos(x) + $e^x$ ·(-sin(x)) = $e^x$ (cos(x) - sin(x)). A tempting error would be to differentiate each factor separately without using the product rule, yielding $e^x$ ·(-sin(x)). When differentiating products of functions, always use the product rule: (fg)' = f'g + fg'.
A signal is $S(x)=\cos x + e^{-x}$ . What is $S'(x)$ ?
Answer: $-\sin x - e^{-x}$ This question tests basic function differentiation combining trigonometric and exponential functions. To find $S'(x)$ , we differentiate each term: the derivative of $\cos(x)$ is $-\sin(x)$ , and the derivative of $e^{-x}$ is $e^{-x} \cdot (-1) = -e^{-x}$ . Therefore, $S'(x) = -\sin(x) - e^{-x}$ . A tempting mistake would be to forget the negative sign from the chain rule when differentiating $e^{-x}$ . When differentiating $e^{-x}$ , multiply by the derivative of the exponent, which is -1.
The function $g(x)=5\ln x+4\sin x$ models a rate. What is $g'(x)$ ?
Answer: $\dfrac{5}{x}+4\cos x$ This problem requires differentiating logarithmic and trigonometric functions. The derivative of ln(x) is 1/x, so the derivative of 5ln(x) is 5/x. The derivative of sin(x) is cos(x), so the derivative of 4sin(x) is 4cos(x). Therefore, g'(x) = 5/x + 4cos(x). A common mistake would be to confuse the derivative of sin(x) with sin(x) itself, leading to answer C. Remember the fundamental derivatives: d/dx[ln(x)] = 1/x and d/dx[sin(x)] = cos(x).
A periodic model is $p(t)=\sin t-\cos t+e^{t}$ . What is $p'(t)$ ?
Answer: $\cos t+\sin t+e^{t}$ This question involves basic function differentiation combining trigonometric and exponential functions. To find p'(t), we differentiate each term: the derivative of sin(t) is cos(t), the derivative of -cos(t) is -(-sin(t)) = sin(t), and the derivative of $e^t$ is $e^t$ . Therefore, p'(t) = cos(t) + sin(t) + $e^t$ . A tempting error would be to get the sign wrong when differentiating -cos(t). When differentiating sums with multiple trigonometric terms, track the signs carefully using standard derivative formulas.
A position is $p(x)=9\sin x+e^{x}+\ln x$ . What is $p'(x)$ for $x>0$ ?
Answer: $9\cos x+e^{x}+\dfrac{1}{x}$ This problem involves differentiating trigonometric, exponential, and logarithmic functions. The derivative of sin(x) is cos(x), the derivative of $e^x$ is $e^x$ , and the derivative of ln(x) is 1/x. For p(x) = 9sin(x) + $e^x$ + ln(x), we get p'(x) = 9cos(x) + $e^x$ + 1/x. A student might incorrectly apply a negative sign to the derivative of sine, resulting in -9sin(x) + $e^x$ + 1/x. When differentiating sums of multiple functions, differentiate each term independently and then combine the results.
A cooling model uses $T(x)=2e^{-x}+\cos x$ ; what is the derivative $T'(x)$ ?
Answer: $-2e^{-x}-\sin x$ This problem tests the skill of differentiating exponential and trigonometric functions. The function is $T(x) = 2 e^{-x} + \cos x$ , a sum of an exponential and cosine. The derivative of $e^{-x}$ is $e^{-x} * (-1)$ , so for $2 e^{-x}$ it is $-2 e^{-x}$ . The derivative of $\cos x$ is $-\sin x$ . Adding these gives $T'(x) = -2 e^{-x} - \sin x$ . A tempting distractor is B, $2 e^{-x} - \sin x$ , which forgets the negative sign from the chain rule in differentiating $e^{-x}$ . A transferable strategy is to apply the chain rule for exponentials like $e^{u(x)}$ as $e^u * u'$ , ensuring the sign of $u'$ is included correctly.
A rotating wheel height is $h(t)=8\cos(2t)-1$ . What is $h'(t)$ ?
Answer: $-16\sin(2t)$ This question tests the skill of differentiating basic trigonometric functions. The derivative of a cosine function of the form $a \cos(kt)$ is $-a k \sin(kt)$ , incorporating the negative sign from the cosine derivative rule. In $h(t) = 8 \cos(2t) - 1$ , differentiate to get $8 \times (-2) \sin(2t) = -16 \sin(2t)$ , with the constant -1 becoming zero. Thus, $h'(t) = -16 \sin(2t)$ . A tempting distractor is choice A, $16 \sin(2t)$ , which omits the negative sign from the cosine derivative. Always include the appropriate sign and chain rule factor when differentiating cosine functions with composite arguments.
A periodic current is $i(t)=8\sin(2t)$ . What is $i'(t)$ ?
Answer: $16\cos(2t)$ This question tests basic function differentiation with trigonometric functions using chain rule. To find i'(t), we differentiate 8sin(2t): the derivative of sin(2t) is cos(2t) multiplied by the derivative of 2t, which is 2. Therefore, i'(t) = 8·2·cos(2t) = 16cos(2t). A tempting mistake would be to forget the chain rule factor of 2, yielding 8cos(2t) instead of 16cos(2t). When differentiating sin(ax), multiply by the coefficient a.
A signal is modeled by $g(x)=\dfrac{e^x}{\sin x}$ where defined; what is $g'(x)$ ?
Answer: $\dfrac{e^x\sin x-e^x\cos x}{\sin^2 x}$ This problem tests the skill of differentiating quotients involving exponential and trigonometric functions. The function is g(x) = $e^x$ / sin x, requiring the quotient rule. The quotient rule is (v u' - u v') / $v^2$ , where u = $e^x$ (u' = $e^x$ ), v = sin x (v' = cos x). This gives [sin x * $e^x$ - $e^x$ * cos x] / $\sin^2$ x = $e^x$ (sin x - cos x) / $\sin^2$ x. This is the correct g'(x). A tempting distractor is C, $e^x$ (sin x + cos x) / $\sin^2$ x, which switches the sign by subtracting incorrectly in the numerator. A transferable strategy is to memorize the quotient rule order as 'bottom times top derivative minus top times bottom derivative' over bottom squared, and factor common terms afterward.
An intensity function is $F(t)=2\sin t+\ln t$ . What is $F'(t)$ ?
Answer: $2\cos t+\dfrac{1}{t}$ This problem involves differentiating trigonometric and logarithmic functions. The derivative of 2sin(t) is 2cos(t), since the derivative of sin(t) is cos(t). The derivative of ln(t) is 1/t, following the fundamental logarithmic derivative rule. Combining these gives F'(t) = 2cos(t) + 1/t. Choice B incorrectly keeps ln(t) in the derivative, confusing the function with its derivative. When differentiating mixed functions, apply each rule separately: sin(x)→cos(x) and ln(x)→1/x, then combine the results.
A particle's position is $s(t)=3\sin t-2e^{t}$ . What is $s'(t)$ ?
Answer: $3\cos t-2e^{t}$ This problem requires differentiating a combination of trigonometric and exponential functions. The derivative of sin(t) is cos(t), following the standard trigonometric derivative rule. The derivative of - $2e^t$ is - $2e^t$ , since the derivative of $e^t$ is $e^t$ and constants factor out. Therefore, s'(t) = 3cos(t) - $2e^t$ . A common error would be thinking the derivative of sin(t) is -cos(t), which would incorrectly lead to answer C. Remember that derivatives of basic trigonometric and exponential functions follow specific patterns: sin→cos, cos→-sin, and $e^x$ → $e^x$ .
A sensor outputs $g(x)=\cos(5x)$ . What is $g'(x)$ ?
Answer: $-5\sin(5x)$ This question tests basic function differentiation using the chain rule with trigonometric functions. To find g'(x), we differentiate cos(5x): the derivative is -sin(5x) multiplied by the derivative of the inner function 5x, which is 5. Therefore, g'(x) = -sin(5x)·5 = -5sin(5x). A tempting mistake would be to forget the chain rule factor of 5, yielding just -sin(5x). When differentiating composite trigonometric functions, always multiply by the derivative of the argument.
A signal is modeled by $f(x)=3\sin x-2e^x$ . What is $f'(x)$ ?
Answer: $3\cos x-2e^x$ This problem involves differentiating trigonometric and exponential functions. The derivative of sin(x) is cos(x), so the derivative of 3sin(x) is 3cos(x). The derivative of $e^x$ is $e^x$ , so the derivative of - $2e^x$ is - $2e^x$ . Therefore, f'(x) = 3cos(x) - $2e^x$ . Students might confuse the derivative of sin(x) with -sin(x), which would lead to the incorrect answer C. When differentiating, remember that d/dx[sin(x)] = cos(x) and d/dx[ $e^x$ ] = $e^x$ .
A vibration is $q(t)=\sin t\, +\, e^{t}$ . What is $q'(t)$ ?
Answer: $\cos t+e^{t}$ This question tests basic function differentiation combining trigonometric and exponential functions. To find q'(t), we differentiate each term: the derivative of sin(t) is cos(t), and the derivative of $e^t$ is $e^t$ . Therefore, q'(t) = cos(t) + $e^t$ . A tempting error would be to confuse sine and cosine derivatives, perhaps writing sin(t) + $e^t$ instead. When differentiating sums of basic transcendental functions, apply standard formulas systematically.
An intensity function is $I(t)=\cos t+2\ln t$ . What is $I'(t)$ for $t>0$ ?
Answer: $-\sin t+\dfrac{2}{t}$ This problem requires differentiating trigonometric and logarithmic functions. The derivative of cos(t) is -sin(t), and the derivative of ln(t) is 1/t. For I(t) = cos(t) + 2ln(t), we get I'(t) = -sin(t) + 2(1/t) = -sin(t) + 2/t. A common error would be to forget the negative sign when differentiating cosine, giving the incorrect answer sin(t) + 2/t. When differentiating logarithmic functions with coefficients, multiply the coefficient by the derivative of the logarithm.
A model uses $m(t)=\cos t+\ln t$ . What is $m'(t)$ for $t>0$ ?
Answer: $-\sin t+\dfrac{1}{t}$ This question involves basic function differentiation combining trigonometric and logarithmic functions. To find $m'(t)$ , we differentiate each term: the derivative of $cos(t)$ is $-sin(t)$ , and the derivative of $ln(t)$ is $1/t$ . Therefore, $m'(t)$ = $-sin(t)$ + $1/t$ . A common mistake would be to forget the negative sign when differentiating cosine, leading to $sin(t)$ instead of $-sin(t)$ . When combining trigonometric and logarithmic derivatives, apply standard formulas to each term independently.
In economics, $D(x)=\ln x+e^{x}$ . What is $D'(x)$ for $x>0$ ?
Answer: $\dfrac{1}{x}+e^{x}$ This question involves basic function differentiation combining logarithmic and exponential functions. To find D'(x), we differentiate each term: the derivative of ln(x) is 1/x, and the derivative of $e^x$ is $e^x$ . Therefore, D'(x) = 1/x + $e^x$ . A common error would be to confuse the derivative of ln(x) with ln(x) itself or with x. When working with combinations of logarithmic and exponential functions, apply the standard derivative formulas to each term.
In acoustics, $B(x)=\sin x-\ln x$ . What is $B'(x)$ for $x>0$ ?
Answer: $\cos x-\dfrac{1}{x}$ This question involves basic function differentiation combining trigonometric and logarithmic functions. To find B'(x), we differentiate each term: the derivative of sin(x) is cos(x), and the derivative of -ln(x) is -1/x. Therefore, B'(x) = cos(x) - 1/x. A common mistake would be to forget the negative sign on the second term or confuse the derivatives of sine and cosine. When differentiating differences of functions, preserve the signs and apply derivative rules to each term.

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