This problem tests the skill of differentiating logarithmic functions composed with trigonometric functions. The function is h(x) = ln(cos x), requiring the chain rule for the composition. The derivative of ln(u) is (1/u) u', where u = cos x and u' = -sin x. Thus, h'(x) = (1/cos x) * (-sin x) = -sin x / cos x. This simplifies to -tan x, but the form matches the choice. A tempting distractor is A, 1/cos x, which forgets the chain rule's multiplication by u' = -sin x. A transferable strategy is to apply the chain rule for ln(u(x)) by computing (1/u) times u', always including the inner derivative.

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 $d/dx[e^x$] = $e^x$.

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).

This problem requires differentiating logarithmic and trigonometric functions. The derivative of ln(x) is 1/x. For the term -6cos(x), the derivative of cos(x) is -sin(x), so the derivative of -6cos(x) is -6(-sin(x)) = 6sin(x). Therefore, C'(x) = 1/x + 6sin(x). A common error would be to forget the negative sign when differentiating cos(x), leading to answer A with -6cos(x). Remember that d/dx[cos(x)] = -sin(x), and when combined with a negative coefficient, the result becomes positive.

This problem requires differentiating a combination of trigonometric and polynomial functions. The derivative of sin(t) is cos(t), so the derivative of 4sin(t) is 4cos(t). For the polynomial term -3t², we apply the power rule: the derivative is -3(2t) = -6t. Combining these results, s'(t) = 4cos(t) - 6t. A common error would be to think the derivative of sin(t) is -sin(t), which would incorrectly lead to answer C. When differentiating, remember that d/dx[sin(x)] = cos(x) and apply the constant multiple rule to handle coefficients.

This question tests the skill of differentiating basic logarithmic functions. The derivative of ln(kt), where k is a constant, is (1/(kt)) * k = 1/t, simplifying due to the chain rule. For C(t) = ln(4t) + 7, the derivative is 1/t, and the constant +7 has a derivative of zero. Therefore, C'(t) = 1/t. A tempting distractor is choice A, 1/(4t), which forgets to apply the chain rule by multiplying by the derivative of 4t, which is 4. When differentiating natural logarithms of linear functions, apply the chain rule to account for the inner derivative.

This question tests the skill of differentiating combinations of trigonometric, exponential, and logarithmic functions. The derivative of sin t is cos t, of $e^t$ is $e^t$, and of ln t is 1/t. For p(t) = sin t + $e^t$ + ln t, the result is cos t + $e^t$ + 1/t. Therefore, p'(t) = cos t + $e^t$ + 1/t. A tempting distractor is choice C, -cos t + $e^t$ + 1/t, which incorrectly uses the derivative of cosine instead of sine. For sums of diverse functions, differentiate each component using its fundamental rule and sum the results.

This question tests the skill of differentiating basic trigonometric functions. The derivative of a sine function of the form a sin(kt) is a k cos(kt), where a is a constant and k is the coefficient inside the sine. For s(t) = 5 sin(3t) + 2, the derivative is 5 * 3 cos(3t) = 15 cos(3t), and the constant +2 differentiates to zero. Therefore, s'(t) = 15 cos(3t). A tempting distractor is choice A, 5 cos(3t) + 2, which neglects the chain rule multiplication by 3 and incorrectly keeps the constant. To differentiate trigonometric functions with linear arguments, remember to use the chain rule and multiply by the inner derivative.

This question tests basic function differentiation with logarithmic and exponential functions. To find P'(t), we differentiate each term: the derivative of 5ln(t) is 5·(1/t) = 5/t, and the derivative of $-2e^t$ is $-2e^t$. Therefore, P'(t) = 5/t - $2e^t$. A tempting error would be to confuse the derivative of ln(t) with the function itself, perhaps writing 5ln(t) instead of 5/t. When differentiating functions involving natural logarithms, remember that d/dx[ln(x)] = 1/x.

This question tests basic function differentiation with trigonometric functions using chain rule. To find d'(x), we differentiate sin(πx): the derivative is cos(πx) multiplied by the derivative of πx, which is π. Therefore, d'(x) = π·cos(πx). A tempting error would be to forget the chain rule factor π, yielding just cos(πx). When differentiating sin(ax) where a is a constant, multiply by that constant.