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 tests the skill of differentiating basic trigonometric and exponential functions. The function is f(x) = 3 sin x - 2 $e^x$, which is a linear combination of sine and exponential terms. The derivative of sin x is cos x, so the derivative of 3 sin x is 3 cos x. The derivative of $e^x$ is $e^x$ itself, so the derivative of -2 $e^x$ is -2 $e^x$. Combining these results gives f'(x) = 3 cos x - 2 $e^x$. A tempting distractor is C, -3 sin x - 2 $e^x$, which might occur if one incorrectly applies a negative sign to the derivative of sin x and forgets the correct cos x form. A transferable strategy is to differentiate each term in a sum or difference individually using the appropriate basic derivative rules before combining them.

This problem tests the skill of differentiating trigonometric functions with chain rule and logarithmic functions. The function is w(x) = sin(3x) - ln x, a difference of a composed sine and a logarithm. The derivative of sin(3x) is cos(3x) * 3 by the chain rule, giving 3 cos(3x). The derivative of -ln x is -1/x. Thus, w'(x) = 3 cos(3x) - 1/x. A tempting distractor is A, 3 sin(3x) - 1/x, which neglects differentiating the sin(3x) term entirely. A transferable strategy is to apply the chain rule to compositions like sin(u(x)) by multiplying the outer derivative by u', and handle each term in sums independently.

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 exponential and logarithmic functions. The derivative of $e^t$ is $e^t$ (the exponential function is its own derivative). For the logarithmic term 5ln(t), we use the fact that d/dt[ln(t)] = 1/t, so the derivative of 5ln(t) is 5/t. Therefore, P'(t) = $e^t$ + 5/t. A common mistake would be to think the derivative of ln(t) is ln(t) itself, leading to answer C. Remember that $d/dx[e^x$] = $e^x$ and d/dx[ln(x)] = 1/x for these fundamental transcendental functions.

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 involves differentiating trigonometric and logarithmic functions. The derivative of sin(x) is cos(x). For the term 3ln(x), the derivative of ln(x) is 1/x, so the derivative of 3ln(x) is 3/x. Therefore, D'(x) = cos(x) + 3/x. Students might confuse the derivative of sin(x) with -sin(x), which would lead to the incorrect answer D. When differentiating these fundamental functions, remember that d/dx[sin(x)] = cos(x) and d/dx[ln(x)] = 1/x.

This problem involves differentiating trigonometric and linear functions. The derivative of cos(t) is -sin(t), so the derivative of 9cos(t) is -9sin(t). The derivative of 2t is simply 2 (using the power rule). Therefore, h'(t) = -9sin(t) + 2. Students might forget the negative sign when differentiating cos(t), incorrectly choosing answer A. When differentiating trigonometric functions, remember that d/dt[cos(t)] = -sin(t), not sin(t).

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.