Calculus Problems and Solutions

Derivatives of Logarithmic and Exponential Functions

Problem 4

• Problem Statement: Given the function y = \ln(\sin(x)), find \frac{dy}{dx}.

• Solution:

  • We need to find the derivative of y with respect to x.

  • Using the chain rule, we have: \frac{dy}{dx} = \frac{d}{dx} [\ln(\sin(x))]

  • The derivative of \ln(u) is \frac{1}{u} \cdot \frac{du}{dx}, where u = \sin(x).

  • So, \frac{dy}{dx} = \frac{1}{\sin(x)} \cdot \frac{d}{dx}(\sin(x)).

  • The derivative of \sin(x) is \cos(x).

  • Therefore, \frac{dy}{dx} = \frac{\cos(x)}{\sin(x)}.

  • Since \frac{\cos(x)}{\sin(x)} = \cot(x), the derivative is \cot(x).

• Answer Options:

  • a. \frac{\cos(x)}{\sin(x)} (This is the correct derivative)

  • b. \frac{\cos(x)}{\sin(x)} (Repeated option)

  • c. \frac{1}{\sin(x)} (Incorrect)

  • d. \frac{1}{\sin(x)} (Incorrect)

Problem 5

• Problem Statement: Given the function y = (mx)^3 e^x, find \frac{dy}{dx}.

• Solution:

  • We need to find the derivative of y with respect to x.

  • Using the product rule, \frac{d}{dx}(uv) = u'v + uv', where u = (mx)^3 and v = e^x.

  • First, find the derivative of u = (mx)^3 with respect to x.

    • \frac{du}{dx} = 3(mx)^2 \cdot m = 3m^3x^2

  • Next, find the derivative of v = e^x with respect to x.

    • \frac{dv}{dx} = e^x

  • Now, apply the product rule:

    • \frac{dy}{dx} = (3m^3x^2)(e^x) + (mx)^3(e^x)

    • \frac{dy}{dx} = 3m^3x^2e^x + m^3x^3e^x

    • \frac{dy}{dx} = m^3x^2e^x(3 + x)

• Answer Options:

  • a. 3(mx)^{-1} (Incorrect)

  • b. \frac{3(mx) - 1}{x} (Incorrect)

  • c. 3(x)e^x (Incorrect, missing terms and constants)

  • d. 3(x) (Incorrect)

It's important to notice that the provided options do not match the correct derivative computed. We need to compare the computed derivative with the intended options.