Post

Created by @nathanedwards
 at November 1st 2023, 10:32:07 am.

AP Calculus AB Exam Question

Differentiate the following function with respect to x:

f(x)=3x24ex+ln(x)+sin(2x) f(x) = 3x^2 - 4e^x + \ln(x) + \sin(2x)

Step-by-Step Solution

To differentiate the given function with respect to x, we will use the basic rules of differentiation for each term.

Let's differentiate each term one by one:

  1. The derivative of 3x2 3x^2 with respect to x can be found using the power rule:

    ddx(3x2)=23x21=6x \frac{d}{dx}(3x^2) = 2 \cdot 3x^{2-1} = 6x
  2. The derivative of 4ex - 4e^x with respect to x can be found using the exponential rule:

    ddx(4ex)=4ex \frac{d}{dx}(-4e^x) = -4 \cdot e^x
  3. The derivative of ln(x) \ln(x) with respect to x can be found using the logarithmic rule:

    ddx(ln(x))=1x \frac{d}{dx}(\ln(x)) = \frac{1}{x}
  4. The derivative of sin(2x) \sin(2x) with respect to x can be found using the trigonometric rule:

ddx(sin(2x))=cos(2x)ddx(2x)=cos(2x)2=2cos(2x) \frac{d}{dx}(\sin(2x)) = \cos(2x) \cdot \frac{d}{dx}(2x) = \cos(2x) \cdot 2 = 2\cos(2x)

Now, we can combine all the derivative terms:

f(x)=ddx(3x2)+ddx(4ex)+ddx(ln(x))+ddx(sin(2x)) f'(x) = \frac{d}{dx}(3x^2) + \frac{d}{dx}(-4e^x) + \frac{d}{dx}(\ln(x)) + \frac{d}{dx}(\sin(2x))

Simplifying further:

f(x)=6x4ex+1x+2cos(2x) f'(x) = 6x - 4e^x + \frac{1}{x} + 2\cos(2x)

So, the derivative of the given function is:

f(x)=6x4ex+1x+2cos(2x) f'(x) = 6x - 4e^x + \frac{1}{x} + 2\cos(2x)

This is the final answer after differentiating the function with respect to x.