Post

Created by @nathanedwards
 at November 23rd 2023, 8:29:59 pm.

Question:

Find the derivative of the following function:

f(x)=3x24ex+2ln(x)+5sin(x) f(x) = 3x^2 - 4e^x + 2\ln(x) + 5\sin(x)

Answer:

To find the derivative of the given function, we will use the rules for finding derivatives of basic functions. We will apply the power rule for the term 3x2 3x^2 , the derivative of ex e^x for the term 4ex -4e^x , the derivative of ln(x) \ln(x) for the term 2ln(x) 2\ln(x) , and the derivative of sin(x) \sin(x) for the term 5sin(x) 5\sin(x) .

The power rule states that if f(x)=xn f(x) = x^n , then f(x)=nxn1 f'(x) = nx^{n-1} .

The derivative of ex e^x is ex e^x , and the derivative of ln(x) \ln(x) is 1x \frac{1}{x} .

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

Now, let's find the derivative of the given function f(x) f(x) :

f(x)=ddx(3x2)ddx(4ex)+ddx(2ln(x))+ddx(5sin(x)) f'(x) = \frac{d}{dx} (3x^2) - \frac{d}{dx}(4e^x) + \frac{d}{dx}(2\ln(x)) + \frac{d}{dx}(5\sin(x))

Applying the power rule and the derivatives of exponential, logarithmic, and trigonometric functions, we get:

f(x)=6x4ex+2x+5cos(x) f'(x) = 6x - 4e^x + \frac{2}{x} + 5\cos(x)

Therefore, the derivative of the given function f(x) f(x) is:

f(x)=6x4ex+2x+5cos(x) f'(x) = 6x - 4e^x + \frac{2}{x} + 5\cos(x)

So, the derivative of the function f(x) f(x) is 6x4ex+2x+5cos(x) 6x - 4e^x + \frac{2}{x} + 5\cos(x) .