Post

Created by @nathanedwards
 at November 5th 2023, 1:12:44 am.

Question:

Let f(x)f(x) and g(x)g(x) be differentiable functions, and let h(x)=f(x)g(x)h(x) = f(x) \cdot g(x). Suppose f(x)=3x22f'(x) = 3x^2 - 2 and g(x)=4xg'(x) = \frac{4}{x} for all x0x \neq 0.

a) Using the product rule, find an expression for h(x)h'(x) in terms of f(x)f(x), f(x)f'(x), g(x)g(x), and g(x)g'(x).

b) Evaluate h(2)h'(2).

Answer:

a) According to the product rule, the derivative of the product of two functions is given by:

h(x)=f(x)g(x)+g(x)f(x)h'(x) = f(x) \cdot g'(x) + g(x) \cdot f'(x)

Substituting the given functions, we have:

h(x)=(f(x)g(x))+(g(x)f(x))=(f(x)4x)+(g(x)(3x22))h'(x) = (f(x) \cdot g'(x)) + (g(x) \cdot f'(x)) = (f(x) \cdot \frac{4}{x}) + (g(x) \cdot (3x^2 - 2))

Therefore, we can express h(x)h'(x) in terms of f(x)f(x), f(x)f'(x), g(x)g(x), and g(x)g'(x) as:

h(x)=4f(x)x+g(x)(3x22)h'(x) = \frac{4f(x)}{x} + g(x)(3x^2 - 2)

b) To evaluate h(2)h'(2), we substitute x=2x = 2 into the expression for h(x)h'(x) from part a:

h(2)=4f(2)2+g(2)(3(2)22)h'(2) = \frac{4f(2)}{2} + g(2)(3(2)^2 - 2)

Using the given information that f(x)=3x22f'(x) = 3x^2 - 2 and g(x)=4xg'(x) = \frac{4}{x}, we can find f(2)f(2) and g(2)g(2).

Since f(x)=3x22f'(x) = 3x^2 - 2, we can integrate f(x)f'(x) to find f(x)f(x). Integrating, we get:

f(x)=(3x22)dx=x32x+Cf(x) = \int (3x^2 - 2)dx = x^3 - 2x + C

where CC is the constant of integration.

To find the constant of integration, we need additional information. Since no other information is given, we assume C=0C = 0 for simplicity. Therefore, f(x)=x32xf(x) = x^3 - 2x.

Next, we find g(2)g(2) using the given information g(x)=4xg'(x) = \frac{4}{x}. Integrating, we get:

g(x)=4xdx=4lnx+Cg(x) = \int \frac{4}{x}dx = 4\ln|x| + C

Again, assuming C=0C = 0 for simplicity, we have g(x)=4lnxg(x) = 4\ln|x|.

Substituting the found values into h(2)h'(2), we get:

h(2)=4f(2)2+g(2)(3(2)22)=4(f(2))2+g(2)(122)h'(2) = \frac{4f(2)}{2} + g(2)(3(2)^2 - 2) = \frac{4(f(2))}{2} + g(2)(12 - 2)
h(2)=2((2322)+4ln210)h'(2) = 2((2^3 - 2\cdot 2) + 4\ln|2| \cdot 10)
h(2)=2(4+4ln210)h'(2) = 2(4 + 4\ln|2| \cdot 10)
h(2)=2(4+40ln2)h'(2) = 2(4 + 40\ln|2|)

Therefore, h(2)=8+80ln2h'(2) = 8 + 80\ln|2|.

The final result is h(2)=8+80ln2h'(2) = 8 + 80\ln|2|.