Question:
Let f(x) and g(x) be differentiable functions, and let h(x)=f(x)⋅g(x). Suppose f′(x)=3x2−2 and g′(x)=x4 for all x=0.
a) Using the product rule, find an expression for h′(x) in terms of f(x), f′(x), g(x), and g′(x).
b) Evaluate 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)Substituting the given functions, we have:
h′(x)=(f(x)⋅g′(x))+(g(x)⋅f′(x))=(f(x)⋅x4)+(g(x)⋅(3x2−2))Therefore, we can express h′(x) in terms of f(x), f′(x), g(x), and g′(x) as:
h′(x)=x4f(x)+g(x)(3x2−2)b) To evaluate h′(2), we substitute x=2 into the expression for h′(x) from part a:
h′(2)=24f(2)+g(2)(3(2)2−2)Using the given information that f′(x)=3x2−2 and g′(x)=x4, we can find f(2) and g(2).
Since f′(x)=3x2−2, we can integrate f′(x) to find f(x). Integrating, we get:
f(x)=∫(3x2−2)dx=x3−2x+Cwhere C is the constant of integration.
To find the constant of integration, we need additional information. Since no other information is given, we assume C=0 for simplicity. Therefore, f(x)=x3−2x.
Next, we find g(2) using the given information g′(x)=x4. Integrating, we get:
g(x)=∫x4dx=4ln∣x∣+CAgain, assuming C=0 for simplicity, we have g(x)=4ln∣x∣.
Substituting the found values into h′(2), we get:
h′(2)=24f(2)+g(2)(3(2)2−2)=24(f(2))+g(2)(12−2)h′(2)=2((23−2⋅2)+4ln∣2∣⋅10)h′(2)=2(4+4ln∣2∣⋅10)h′(2)=2(4+40ln∣2∣)Therefore, h′(2)=8+80ln∣2∣.
The final result is h′(2)=8+80ln∣2∣.