Question

If f and g are twice differentiable functions such that g(x) = eᶠ⁽ˣ⁾ and g"(x) = h(x)eᶠ⁽ˣ⁾, then h(x) =

a. f'(x) + f"(x)
b. f'(x) + (f"(x))²
c. (f'(x) + f"(x))²
d. (f'(x))² + f"(x)
e. 2f'(x) + f"(x)

Answer 1

The function h(x) is determined by differentiating g(x) and applying the chain rule and product rule. The correct expression for h(x) is (f'(x))2 + f"(x), which corresponds to option (d) in the question.

Step-by-step explanation:

The question is asking to find the function h(x) given that g(x) is an exponential function where g(x) = ef(x) and its second derivative g"(x) is given as h(x)ef(x). To solve this problem, we need to apply the chain rule and product rule of differentiation.

First, we find g'(x) by differentiating g(x) with respect to x using the chain rule: g'(x) = f'(x)ef(x). Then, we differentiate g'(x) to find g"(x). Applying the product rule, g"(x) = (f'(x))2ef(x) + f"(x)ef(x).

Since we are given that g"(x) = h(x)ef(x), we can compare this with the expression for g"(x) to solve for h(x). So h(x) = (f'(x))2 + f"(x), which corresponds to Answer (d) in the provided options.