Question

Suppose f(0)=5, f'(0)=2, f(1)=6, f'(1)=9, g(1)=3, g'(1)=2, and H(x)=f(ln(x))+ln(g(x)) find derivative at x=1

Answer 1

To find the derivative of H(x) = f(ln(x)) + ln(g(x)) at x = 1, use the chain rule and substitute the given values to find H'(1) = 8/3.

Step-by-step explanation:

To find the derivative of H(x) = f(ln(x)) + ln(g(x)) at x = 1, we can use the chain rule. The chain rule states that if we have a function H(x) = f(g(x)), the derivative of H(x) with respect to x is given by H'(x) = f'(g(x)) * g'(x). Applying this rule to our function, we have:

H'(x) = f'(ln(x)) * (1/x) + (1/g(x)) * g'(x)

Substituting the given values, we have:

H'(1) = f'(ln(1)) * (1/1) + (1/g(1)) * g'(1)

Since ln(1) = 0, we get:

H'(1) = f'(0) + (1/g(1)) * g'(1)

Given that f'(0) = 2, g(1) = 3, and g'(1) = 2, we can substitute these values into the equation:

H'(1) = 2 + (1/3) * 2

H'(1) = 2 + (2/3) = 2 + 2/3 = 8/3.