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Let f(1) = e^(π/4), f'(1) = 5e^(π/4), g(1) = -3, and g'(1) = 1.

Find h'(1), where h(x) = g(x)arctan(ln(f(x))).
Please enter your answer in decimal format with three significant digits after the decimal point.

2 Answers

5 votes

Final answer:

To find h'(1), we need to take the derivative of the function h(x) = g(x)arctan(ln(f(x))) and evaluate it at x = 1. Substituting the given values and using the chain rule, the value of h'(1) is -0.0316.

Step-by-step explanation:

To find h'(1), we need to take the derivative of the function h(x) = g(x)arctan(ln(f(x))) and evaluate it at x = 1.

First, let's find the derivative of the function. Applying the chain rule, we get:

h'(x) = g'(x) * arctan(ln(f(x))) + g(x) * (1/(1+ln(f(x))^2) * (1/f(x)) * f'(x)

Now, substitute the given values at x = 1:

h'(1) = g'(1) * arctan(ln(f(1))) + g(1) * (1/(1+ln(f(1))^2) * (1/f(1)) * f'(1)

Substituting the given values, we get: h'(1) = 1 * arctan(ln(e^(π/4))) + (-3) * (1/(1+ln(e^(π/4))^2) * (1/e^(π/4))) * 5e^(π/4)

Calculating this expression gives us the value of h'(1) as -0.0316 (rounded to three significant digits).

User Ferrard
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8.6k points
3 votes

Let f(1) =
e^{(\pi )/(4) }, f'(1) = 5
e^{(\pi )/(4) }, g(1) = -3, and g'(1) = 1, find h'(1), where h(x) = g(x)arctan(ln(f(x))). So, h'(1) = 0.396

Let f(1) =
e^{(\pi )/(4) }, f'(1) = 5
e^{(\pi )/(4) }, g(1) = -3, and g'(1) = 1.

To find h'(1), where h(x) = g(x)arctan(ln(f(x))), we proceed as follows

To find h'(x), since h(x) = g(x)arctan(ln(f(x))), we need to find h'(x). using the product rule of differentiation

duv/dx = uv' + vu'

Now, u = g(x) and v = arctan(ln(f(x)))

Let ln(f(x)) = w

Differentiating, we have

u'(x) = g'(x) and

v'(x) = (arctan(w))' = (arctan(w))/dw × dw/df × df(x)/dx = 1/[1 + w²] × 1/f(x) × f'(x)

= 1/[1 + (ln(f(x))²] × 1/f(x) × f'(x)

= f'(x)/{[1 + (ln(f(x))²]f(x)}

h'(x) = uv' + vu'

So, substituting the values of the variables into the equation, we have that

= g(x)f'(x)/{[1 + (ln(f(x))²]f(x)} + arctan(ln(f(x)))(g'(x))

Since x = 1

h'(1) = g(1)f'(1)/{[1 + (ln(f(1))²]f(1)} + arctan(ln(f(1)))(g'(1))

Given that

  • f(1) =
    e^{(\pi )/(4) },
  • f'(1) = 5
    e^{(\pi )/(4) },
  • g(1) = -3, and
  • g'(1) = 1.

Substituting the values of the variables into the equation, we have that

h'(1) = g(1)f'(1)/{[1 + (ln(f(1))²]f(1)} + arctan(ln(f(1)))(g'(1))

h'(1) = (-3)
e^{(\pi )/(4) },/{[1 + (ln(5
e^{(\pi )/(4) })²]e^(π/4)} + arctan(ln(
e^{(\pi )/(4) })(1)

h'(1) = -3/{[1 + (ln(5
e^{(\pi )/(4) })²]} + arctan(ln(
e^{(\pi )/(4) })(1)

h'(1) = -3/{[1 + (ln(5
e^{(\pi )/(4) })²]} + arctan(π/4ln(e))

h'(1) = -3/{[1 + (ln5 + ln
e^{(\pi )/(4) })²]} + arctan(π/4 × 1)

h'(1) = -3/{[1 + (ln5 + π/4 × lne)²]} + arctan(π/4 × 1)

h'(1) = -3/{[1 + (ln5 + π/4 × 1)²]} + arctan(π/4)

h'(1) = -3/(1 + (ln5 + π/2)²)+ arctan(π/4)

h'(1) = -3/[1 + (1.6094 + π/2)²]+ arctan(0.7854)

h'(1) = -3/(1 + (1.6094 + 1.5708)² + 0.6658

h'(1) = -3/(1 + 3.1802²) + 0.6658

h'(1) = -3/(1 + 10.1137) + 0.6658

h'(1) = -3/11.1137 + 0.6658

h'(1) = -0.2699 + 0.6658

h'(1) = 0.3959

h'(1) ≅ 0.396

User Ganpaan
by
8.1k points