Question

Determine the domain and differentiate $f(x) = \ln \sqrt{4 + 7x^2}$. Domain of $f$: $(-\infty,\infty)$; $f'(x) = \frac{7x}{7x^2 + 4}$ Domain of $f$: $(-\infty,\infty)$; $f'(x) = \frac{1}{14x^2 + 8}$ Domain of $f$: $(-\infty,\infty)$; $f'(x) = \frac{1}{7x^2 + 4}$ Domain of $f$: $(0,\infty)$; $f'(x) = \frac{7x}{7x^2 + 4}$ Domain of $f$: $(0,\infty)$; $f'(x) = \frac{1}{7x^2 + 4}$

          Determine the domain and differentiate $f(x) = \ln \sqrt{4 + 7x^2}$.
Domain of $f$: $(-\infty,\infty)$; $f'(x) = \frac{7x}{7x^2 + 4}$
Domain of $f$: $(-\infty,\infty)$; $f'(x) = \frac{1}{14x^2 + 8}$
Domain of $f$: $(-\infty,\infty)$; $f'(x) = \frac{1}{7x^2 + 4}$
Domain of $f$: $(0,\infty)$; $f'(x) = \frac{7x}{7x^2 + 4}$
Domain of $f$: $(0,\infty)$; $f'(x) = \frac{1}{7x^2 + 4}$

Determine the domain and differentiate f(x) = ln√(4 + 7x^2).
Domain of f: (-∞,∞); f'(x) = (7x)/(7x^2 + 4)
Domain of f: (-∞,∞); f'(x) = (1)/(14x^2 + 8)
Domain of f: (-∞,∞); f'(x) = (1)/(7x^2 + 4)
Domain of f: (0,∞); f'(x) = (7x)/(7x^2 + 4)
Domain of f: (0,∞); f'(x) = (1)/(7x^2 + 4)

Added by Tommy S.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Madhur L

04/21/2023

Step 1

The square root function inside the natural logarithm must have a non-negative argument, so 4 + 7x^2 ≥ 0. This inequality simplifies to x^2 ≥ -4/7, which is true for all real numbers x. Therefore, the domain of f is (-∞, ∞). Show more…

Show all steps

Thanks for your feedback!

Determine the domain and differentiate f(x) = ln(sqrt(4 + 7x^(2)). Domain of f: (-∞, ∞); f'(x) = (7x)/(7x^(2) + 4) Domain of f: (-∞, ∞); f'(x) = (1)/(14x^(2) + 8) Domain of f: (-∞, ∞); f'(x) = (1)/(7x^(2) + 4) Domain of f: (0, ∞); f'(x) = (7x)/(7x^(2) + 4) Domain of f: (0, ∞); f'(x) = (1)/(7x^(2) + 4) Determine the domain and differentiate f(x) = ln(sqrt(4 + 7x^(2)). Domain of f: (-∞, ∞); f(x) = 7x^(2) + 4 Domain of f: (-∞, ∞); f'(x) = 14x^(2) + 8 Domain of f: (0, ∞); f'(x) = 7x^(2) + 4