Let g(x)=ln(x^(4)). Find g^(')(x). Choose 1 answer: (A) (4)/(x) (B) (1)/(ln(x))+4x^(3) (C) (1)/(x^(4)) (D) 4ln(x^(3))

Apply Chain Rule: Apply the chain rule to differentiate g(x) = ln(x^(4)). The chain rule states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function times the derivative of the inner function. Here, the outer function is ln(u) and the inner function is u = x^(4). The derivative of ln(u) with respect to u is 1/u, and the derivative of u = x^(4) with respect to x is 4x^(3). So, g'(x) = (1/u) * (du/dx) = (1/x^(4)) * (4x^(3)).Use Derivatives: Simplify the expression for g'(x). We have g'(x) = (1/x^(4)) * (4x^(3)). Simplify by canceling out x^(3) from the numerator and denominator. g'(x) = 4/x.

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Let
g(x)=ln(x^(4)).
Find
g^(')(x).
Choose 1 answer:
(A)
(4)/(x)
(B)
(1)/(ln(x))+4x^(3)
(C)
(1)/(x^(4))
(D)
4ln(x^(3))

Let $g(x)=\ln \left(x^{4}\right)$ . $\newline$ Find $g^{\prime}(x)$ . $\newline$ Choose $1$ answer: $\newline$ (A) $\frac{4}{x}$ $\newline$ (B) $\frac{1}{\ln (x)}+4 x^{3}$ $\newline$ (C) $\frac{1}{x^{4}}$ $\newline$ (D) $4 \ln \left(x^{3}\right)$

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Math Problems

Algebra 1

Multiplication with rational exponents

Full solution

Q. Let

g(x)=\ln \left(x^{4}\right)

\newline

Find

g^{\prime}(x)

\newline

Choose

1

answer:

\newline

(A)

\frac{4}{x}

\newline

(B)

\frac{1}{\ln (x)}+4 x^{3}

\newline

(C)

\frac{1}{x^{4}}

\newline

(D)

4 \ln \left(x^{3}\right)

Apply Chain Rule: Apply the chain rule to differentiate $g(x) = \ln(x^{4})$ . The chain rule states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function times the derivative of the inner function. Here, the outer function is $\ln(u)$ and the inner function is $u = x^{4}$ . The derivative of $\ln(u)$ with respect to $u$ is $1/u$ , and the derivative of $u = x^{4}$ with respect to $x$ is $4x^{3}$ . So, $g'(x) = (1/u) \cdot (du/dx) = (1/x^{4}) \cdot (4x^{3})$ .
Use Derivatives: Simplify the expression for $g'(x)$ . We have $g'(x) = \frac{1}{x^{4}} \cdot (4x^{3})$ . Simplify by canceling out $x^{3}$ from the numerator and denominator. $g'(x) = \frac{4}{x}$ .