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

Let g(x)=ln(x4) g(x)=\ln \left(x^{4}\right) .\newlineFind g(x) g^{\prime}(x) .\newlineChoose 11 answer:\newline(A) 4x \frac{4}{x} \newline(B) 1x4 \frac{1}{x^{4}} \newline(C) 4ln(x3) 4 \ln \left(x^{3}\right) \newline(D) 1ln(x)+4x3 \frac{1}{\ln (x)}+4 x^{3}

Full solution

Q. Let g(x)=ln(x4) g(x)=\ln \left(x^{4}\right) .\newlineFind g(x) g^{\prime}(x) .\newlineChoose 11 answer:\newline(A) 4x \frac{4}{x} \newline(B) 1x4 \frac{1}{x^{4}} \newline(C) 4ln(x3) 4 \ln \left(x^{3}\right) \newline(D) 1ln(x)+4x3 \frac{1}{\ln (x)}+4 x^{3}
  1. Apply Chain Rule: Apply the chain rule to differentiate g(x)=ln(x4)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)\ln(u) and the inner function is u=x4u = x^{4}. The derivative of ln(u)\ln(u) with respect to uu is 1u\frac{1}{u}, and the derivative of u=x4u = x^{4} with respect to xx is 4x34x^{3}. So, g'(x) = \left(\frac{\(1\)}{u}\right) \cdot \left(\frac{du}{dx}\right) = \left(\frac{\(1\)}{x^{\(4\)}}\right) \cdot \left(\(4x^{33}\right)\.
  2. Simplify Expression: Simplify the expression for g(x)g'(x). We have g(x)=1x4(4x3)g'(x) = \frac{1}{x^{4}} \cdot (4x^{3}). When we multiply the fractions, we can cancel out x3x^{3} from the numerator and denominator. This gives us g(x)=4xg'(x) = \frac{4}{x}.
  3. Match Answer Choices: Match the simplified expression to the given answer choices.\newlineThe simplified expression for the derivative is g(x)=4xg'(x) = \frac{4}{x}, which corresponds to answer choice (A) 4x\frac{4}{x}.

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