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let $g(x)=\\sqrt4{x^{3}}$. $g(1)=\\square$

Question

let $g(x)=\sqrt4{x^{3}}$.
$g(1)=\square$

Explanation:

Step1: Rewrite the function

Rewrite \(g(x)=\sqrt[4]{x^{3}}\) as \(g(x)=x^{\frac{3}{4}}\) using the rule \(\sqrt[n]{x^{m}} = x^{\frac{m}{n}}\).

Step2: Differentiate using the power rule

The power rule is \((x^{n})^\prime=nx^{n - 1}\). For \(y = x^{\frac{3}{4}}\), \(g^\prime(x)=\frac{3}{4}x^{\frac{3}{4}-1}=\frac{3}{4}x^{-\frac{1}{4}}=\frac{3}{4x^{\frac{1}{4}}}\).

Step3: Evaluate at \(x = 1\)

Substitute \(x = 1\) into \(g^\prime(x)\). When \(x = 1\), \(g^\prime(1)=\frac{3}{4\times1^{\frac{1}{4}}}\). Since \(1^{\frac{1}{4}}=1\), then \(g^\prime(1)=\frac{3}{4}\).

Answer:

\(\frac{3}{4}\)