Question

the derivative of $f(x)=7sqrt4{x^3}$ is $f(x)=\frac{b}{sqrt4{x}}$. find the value of $b$. give an exact answer either as a decimal or as a fully simplified fraction (in the form n/m).

Explanation:

Step1: Rewrite the function using exponent rules

We know that \(\sqrt[4]{x^{3}}=x^{\frac{3}{4}}\), so \(f(x) = 7x^{\frac{3}{4}}\).

Step2: Apply the power rule for differentiation

The power rule is \((x^{n})^\prime=nx^{n - 1}\). For \(y = ax^{n}\), \(y^\prime=anx^{n-1}\). Here \(a = 7\) and \(n=\frac{3}{4}\). Then \(f^\prime(x)=7\times\frac{3}{4}x^{\frac{3}{4}-1}\).

Step3: Simplify the exponent and the coefficient

\(f^\prime(x)=\frac{21}{4}x^{-\frac{1}{4}}\). Since \(x^{-m}=\frac{1}{x^{m}}\) and \(\sqrt[4]{x}=x^{\frac{1}{4}}\), we can rewrite \(f^\prime(x)\) as \(f^\prime(x)=\frac{21}{4\sqrt[4]{x}}\).

Answer:

\(b = 21/4\) (or \(b = 5.25\))