the derivative of ( f(x)=\frac{sqrt3{x}}{sqrt{x}} ) is ( f^{prime}(x)=-\frac{x^{n}}{6} ).\nfind the value of…

the derivative of ( f(x)=\frac{sqrt3{x}}{sqrt{x}} ) is ( f^{prime}(x)=-\frac{x^{n}}{6} ).\nfind the value of ( n ).\ngive an exact answer as a fully simplified fraction (in the form ( n / m )).

the derivative of ( f(x)=\frac{sqrt3{x}}{sqrt{x}} ) is ( f^{prime}(x)=-\frac{x^{n}}{6} ).\nfind the value of ( n ).\ngive an exact answer as a fully simplified fraction (in the form ( n / m )).

Answer

Explanation:

Step1: Rewrite the function using exponent rules

We know that (\sqrt[3]{x}=x^{\frac{1}{3}}) and (\sqrt{x}=x^{\frac{1}{2}}). So (f(x)=\frac{\sqrt[3]{x}}{\sqrt{x}}=\frac{x^{\frac{1}{3}}}{x^{\frac{1}{2}}}). Using the rule (\frac{a^{m}}{a^{n}} = a^{m - n}), we get (f(x)=x^{\frac{1}{3}-\frac{1}{2}}=x^{\frac{2 - 3}{6}}=x^{-\frac{1}{6}}).

Step2: Differentiate using the power rule

The power rule for differentiation is ((x^{k})^\prime=kx^{k - 1}). For (y = x^{-\frac{1}{6}}), then (y^\prime=-\frac{1}{6}x^{-\frac{1}{6}-1}=-\frac{1}{6}x^{-\frac{7}{6}}).

Answer:

(n = 7)