The given expression is
[tex]\begin{gathered} f(x)\text{ = (1 + 12}\sqrt[]{x})(4-x^2) \\ By\text{ applying the laws of exponents, } \\ \sqrt[]{x}^{}=x^{\frac{1}{2}} \\ \text{The expression would be} \\ f(x)=(1+12x^{\frac{1}{2}})(4-x^2) \\ \text{Let p(x) = (1 + 12x}^{\frac{1}{2}})andq(x)=(4-x^2) \\ By\text{ applying the product rule for p(x)q(x), we have} \\ p(x)q^{\prime}(x)\text{ + q(x)p'(x)} \end{gathered}[/tex]
Recall, if we differentiate y = x^n, it becomes
y' = nx^n - 1
By applying this rule, we have
[tex]\begin{gathered} p^{\prime}(x)\text{ = }\frac{1}{2}\times12x^{\frac{1}{2}-1}=6x^{-\frac{1}{2}} \\ q^{\prime}(x)=-2x^{2-1}=-\text{ 2x} \\ \text{Thus, } \\ f^{\prime}(x)\text{ = - 2x(1 + 12}\sqrt[]{x})+6x^{-\frac{1}{2}}(4-x^2) \\ Note,x^{-\frac{1}{2}}\text{ = }\frac{1}{\sqrt[]{x}} \\ \text{Therefore,} \\ f^{\prime}(x)\text{ = }\frac{6(4-x^2)}{\sqrt[]{x}}\text{ - 2x(1 + 12}\sqrt[]{x}) \end{gathered}[/tex]
To find f'(3), we would substitute x = 3 into f'(x). We have
[tex]\begin{gathered} f^{\prime}(3)\text{ = }\frac{6(4-3^2)}{\sqrt[]{3}}\text{ - 2}\times3(1\text{ + 12}\sqrt[]{3)} \\ f^{\prime}(3)\text{ = - 17.32 - 130.71} \\ f^{\prime}(3)\text{ = - 148.03} \end{gathered}[/tex]
