Question

What are the steps that I need to find the solution

Answer 1

The given expression is

`\begin{gathered} f(x)\text{ = (1 + 12}\sqrt[]{x})(4-x^2) \\ By\text{ applying the laws of exponents, } \\ \sqrt[]{x}^{}=x^{(1)/(2)} \\ \text{The expression would be} \\ f(x)=(1+12x^{(1)/(2)})(4-x^2) \\ \text{Let p(x) = (1 + 12x}^{(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}`

Recall, if we differentiate y = x^n, it becomes

y' = nx^n - 1

By applying this rule, we have

`\begin{gathered} p^(\prime)(x)\text{ = }(1)/(2)*12x^{(1)/(2)-1}=6x^{-(1)/(2)} \\ q^(\prime)(x)=-2x^(2-1)=-\text{ 2x} \\ \text{Thus, } \\ f^(\prime)(x)\text{ = - 2x(1 + 12}\sqrt[]{x})+6x^{-(1)/(2)}(4-x^2) \\ Note,x^{-(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}`

To find f'(3), we would substitute x = 3 into f'(x). We have

`\begin{gathered} f^(\prime)(3)\text{ = }\frac{6(4-3^2)}{\sqrt[]{3}}\text{ - 2}*3(1\text{ + 12}\sqrt[]{3)} \\ f^(\prime)(3)\text{ = - 17.32 - 130.71} \\ f^(\prime)(3)\text{ = - 148.03} \end{gathered}`