Questions: y=(x^2+x+4)(x^(1/3)-2x^(1/2)+9) y'=

Solution Steps

To find the derivative of the given function, we will use the product rule. The product rule states that if you have a function \( y = u(x) \cdot v(x) \), then the derivative \( y' \) is given by \( y' = u'(x) \cdot v(x) + u(x) \cdot v'(x) \). Here, \( u(x) = x^2 + x + 4 \) and \( v(x) = \sqrt[3]{x} - 2\sqrt{x} + 9 \). We will find the derivatives \( u'(x) \) and \( v'(x) \) separately and then apply the product rule.

We are given the function \( y = (x^2 + x + 4)(x^{1/3} - 2\sqrt{x} + 9) \). To find the derivative \( y' \), we will use the product rule. The product rule states that if \( y = u(x) \cdot v(x) \), then \( y' = u'(x) \cdot v(x) + u(x) \cdot v'(x) \).

First, we differentiate \( u(x) = x^2 + x + 4 \): \[ u'(x) = 2x + 1 \]

Next, we differentiate \( v(x) = x^{1/3} - 2\sqrt{x} + 9 \): \[ v'(x) = \frac{1}{3}x^{-2/3} - \frac{1}{\sqrt{x}} \]

Substitute the derivatives into the product rule formula: \[ y' = (2x + 1)(x^{1/3} - 2\sqrt{x} + 9) + (x^2 + x + 4)\left(\frac{1}{3}x^{-2/3} - \frac{1}{\sqrt{x}}\right) \]

Final Answer