Questions: Differentiate. G(x)=(6 x^2+5)(5 x+sqrtx) G'(x)=□

Transcript text: Differentiate. \[ \begin{array}{l} G(x)=\left(6 x^{2}+5\right)(5 x+\sqrt{x}) \\ G^{\prime}(x)=\square \end{array} \]

Solution

Solution Steps

To differentiate the function \( G(x) = (6x^2 + 5)(5x + \sqrt{x}) \), we will use the product rule. The product rule states that if you have a function \( G(x) = u(x) \cdot v(x) \), then the derivative \( G'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x) \). Here, \( u(x) = 6x^2 + 5 \) and \( v(x) = 5x + \sqrt{x} \). We will find the derivatives \( u'(x) \) and \( v'(x) \) and then apply the product rule.

Step 1: Identify the Functions

We have the function \( G(x) = (6x^2 + 5)(5x + \sqrt{x}) \). We identify the two parts of the product as:

\( u(x) = 6x^2 + 5 \)
\( v(x) = 5x + \sqrt{x} \)

Step 2: Differentiate Each Function

Differentiate \( u(x) \) and \( v(x) \) with respect to \( x \):

\( u'(x) = \frac{d}{dx}(6x^2 + 5) = 12x \)
\( v'(x) = \frac{d}{dx}(5x + \sqrt{x}) = 5 + \frac{1}{2\sqrt{x}} \)

Step 3: Apply the Product Rule

The product rule states that \( G'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x) \). Substituting the derivatives and original functions, we get: \[ G'(x) = 12x(5x + \sqrt{x}) + (6x^2 + 5)\left(5 + \frac{1}{2\sqrt{x}}\right) \]

Step 4: Simplify the Expression

Simplify the expression for \( G'(x) \): \[ G'(x) = 12x(5x + \sqrt{x}) + (6x^2 + 5)\left(5 + \frac{1}{2\sqrt{x}}\right) \] \[ = 60x^2 + 12x\sqrt{x} + 30x^2 + \frac{6x^2}{2\sqrt{x}} + 25 + \frac{5}{2\sqrt{x}} \]