Question

determine the derivative of the following

$(3 {x}^{2} - \sqrt{x} ) {}^{2}$
​

Answer 1

Answer:

$\huge{ \boxed{36 {x}^{3} - \frac{6 {x}^{2} }{ \sqrt{x} } - 12x \sqrt{x} + 2 \sqrt{x} }}$

Step-by-step explanation:

To find the derivative of the function $(3 {x}^{2} - \sqrt{x} )^{2}$ , the chain rule can be applied.

• First let's define the function as $f(x) = (3 {x}^{2} - \sqrt{x} )^{2}$
• Next, the function can be rewritten as f(x) = u(x)² , where $u(x) = (3 {x}^{2} - \sqrt{x})$

According to the chain rule, the derivative of f(x) with respect to x is given by:

$\dfrac{df}{dx} = \dfrac{df}{du} \times \: \dfrac{du}{dx}$

where

$\dfrac{df}{du}$ represents the derivative f(x) with respect to u.

• Since f(x) = u(x)² , it can be differentiated as:

$\dfrac{df}{du} = 2u(x)$

• Next we differentiate $u(x) = (3 {x}^{2} - \sqrt{x})$ which is given as:

$\frac{du}{dx} = \frac{d}{dx} ( {3x}^{2} ) - \frac{d}{dx} ( \sqrt{x} ) \\ \\ \frac{du}{dx} = {6x} - \frac{1}{2 \sqrt{x} }$

• Next we substitute the calculated values back into the chain rule formula given above, we have:

$\frac{df}{dx} = 2u(x) \times (6x - \frac{1}{2 \sqrt{x} } )$

• Lastly we substitute $u(x) = (3 {x}^{2} - \sqrt{x})$ back into the equation and simplify:

$\frac{df}{dx} = 2(3 {x}^{2} - \sqrt{x} ) \times (6x - \frac{1}{2 \sqrt{x} } ) \\ = (6 {x}^{2} - 2 \sqrt{x} )( 6x - \frac{1}{2 \sqrt{x} } ) \\ = 36 {x}^{3} - \frac{6 {x}^{2} }{ \sqrt{x} } - 12x \sqrt{x} + 2 \sqrt{x}$

We have the final answer as

$36 {x}^{3} - \frac{6 {x}^{2} }{ \sqrt{x} } - 12x \sqrt{x} + 2 \sqrt{x}$