let ( f(x) = (6x - 5x^3)(2 + sqrt{x}) ). find ( f(x) ).

Explanation:

Step1: Identify the product rule

Given ( f(x) = (6x - 5x^3)(2 + \sqrt{x}) ), we use the product rule: if ( f(x) = u(x)v(x) ), then ( f'(x) = u'(x)v(x) + u(x)v'(x) ). Let ( u(x) = 6x - 5x^3 ) and ( v(x) = 2 + \sqrt{x}=2 + x^{\frac{1}{2}} ).

Step2: Find ( u'(x) )

Differentiate ( u(x) = 6x - 5x^3 ) with respect to ( x ). Using the power rule ( \frac{d}{dx}(x^n)=nx^{n - 1} ), we get: ( u'(x)=\frac{d}{dx}(6x)-\frac{d}{dx}(5x^3)=6 - 15x^2 )

Step3: Find ( v'(x) )

Differentiate ( v(x)=2 + x^{\frac{1}{2}} ) with respect to ( x ). Using the power rule: ( v'(x)=\frac{d}{dx}(2)+\frac{d}{dx}(x^{\frac{1}{2}})=0+\frac{1}{2}x^{-\frac{1}{2}}=\frac{1}{2\sqrt{x}} )

Step4: Apply the product rule

Now, substitute ( u(x), u'(x), v(x), v'(x) ) into the product rule formula: ( f'(x)=u'(x)v(x)+u(x)v'(x)=(6 - 15x^2)(2 + \sqrt{x})+(6x - 5x^3)\left(\frac{1}{2\sqrt{x}}\right) )

Step5: Simplify the second term

Simplify ( (6x - 5x^3)\left(\frac{1}{2\sqrt{x}}\right) ): [ \begin{align*} (6x - 5x^3)\left(\frac{1}{2\sqrt{x}}\right)&=\frac{6x}{2\sqrt{x}}-\frac{5x^3}{2\sqrt{x}}\ &= 3\sqrt{x}-\frac{5}{2}x^{\frac{5}{2}} \end{align*} ]

Step6: Expand the first term

Expand ( (6 - 15x^2)(2 + \sqrt{x}) ): [ \begin{align*} (6 - 15x^2)(2 + \sqrt{x})&=6\times2+6\sqrt{x}-15x^2\times2-15x^2\sqrt{x}\ &=12 + 6\sqrt{x}-30x^2-15x^{\frac{5}{2}} \end{align*} ]

Step7: Combine the two terms

Add the simplified first and second terms together: [ \begin{align*} f'(x)&=(12 + 6\sqrt{x}-30x^2-15x^{\frac{5}{2}})+\left(3\sqrt{x}-\frac{5}{2}x^{\frac{5}{2}}\right)\ &=12+(6\sqrt{x}+3\sqrt{x})-30x^2+\left(-15x^{\frac{5}{2}}-\frac{5}{2}x^{\frac{5}{2}}\right)\ &=12 + 9\sqrt{x}-30x^2-\frac{35}{2}x^{\frac{5}{2}} \end{align*} ]

Answer:

( f'(x)=12 + 9\sqrt{x}-30x^2-\frac{35}{2}x^{\frac{5}{2}} )