Mathematics: Urgent! will give brainliest to best derive this question using the product rule and factor it! y= √x(x^2+4)

Urgent! will give brainliest to best derive this

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imsmart916

04.08.2019

Answer is $y' = \dfrac{\left(5x^2 + 4\right)}{2\sqrt{x}}$ ======product rule: $\dfrac{d}{dx} \left[f(x) \cdot g(x)\right] = f(x)g'(x) + g(x)f'(x)$ get the function in front and leave it alone.multiply by derivative of the second function.add on to that the second function left alone multiplied by the derivative of the first function.so applying this $\begin{aligned} y & = \sqrt{x}\left(x^2+4\right) \\ y' & = \sqrt{x}\left(x^2+4\right)' + \left(x^2+4\right) \left(\sqrt{x}\right)' \end{aligned}$ i will deal with $\left(x^2+4\right)'$ and $\left(\sqrt{x}\right)'$ belowdifferentiating $\left(x^2+4\right)'$ , we get $\begin{aligned} \left(x^2+4\right)' & = \left(x^2\right)' + (4)' \\ & = 2x^{2-1} + 0 \\ & = 2x \end{aligned}$ for $\left(\sqrt{x}\right)'$ , we can rewrite as $\left(x^{1/2}\right)'$ and use power rule. $\begin{aligned} \left(x^{1/2}\right)' & = \frac{1}{2}x^{\frac{1}{2}-1} \\ & = \frac{1}{2}x^{-\frac{1}{2}}\end{aligned}$ bringing these back together, we get $\begin{aligned} y' & = \sqrt{x}\left(x^2+4\right)' + \left(x^2+4\right) \left(\sqrt{x}\right)' \\ & = \sqrt{x} \left(2x\right) + \left(x^2+4\right)\tfrac{1}{2}x^{-\frac{1}{2}} \\ & = x^{\frac{1}{2}} \left(2x\right) + \tfrac{1}{2}\left(x^2+4\right)x^{-\frac{1}{2}} \\ & = 2x^{\frac{1}{2}+1}+ \tfrac{1}{2}x^{-\frac{1}{2}} \left(x^2+4\right)\\ & = 2x^{\frac{3}{2}} + \tfrac{1}{2}x^{-\frac{1}{2}} \left(x^2+4\right) \end{aligned}$ distribute $\tfrac{1}{2}x^{-\frac{1}{2}}$ into the parentheses $\begin{aligned} y' & = 2x^{\frac{3}{2}} + \tfrac{1}{2}x^{-\frac{1}{2}} \left(x^2+4\right) \\ & = 2x^{\frac{3}{2}} + \left(\tfrac{1}{2}x^{-\frac{1}{2}}\right)\left(x^2\right) + \left(\tfrac{1}{2}x^{-\frac{1}{2}}\right)(4) \\ & = 2x^{\frac{3}{2}} + \tfrac{1}{2}x^{-\frac{1}{2}+2} + \tfrac{4}{2}x^{-\frac{1}{2}} \\ & = 2x^{\frac{3}{2}} + \tfrac{1}{2}x^{\frac{3}{2}} + 2x^{-\frac{1}{2}} \end{aligned}$ factor out the smallest power of $x$ , which is $x^{-\frac{1}{2}}$ when you take $x^{-\frac{1}{2}}$ out, we need to divide every single term by $x^{-\frac{1}{2}}$ . hence why we add 1/2 to all the exponents (exponent properties) $\begin{aligned} y' & = 2x^{\frac{3}{2}} + \tfrac{1}{2}x^{\frac{3}{2}} + 2x^{-\frac{1}{2}} \\ & = x^{-\frac{1}{2}} \left( 2x^{\frac{3}{2} + \frac{1}{2}} + \tfrac{1}{2}x^{\frac{3}{2}+\frac{1}{2}} + 2x^{-\frac{1}{2}+\frac{1}{2}}\right) \\ & = x^{-\frac{1}{2}} \left(2x^{\frac{4}{2} } + \tfrac{1}{2}x^{\frac{4}{2}} + 2x^0\right) \\ & = x^{-\frac{1}{2}} \left(2x^2+ \tfrac{1}{2}x^2 + 2\right) \end{aligned}$ factor out $\tfrac{1}{2}$ . divide everything inside by 1/2 as we take it out.then convert back to radical forms and clean it up. $\begin{aligned} y' & = x^{-\frac{1}{2}} \left(2x^2+ \tfrac{1}{2}x^2 + 2\right) \\ & = \tfrac{1}{2}x^{-\frac{1}{2}} \left(4x^2+ x^2 + 4\right) \\ & = \tfrac{1}{2} \cdot \tfrac{1}{\sqrt{x}}\left(5x^2 + 4\right) \\ & = \frac{1}{2\sqrt{x}} \left(5x^2 + 4\right) \\ & = \frac{\left(5x^2 + 4\right)}{2\sqrt{x}} \end{aligned}$

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Urgent! will give brainliest to best derive this question using the product rule and factor it! y= √x(x^2+4)

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