How do you differentiate y=(x^2+4x+3)/sqrtx?

Aug 30, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3 {x}^{2} + 4 x - 3}{2 {x}^{\frac{3}{2}}}$

Explanation:

What I'd prefer is to do is split up the fraction into its three parts by dividing each term by $\sqrt{x} = {x}^{\frac{1}{2}}$:

$y = {x}^{2} / {x}^{\frac{1}{2}} + \frac{4 x}{x} ^ \left(\frac{1}{2}\right) + \frac{3}{x} ^ \left(\frac{1}{2}\right)$

$y = {x}^{\frac{3}{2}} + 4 {x}^{\frac{1}{2}} + 3 {x}^{- \frac{1}{2}}$

Now instead of having to use the quotient rule, we can simply differentiate term-by-term using the power rule:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3}{2} {x}^{\frac{1}{2}} + 4 \cdot \frac{1}{2} {x}^{- \frac{1}{2}} + 3 \cdot - \frac{1}{2} {x}^{- \frac{3}{2}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3 {x}^{\frac{1}{2}}}{2} + \frac{2}{x} ^ \left(\frac{1}{2}\right) - \frac{3}{2 {x}^{\frac{3}{2}}}$

Finding a common denominator of $2 {x}^{\frac{3}{2}}$:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3 {x}^{2} + 4 x - 3}{2 {x}^{\frac{3}{2}}}$