How do you differentiate (x^2+4)sqrt(x-3)?

Feb 5, 2015

The answer is: $y ' = \frac{\left(5 x - 2\right) \left(x - 2\right)}{2 \sqrt{x - 3}}$.

The rule of the derivative of a product is:

$y = f \left(x\right) \cdot g \left(x\right) \Rightarrow y ' = f ' \left(x\right) \cdot g \left(x\right) + f \left(x\right) \cdot g ' \left(x\right)$;

so:

$y ' = 2 x \sqrt{x - 3} + \left({x}^{2} + 4\right) \cdot \frac{1}{2 \sqrt{x - 3}} =$

$= \frac{2 x \sqrt{x - 3} \cdot 2 \sqrt{x - 3} + {x}^{2} + 4}{2 \sqrt{x - 3}} = \frac{4 x \left(x - 3\right) + {x}^{2} + 4}{2 \sqrt{x - 3}} =$

$= \frac{5 {x}^{2} - 12 x + 4}{2 \sqrt{x - 3}}$

We can factor $5 {x}^{2} - 12 x + 4$ remembering that if we consider that polynomial like a square equation, we can find the two solutions ${x}_{1} \mathmr{and} {x}_{2}$, than:

$a {x}^{2} + b x + c = a \left(x - {x}_{1}\right) \left(x - {x}_{2}\right)$;

so, we can use the reduced formula because $b$ is even:

$5 {x}^{2} - 12 x + 4 = 0 \Rightarrow$ $\frac{\Delta}{4} = {\left(\frac{b}{2}\right)}^{2} - a c = 36 - 20 = 16$ and

${x}_{1 , 2} = \frac{- \frac{b}{2} \pm \sqrt{\frac{\Delta}{4}}}{a} = \frac{6 \pm 4}{5} \Rightarrow$

${x}_{1} = \frac{2}{5} \mathmr{and} {x}_{2} = 2$.

So:

$5 {x}^{2} - 12 x + 4 = 5 \left(x - \frac{2}{5}\right) \left(x - 2\right) = \left(5 x - 2\right) \left(x - 2\right)$.

Finally:

$y ' = \frac{\left(5 x - 2\right) \left(x - 2\right)}{2 \sqrt{x - 3}}$.