How do you find the first and second derivative of y = 2ln(x)?

Oct 24, 2016

$y ' = \frac{2}{x}$
$y ' ' = - \frac{2}{x} ^ 2$

Explanation:

We will use the product rule and the well-known derivative $\left(\ln x\right) ' = \frac{1}{x}$.

The product rule states that for a function $f \left(x\right) = g \left(x\right) \times h \left(x\right)$, the derivative, $f ' \left(x\right)$, is given by $f ' \left(x\right) = g ' \left(x\right) \times h \left(x\right) + h ' \left(x\right) \times g \left(x\right)$.

$y ' = 0 \times \ln x + 2 \times \frac{1}{x}$

$y ' = \frac{2}{x}$

We will find the second derivative using the quotient rule. The quotient rule states that for a function $f \left(x\right) = g \frac{x}{h} \left(x\right)$, the derivative is given by $f ' \left(x\right) = \frac{g ' \left(x\right) \times h \left(x\right) - g \left(x\right) \times h ' \left(x\right)}{h \left(x\right)} ^ 2$.

$y ' ' = \frac{0 \times x - 2 \times 1}{x} ^ 2$

$y ' ' = - \frac{2}{x} ^ 2$

Hopefully this helps!