Find the first and second derivatives of the function y=4x/ square root (X+1)
Thanks!
Hello,
$\displaystyle f(x)=\frac{4x}{\sqrt{x+1}}=4x \cdot (x+1)^{-1/2}$ (remember that sqrt(x)=x^(1/2))
Note that the constant is not important, because $\displaystyle [af(x)]'=af'(x)$ for any constant.
Apply product rule :
$\displaystyle f'(x)=4 (x+1)^{-1/2}+4x \cdot \tfrac{-1}{2} \cdot (x+1)^{-1/2-1}$$\displaystyle =4(x+1)^{-1/2}-2x (x+1)^{-3/2}=2(x+1)^{-3/2} \big[2(x+1)-x\big]$
$\displaystyle f'(x)=2(x+1)^{-3/2} \big[x+2\big]$
Apply product rule again to get $\displaystyle f''(x)$
You will have to use the quotient rule for this question.
The quotient rule is $\displaystyle \frac{d}{dx} \frac{f(x)}{g(x)} = \frac{g(x)f'(x) - f(x)g'(x)}{(g(x))^2}$.
We have: $\displaystyle y = \frac{4x}{\sqrt{x+1}}$
Therefore, $\displaystyle y' = \frac{\sqrt{x+1} \cdot 4 - 4x \cdot (1/2)(x+1)^{-1/2}}{x + 1} = \frac{4(x + 1) - 2x}{(x + 1)^{3/2}} = \frac{2x + 4}{(x + 1)^{3/2}}$.
This is the first derivative.
To obtain the second derivative, you must use the quotient rule to determine the derivative of the first derivative.