Question

Find the second derivative of: y = sqrt(x + 4)/4; x ≥ -4.

Answer 1

Answer: $\dfrac{-1}{16(x+4)^{\frac{3}{2}}}$

Explanation:

Given

$y=\dfrac{\sqrt{x+4}}{4}\\\text{differentitate w.r.t x}\\\dfrac{\mathrm{d} y}{\mathrm{d} x}=\dfrac{1}{4}\times \dfrac{1}{2\sqrt{x+4}}=\dfrac{1}{8\sqrt{x+4}}$

$\dfrac{\mathrm{d} y}{\mathrm{d} x}=\dfrac{(x+4)^{-0.5}}{8}$

Again differentiate for the second derivative

$\frac{\mathrm{d^2} y}{\mathrm{d} x^2}=\dfrac{1}{8}\times \dfrac{-1}{2(x+4)^{\frac{3}{2}}}=\dfrac{-1}{16(x+4)^{\frac{3}{2}}}$

Answer 2

Answer: $-\dfrac1{16}(x+4)^{-\frac32}$ ; x ≥ -4.

Explanation:

Given: $y=\dfrac{\sqrt{x+4}}{4}$ ; x ≥ -4.

First derivative = $\dfrac{dy}{dx}=\dfrac12\times\dfrac{(x+4)^{-\frac12}}{4}$        $[ \dfrac{d(x^n)}{dx}=nx^{n-1} ]$

$=\dfrac{1}{8}(x+4)^{-\frac12}$ ; x ≥ -4.

Second derivative=  $\dfrac{d^2y}{dx^2}=\dfrac{d}{dx}(\dfrac18(x+4)^{-\frac12})$

$=\dfrac18\times-\dfrac12(x+4)^{-\frac12-1}\\\\=-\dfrac1{16}(x+4)^{-\frac32}$

The second derivative of y = $-\dfrac1{16}(x+4)^{-\frac32}$ ; x ≥ -4.