Question

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

Answer 1

Answer:
`-\frac1{16}(x+4)^(-\frac32)` ; x ≥ -4.

Step-by-step explanation:

Given:
`y=(√(x+4))/(4)` ; x ≥ -4.

First derivative =
`(dy)/(dx)=\frac12*((x+4)^(-\frac12))/(4)`
`[ (d(x^n))/(dx)=nx^(n-1) ]`

`=(1)/(8)(x+4)^(-\frac12)` ; x ≥ -4.

Second derivative=
`(d^2y)/(dx^2)=(d)/(dx)(\frac18(x+4)^(-\frac12))`

`=\frac18*-\frac12(x+4)^(-\frac12-1)\\\\=-\frac1{16}(x+4)^(-\frac32)`

The second derivative of y =
`-\frac1{16}(x+4)^(-\frac32)` ; x ≥ -4.

Answer 2

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

Step-by-step explanation:

Given

`y=(√(x+4))/(4)\\\text{differentitate w.r.t x}\\\frac{\mathrm{d} y}{\mathrm{d} x}=(1)/(4)* (1)/(2√(x+4))=(1)/(8√(x+4))`

`\frac{\mathrm{d} y}{\mathrm{d} x}=((x+4)^(-0.5))/(8)`

Again differentiate for the second derivative

`\frac{\mathrm{d^2} y}{\mathrm{d} x^2}=(1)/(8)* \frac{-1}{2(x+4)^{(3)/(2)}}=\frac{-1}{16(x+4)^{(3)/(2)}}`