Answer: [tex]\dfrac{-1}{16(x+4)^{\frac{3}{2}}}[/tex]
Explanation:
Given
[tex]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}}[/tex]
[tex]\dfrac{\mathrm{d} y}{\mathrm{d} x}=\dfrac{(x+4)^{-0.5}}{8}[/tex]
Again differentiate for the second derivative
[tex]\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}}}[/tex]
Answer: [tex]-\dfrac1{16}(x+4)^{-\frac32}[/tex] ; x ≥ -4.
Explanation:
Given: [tex]y=\dfrac{\sqrt{x+4}}{4}[/tex] ; x ≥ -4.
First derivative = [tex]\dfrac{dy}{dx}=\dfrac12\times\dfrac{(x+4)^{-\frac12}}{4}[/tex] [tex][ \dfrac{d(x^n)}{dx}=nx^{n-1} ][/tex]
[tex]=\dfrac{1}{8}(x+4)^{-\frac12}[/tex] ; x ≥ -4.
Second derivative= [tex]\dfrac{d^2y}{dx^2}=\dfrac{d}{dx}(\dfrac18(x+4)^{-\frac12})[/tex]
[tex]=\dfrac18\times-\dfrac12(x+4)^{-\frac12-1}\\\\=-\dfrac1{16}(x+4)^{-\frac32}[/tex]
The second derivative of y = [tex]-\dfrac1{16}(x+4)^{-\frac32}[/tex] ; x ≥ -4.
