To determine the fourth derivative of [tex]\( y = \frac{1}{\sqrt{x}} \)[/tex], let's follow a detailed, step-by-step process:
1. Rewrite the initial function:
The function [tex]\( y = \frac{1}{\sqrt{x}} \)[/tex] can be rewritten using exponents:
[tex]\[
y = x^{-\frac{1}{2}}
\][/tex]
2. First derivative [tex]\( y' \)[/tex]:
Apply the power rule for differentiation, which states [tex]\(\frac{d}{dx} x^n = nx^{n-1}\)[/tex]:
[tex]\[
y' = \frac{d}{dx} x^{-\frac{1}{2}} = -\frac{1}{2} x^{-\frac{3}{2}}
\][/tex]
3. Second derivative [tex]\( y'' \)[/tex]:
Differentiate [tex]\( y' \)[/tex]:
[tex]\[
y'' = \frac{d}{dx} \left( -\frac{1}{2} x^{-\frac{3}{2}} \right) = -\frac{1}{2} \left( -\frac{3}{2} \right) x^{-\frac{5}{2}} = \frac{3}{4} x^{-\frac{5}{2}}
\][/tex]
4. Third derivative [tex]\( y''' \)[/tex]:
Differentiate [tex]\( y'' \)[/tex]:
[tex]\[
y''' = \frac{d}{dx} \left( \frac{3}{4} x^{-\frac{5}{2}} \right) = \frac{3}{4} \left( -\frac{5}{2} \right) x^{-\frac{7}{2}} = -\frac{15}{8} x^{-\frac{7}{2}}
\][/tex]
5. Fourth derivative [tex]\( y^{(4)} \)[/tex]:
Differentiate [tex]\( y''' \)[/tex]:
[tex]\[
y^{(4)} = \frac{d}{dx} \left( -\frac{15}{8} x^{-\frac{7}{2}} \right) = -\frac{15}{8} \left( -\frac{7}{2} \right) x^{-\frac{9}{2}} = \frac{105}{16} x^{-\frac{9}{2}}
\][/tex]
Thus, the fourth derivative of [tex]\( y = \frac{1}{\sqrt{x}} \)[/tex] is:
[tex]\[
y^{(4)} = \frac{105}{16} \cdot x^{-\frac{9}{2}} = \frac{105}{16 \cdot x^{\frac{9}{2}}}
\][/tex]
In simplified form:
[tex]\[
y^{(4)} = \frac{105}{16} x^{-\frac{9}{2}}
\][/tex]
