Respuesta :
Same question here answered by me .Visit the link below
https://brainly.com/question/24878488?
Answer:
• from first principles:
[tex]{ \rm{ \frac{ \delta y}{ \delta x} = {}^{lim :x } _{h \dashrightarrow0} \: \frac{f(x + h) - f(x)}{h} }} \\ \\ { \rm{\frac{ \delta y}{ \delta x} = {}^{lim :x } _{h \dashrightarrow0} \: \frac{ - { \{2(x + h)}^{ \frac{1}{2} } \} + {(2x)}^{ \frac{1}{2} } }{h} }} \\ [/tex]
• rationalise the numerator:
[tex]{ \rm{\frac{ \delta y}{ \delta x} = {}^{lim :x } _{h \dashrightarrow0} \: \frac{2x - 2(x + h)}{h \{ {(2x)}^{ \frac{1}{2} } + { \{2(x + h)}^{ \frac{1}{2} } \} \}} }} \\ \\ { \rm{\frac{ \delta y}{ \delta x} = {}^{lim :x } _{h \dashrightarrow0} \: \frac{ - 2h}{h \{ {(2x)}^{ \frac{1}{2} } + \{2 {(x + h)}^{ \frac{1}{2} } \} \} } }} \\ \\ { \rm{\frac{ \delta y}{ \delta x} = {}^{lim :x } _{h \dashrightarrow0} \: \frac{ - 2}{ {(2x)}^{ \frac{1}{2} } + \{ {2(x + h)}^{ \frac{1}{2} } \} } }} \\ [/tex]
• when h tends to non-zero, h ≠ 0
[tex]{ \boxed{ \rm{ \frac{dy}{dx} = \frac{ 2}{ - \sqrt{2(x + h)} - \sqrt{2x} } }}} \\ [/tex]
Answer: Third objective