Question

Which expression represents the difference quotient of the function f (x) = negative StartRoot 2 x EndRoot?

Answer 1

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brainly.com/question/24878488?

Answer 2

Answer:

• from first principles:

${ \rm{ \frac{ \delta y}{ \delta x} = {}^{lim :x } _{h \dashrightarrow0} \: \frac{f(x + h) - f(x)}{h} }} \\ \\ { \rm{\frac{ \delta y}{ \delta x} = {}^{lim :x } _{h \dashrightarrow0} \: \frac{ - { \{2(x + h)}^{ \frac{1}{2} } \} + {(2x)}^{ \frac{1}{2} } }{h} }}$

• rationalise the numerator:

${ \rm{\frac{ \delta y}{ \delta x} = {}^{lim :x } _{h \dashrightarrow0} \: \frac{2x - 2(x + h)}{h \{ {(2x)}^{ \frac{1}{2} } + { \{2(x + h)}^{ \frac{1}{2} } \} \}} }} \\ \\ { \rm{\frac{ \delta y}{ \delta x} = {}^{lim :x } _{h \dashrightarrow0} \: \frac{ - 2h}{h \{ {(2x)}^{ \frac{1}{2} } + \{2 {(x + h)}^{ \frac{1}{2} } \} \} } }} \\ \\ { \rm{\frac{ \delta y}{ \delta x} = {}^{lim :x } _{h \dashrightarrow0} \: \frac{ - 2}{ {(2x)}^{ \frac{1}{2} } + \{ {2(x + h)}^{ \frac{1}{2} } \} } }}$

• when h tends to non-zero, h ≠ 0

${ \boxed{ \rm{ \frac{dy}{dx} = \frac{ 2}{ - \sqrt{2(x + h)} - \sqrt{2x} } }}}$

Answer: Third objective