Question

Find f'(7), the derivative of f(x)= Square root x + 2 at X=7, using the LIMIT definition.

Answer 1

Answer:

1/6

Step-by-step explanation:

Derivative Limit Def.

$\displaystyle \large{ f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} }$

We are given the function;

$\displaystyle \large{f(x) = \sqrt{x + 2} }$

Therefore:

$\displaystyle \large{f(x + h) = \sqrt{x + h + 2} }$

Substitute x = 7, f(x) and f(x+h) in.

$\displaystyle \large{ f'(x) = \lim_{h \to 0} \frac{ \sqrt{x + h + 2} - \sqrt{x + 2} }{h} } \\ \displaystyle \large{f'(7) = \lim_{h \to 0} \frac{ \sqrt{7 + h + 2} - \sqrt{7 + 2} }{h} } \\ \displaystyle \large{ \lim_{h \to 0} \frac{ \sqrt{9 + h } - \sqrt{9} }{h} } \\ \displaystyle \large{ \lim_{h \to 0} \frac{ \sqrt{9 + h } - 3 }{h} }$

Multiply both numerator and denominator by √(9+h)+3

$\displaystyle \large{ \lim_{h \to 0} \frac{ (\sqrt{9 + h } - 3)( \sqrt{9 + h} + 3) }{h( \sqrt{9 + h} + 3)} } \\ \displaystyle \large{ \lim_{h \to 0} \frac{ 9 + h - 9}{h( \sqrt{9 + h} + 3)} } \\ \displaystyle \large{ \lim_{h \to 0} \frac{ h}{h( \sqrt{9 + h} + 3)} } \\ \displaystyle \large{ \lim_{h \to 0} \frac{ 1}{ \sqrt{9 + h} + 3} }$

Substitute h= 0 in.

$\displaystyle \large{ \lim_{h \to 0} \frac{ 1}{ \sqrt{9 + 0} + 3} } \\ \displaystyle \large{ \lim_{h \to 0} \frac{ 1}{ \sqrt{9} + 3} } \\ \displaystyle \large{ \lim_{h \to 0} \frac{ 1}{ 3+ 3} } \\ \displaystyle \large{ \lim_{h \to 0} \frac{ 1}{ 6} }\\ \displaystyle \large \boxed{ \frac{1}{6} }$