Question

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

Answer 1

1/6

Explanation:

Derivative Limit Def.

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

We are given the function;

`\displaystyle \large{f(x) = √(x + 2) }`

Therefore:

`\displaystyle \large{f(x + h) = √(x + h + 2) }`

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

`\displaystyle \large{ f'(x) = \lim_(h \to 0) ( √(x + h + 2) - √(x + 2) )/(h) } \\ \displaystyle \large{f'(7) = \lim_(h \to 0) ( √(7 + h + 2) - √(7 + 2) )/(h) } \\ \displaystyle \large{ \lim_(h \to 0) ( √(9 + h ) - √(9) )/(h) } \\ \displaystyle \large{ \lim_(h \to 0) ( √(9 + h ) - 3 )/(h) } \\`

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

`\displaystyle \large{ \lim_(h \to 0) ( (√(9 + h ) - 3)( √(9 + h) + 3) )/(h( √(9 + h) + 3)) } \\ \displaystyle \large{ \lim_(h \to 0) ( 9 + h - 9)/(h( √(9 + h) + 3)) } \\ \displaystyle \large{ \lim_(h \to 0) ( h)/(h( √(9 + h) + 3)) } \\ \displaystyle \large{ \lim_(h \to 0) ( 1)/( √(9 + h) + 3) } \\`

Substitute h= 0 in.

`\displaystyle \large{ \lim_(h \to 0) ( 1)/( √(9 + 0) + 3) } \\ \displaystyle \large{ \lim_(h \to 0) ( 1)/( √(9) + 3) } \\ \displaystyle \large{ \lim_(h \to 0) ( 1)/( 3+ 3) } \\ \displaystyle \large{ \lim_(h \to 0) ( 1)/( 6) }\\ \displaystyle \large \boxed{ (1)/(6) }`