Find $\lim _ { h \rightarrow 0 } \frac { f ( x + h ) - f ( x ) } { h }$ for $f ( x ) = \sqrt { x - 6 }$.
A) $\frac { 1 } { 6 \sqrt { x - 6 } }$
B) $\frac { 1 } { 2 \sqrt { x - 6 } }$
C) $\frac { \sqrt { x - 6 } } { 2 }$
D) $\sqrt { x - 6 }$
E) limit does not exist

Question

Quizplus · Accepted Answer

This limit represents the derivative of $f(x) = \sqrt{x - 6}$ with respect to $x$, calculated using the definition of the derivative. The derivative of $f(x) = \sqrt{x - 6}$ is $f'(x) = \frac{1}{2\sqrt{x - 6}}$, using the chain rule for differentiation.

Find lim⁡h→0f(x+h)−f(x)h\lim _ { h \rightarrow 0 } \frac { f ( x + h ) - f ( x ) } { h }limh→0​hf(x+h)−f(x)​

Find $\lim _ { h \rightarrow 0 } \frac { f ( x + h ) - f ( x ) } { h }$