First Principles Example 3: square root of x
Key Questions
-
Answer:
#f'(x)=1/(2sqrt(x+3))# Explanation:
#f'(x)=lim_(h->0)(f(x+h)-f(x))/h# #f(x)=sqrt(x+3), f(x+h)=sqrt(x+h+3)# , then#f'(x)=lim_(h->0)(sqrt(x+h+3)-sqrt(x+3))/h# If we evaluate this right away, we get
#lim_(h->0)(sqrt(x+h+3)-sqrt(x+3))/h=(sqrt(x+3)-sqrt(x+3))/0=0/0# ,so we need to simplify as this is an indeterminate form.
Multiply the entire limit by the numerator's conjugate, which is
#(sqrt(x+h+3)+sqrt(x+3))/(sqrt(x+h+3)+sqrt(x+3))# . This is the same as multiplying by#1.# #f'(x)=lim_(h->0)(sqrt(x+h+3)-sqrt(x+3))/h*(sqrt(x+h+3)+sqrt(x+3))/(sqrt(x+h+3)+sqrt(x+3))# The numerator becomes
#sqrt(x+h+3)-sqrt(x+3) * [sqrt(x+h+3)+sqrt(x+3)]=x+h+3-(x+3)=x+h+3-x-3=h# #f'(x)=lim_(h->0)(cancelx+h+cancel3-cancelx-cancel3)/(h(sqrt(x+h+3)+sqrt(x+3))# #f'(x)=lim_(h->0)(cancelh)/(cancelh(sqrt(x+h+3)+sqrt(x+3))# #f'(x)=lim_(h->0)1/(sqrt(x+h+3)+sqrt(x+3))# #f'(x)=1/(sqrt(x+3)+sqrt(x+3))# #f'(x)=1/(2sqrt(x+3))# -
Definition
#f'(x)=lim_{h to 0}{f(x+h)-f(x)}/h#
By Definition,
#f'(x)=lim_{h to 0}{sqrt{x+h}-sqrt{x}}/h# by multiplying the numerator and the denominator by
#sqrt{x+h}+sqrt{x}# ,#=lim_{h to 0}{sqrt{x+h}-sqrt{x}}/hcdot{sqrt{x+h}+sqrt{x}}/{sqrt{x+h}+sqrt{x}}# #=lim_{h to 0}{x+h-x}/{h(sqrt{x+h}+sqrt{x})}# by cancelling out
#x# 's and#h# 's,#=lim_{h to 0}1/{sqrt{x+h}+sqrt{x}}=1/{sqrt{x+0}+sqrt{x}}=1/{2sqrt{x}}# Hence,
#f'(x)=1/{2sqrt{x}}# .
I hope that this was helpful.
Questions
Derivatives
-
Tangent Line to a Curve
-
Normal Line to a Tangent
-
Slope of a Curve at a Point
-
Average Velocity
-
Instantaneous Velocity
-
Limit Definition of Derivative
-
First Principles Example 1: x²
-
First Principles Example 2: x³
-
First Principles Example 3: square root of x
-
Standard Notation and Terminology
-
Differentiable vs. Non-differentiable Functions
-
Rate of Change of a Function
-
Average Rate of Change Over an Interval
-
Instantaneous Rate of Change at a Point