Use the definition of derivative to find the derivative of the function

Question

Use the definition of derivative to find the derivative of the function

 

f(x)=sqrt(3+4x)

 

f'(x)=

 

Domain of f(x)=

 

Domain of f'(x)=

 

Answer 1

"Use the definition of derivative to find the derivative of the function"

 

Definition of derivative is:

f'(x) = lim_h->0 [ ( f(x+h) - f(x) ) / h ]

 

if f(x)=√(3+4x)

then f(x+h)=√(3+4(x+h) ) = √(3+4x+4h) )

 

Substitute these into the definition:

f'(x) = lim_h->0 [ ( √(3+4x+4h) - √(3+4x) ) / h ]

 

as it stands, substituting 0 in for h gives  an undefined derivative (BAD) 

so, the "trick" is to multiple by 1 (to keep the value) in the form of conjugate of numerator÷conjugate of numerator

 

f'(x) = lim_h->0 [ ( √(3+4x+4h) - √(3+4x) ) / h × (√(3+4x+4h) + √(3+4x) / (√(3+4x+4h) + √(3+4x) ) ]

f'(x) = lim_h->0 [ ( √(3+4x+4h) - √(3+4x) ) × (√(3+4x+4h) + √(3+4x) ) / ( h(√(3+4x+4h) + √(3+4x) ) ]

f'(x) = lim_h->0 [ ( 3+4x+4h - (3+4x)  ) / ( h(√(3+4x+4h) + √(3+4x) ) ]

f'(x) = lim_h->0 [ ( 3+4x+4h - 3 - 4x) ) / ( h(√(3+4x+4h) + √(3+4x) ) ]

f'(x) = lim_h->0 [ ( 4h ) / ( h(√(3+4x+4h) + √(3+4x) ) ]   now the h's cancel - which was the goal of the "trick"

f'(x) = lim_h->0 [ ( 4 ) /  (√(3+4x+4h) + √(3+4x) ) ]     substitute h=0 to find derivative

f'(x) =  ( 4 ) / (√(3+4x+4(0) ) + √(3+4x) )  **edits start here**

f'(x) = ( 4 ) / (2√(3+4x) )

f'(x) = 2/√(3+4x)   rationalize the denominator to be in most simple form

answer:

f'(x) = (2√(3+4x) )/(3+4x)

 

remember the radicand cannot be negative and a denominator cannot be zero:

3+4x≥0

x≥-3/4 for f(x)

3+4x cannot be 0

x cannot be -3/4 for f'(x)

Domain of f(x): x ∈ {ℜ l x≥-3/4 }

Domain of f'(x): x ∈ {ℜ} -{-3/4 }