Find f'(2) by using the list definition of the derivative at a point? Will reward Best Answer!?

Question

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

I know you'd do the chain/power rule, I'm just wondering how you'd do it using lim(h-->oo) (f(a+h)-f(a)) / h

Answer 1

lim(h-->oo) (f(a+h)-f(a)) / h

h-->0 (not infinity)





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

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

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

f(x+h)-f(x) = sqrt(3x+3h)+x+h - sqrt(3x) - x

f(x+h)-f(x) = sqrt(3x+3h)-sqrt(3x) + h

f(x+h)-f(x) = sqrt(3x+3h) -(sqrt(3x)-h)



Multiply and divide by sqrt(3x+3h) + (sqrt(3x)-h)

f(x+h)-f(x) = [sqrt(3x+3h) -(sqrt(3x)-h)][sqrt(3x+3h) + (sqrt(3x)-h)]/ [sqrt(3x+3h) + (sqrt(3x)-h)]



The numerator is of the form (a-b)(a+b) = a^2-b^2

a = sqrt(3x+3h)

a^2 = 3x+3h

b = (sqrt(3x)-h)

b^2 = 3x+h^2-2hsqrt(3x)



a^2-b^2 = 3x+3h-3x-h^2+2hsqrt(3x)

a^2-b^2 = 3h + 2h sqrt(3x) -h^2



f(x+h)-f(x) = (3h+2hsqrt(3x)-h^2) / [sqrt(3x+3h) + (sqrt(3x)-h)]

(f(x+h)-f(x))/h = (3h+2hsqrt(3x)-h) / (h[sqrt(3x+3h) + (sqrt(3x)-h)])

(f(x+h)-f(x))/h = (3+2sqrt(3x)) / [sqrt(3x+3h) + (sqrt(3x)-h)]



lim h-->0 (f(x+h)-f(x))/h = lim h-->0 (3+2sqrt(3x)) / [sqrt(3x+3h) + (sqrt(3x)-h)]

= (3 +2sqrt(3x)) /(2sqrt(3x))

= 3/(2sqrt(3x)) + 1



substitute x=2

= 3/(2sqrt(6)) + 1

Answer 2

Derivative of sqrt(3x) +x



Start with 3x^(1/2)

Use the power rule: 3* (1/2)(x)^-(1/2)

3* 1/(2(sqrt(x)) is the first part



Derivative of x is 1, so the derivative of f(x) ia 3/2sqrt(x) + 1

Plug in 2: 3/2sqrt 2 +1

Answer 3

.

Calc 1: Definition

lim x→ a { [ f(x) - f(a) ] / ( x - a )}



lim x→ 2 { [ f(x) - f(2) ] / ( x - 2 )}



f(x) = √(3x) + x

f(2) = √(3*2) + 2 = √3√2 + 2



lim x→ 2 { [ f(x) - f(2) ] / ( x - 2 )}

lim x→ 2 { [ √(3x) + x - √3√2 - 2 ] / ( x - 2 )}



this thing here, the numerator [ √(3x) + x - √2√3 - 2 ], let’s simplify it first

[ √(3x) + x - √2√3 - 2 ]

= √3√x - √3√2 + x - 2

= √3√x - √3√2 + √x√x - √2√2

= (√x - √2) ( √3 + √x + √2)

∴ Numerator = (√x - √2) ( √3 + √x + √2)



denominator ( x - 2 ) can be expressed as a difference of two squares

(x - 2) = (√x - √2)(√x + √2)

∴ Denominator = (√x - √2)(√x + √2)



lim x→ 2 { [ f(x) - f(2) ] / ( x - 2 )} same as

lim x→ 2 { Numerator / Denominator}

lim x→ 2 { [ (√x - √2) ( √3 + √x + √2) ] / [ (√x - √2)(√x + √2) ] }

the √x - √2 in the numerator cancels out with the one in the denominator



lim x→ 2 { ( √3 + √x + √2) / (√x + √2) }

= ( √3 + √2 + √2) / (√2 + √2)

= (√3 + 2√2 ) / ( 2√2 )

= ¼ (4 + √6)

━━━━━━━

Answer 4

f'(x) = (3x)^(-1/2) +1

f'(2) = 1/√6 +1

= (6+√6)/6

Answer 5

　

Note: h→0, not ∞



f(x) = √(3x) + x



f'(a) = lim[h→0] (f(a+h) − f(a)) / h

f'(a) = lim[h→0] (√(3(a+h)) + (a+h) − √(3a) − a) / h

f'(a) = lim[h→0] (√(3a+3h) − √(3a) + h) / h

f'(a) = lim[h→0] (√(3a+3h) − √(3a)) / h + 1

f'(a) = lim[h→0] (√(3a+3h)−√(3a))(√(3a+3h)+√(3a)) / [h (√(3a+3h)+√(3a))] + 1

f'(a) = lim[h→0] ((3a+3h)−(3a)) / [h (√(3a+3h)+√(3a))] + 1

f'(a) = lim[h→0] (3h) / [h (√(3a+3h)+√(3a))] + 1

f'(a) = lim[h→0] 3/(√(3a+3h)−√(3a)) + 1

f'(a) = 3/(√(3a+0)+√(3a)) + 1

f'(a) = 3/(2√(3a)) + 1

f'(a) = √3/(2√a) + 1



f'(2) = √3/(2√2) + 1 = √6/4 + 1



- - - - - - - - - - - - - - - - - - - - - - - - - - - - - -



Here is an alternate method using limits. I usually find this method a little easier to work with:



f'(a) = lim[x→a] (f(x) − f(a)) / (x − a)

f'(a) = lim[x→a] (√(3x) + x − √(3a) − a) / (x − a)

f'(a) = lim[x→a] (√(3x)−√(3a))/(x−a) + (x−a)/(x−a)

f'(a) = lim[x→a] √3(√x−√a)/((√x−√a)(√x+√a)) + 1

f'(a) = lim[x→a] √3/(√x+√a) + 1

f'(a) = √3/(√a+√a) + 1

f'(a) = √3/(2√a) + 1



f'(2) = √3/(2√2) + 1 = √6/4 + 1

Answer 6

hint:

f'(2) =lim(h-->oo) (f(2+h)-f(2)) / h

and

f(2+h)= √(3(2+h))+(2+h)