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Derivativesfind the derivative of y=(2x+4)^1/2 using the first principle

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bulbul | Student, Undergraduate | eNoter

Posted May 5, 2011 at 2:20 AM via web

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Derivatives

find the derivative of y=(2x+4)^1/2 using the first principle

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giorgiana1976 | College Teacher | Valedictorian

Posted May 5, 2011 at 6:08 AM (Answer #2)

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First principle states:

lim [f(x+h) - f(x)]/h, for h->0

We'll apply the principle to the given polynomial:

lim {sqrt [2(x+h)+4] - sqrt(2x+4)}/h

The next step is to remove the brackets under the square root:

lim [sqrt (2x+2h+4) - sqrt(2x+4)]/h

We'll remove multiply both, numerator and denominator, by the conjugate of numerator:

lim [sqrt (2x+2h+4) - sqrt(2x+4)][sqrt (2x+2h+4)+sqrt(2x+4)]/h*[sqrt (2x+2h+4)+sqrt(2x+4)]

We'll substitute the numerator by the difference of squares:

lim [(2x+2h+4) - (2x+4)]/h*[sqrt (2x+2h+4)+sqrt(2x+4)]

We'll eliminate like terms form numerator:

lim 2h/h*[sqrt (2x+2h+4)+sqrt(2x+4)]

We'll simplify and we'll get:

lim 2/[sqrt (2x+2h+4)+sqrt(2x+4)]

We'll substitute h by 0:

lim 2/[sqrt (2x+2h+4)+sqrt(2x+4)] = 2/[sqrt(2x+4)+sqrt(2x+4)]

We'll combine like terms from denominator:

f'(x)=1/sqrt(2x+4)

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted May 7, 2011 at 6:55 AM (Answer #3)

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The derivative of a function f(x) from first principles is the value:

lim h-->0 [(f(x + h) - f(x))/h]

Here y =  f(x) = (2x+4)^1/2

dy/dx = lim h-->0 [((2(x + h)+4)^1/2 - (2x+4)^1/2)/h]

=> lim h-->0 [((2x + 2h + 4)^1/2 - (2x+4)^1/2)/h]

=> sqrt 2*lim h-->0[((x + h + 2)^1/2 - (x + 2)^1/2)/h]

=> sqrt 2*lim h-->0[((x + h + 2)^1/2 - (x + 2)^1/2)/((x + h + 2) - (x + 2))]

expand the denominator using a^2 - b^2 = (a - b)(a + b)

=> sqrt 2*lim h-->0[((x + h + 2)^1/2 - (x + 2)^1/2)/((x + h + 2)^1/2 - (x + 2)^1/2)((x + h + 2)^1/2 + (x + 2)^1/2)]

=> sqrt 2*lim h-->0[1 /((x + h + 2)^1/2 + (x + 2)^1/2)]

let h = 0

=> sqrt 2*(1 / 2*(x + 2)^1/2)

=> 1/sqrt 2(x + 2)

The value of dy/dx = 1/sqrt 2(x + 2)

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