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Find the derivative of y = sqrt x, from the first principles.

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alicesmith | Student, Grade 9 | Salutatorian

Posted December 6, 2010 at 4:49 PM via web

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Find the derivative of y = sqrt x, from the first principles.

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted December 6, 2010 at 4:50 PM (Answer #1)

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To find the derivative of a function f(x) from the first principles we have to find the value of lim h--> 0 [(f(x +h) – f(x))/h]

We have y = sqrt (x)

So y’ = lim h--> 0 [(sqrt (x +h) – sqrt (x))/h]

=> lim h--> 0 [((sqrt (x +h) – sqrt x)/h)* (sqrt (x + h) + sqrt x) / (sqrt (x + h) + sqrt x)]

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

=> lim h--> 0 [(x +h – x)/ (h* (sqrt (x + h) + sqrt x))]

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

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

=> 1/ (sqrt (x + 0) + sqrt x)

=> 1/ 2*sqrt x

Therefore the required result is 1/2*sqrt x.

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neela | High School Teacher | Valedictorian

Posted December 6, 2010 at 6:29 PM (Answer #2)

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y = sqrtx

To find the derivative by first pinciple.

If y = f(x), then the derivative dy/dx by first principle is given by:

dy/dx = Lt{f(x+h)-f(x)}/h, as x-->0.

Therefore here, y = x^(1/2).

dy/dx = Lt {(x+h)^(1/2 - x^(1/2)}/h, as x-->0.

dy/dx = Lt {(x+h)^(1/2)-x^(1/2)}{(x+h)^(1/2)+x^(1/2)}/h{(x+h)^(1/2)+x^(1/2)}. Where both numerator and denominators are multiplied by {(x+h)^(1/2)+x^(1/2)} to rationalise the numerator.

dy/dx = Lt {(x+h)-x}/h{(x+h)^(1/2)+x^(1/2)},= h/ h{(x+h)^(1/2)+x^(1/2)} in numerator and denominator get cancelled amd now we set h = 0.

dy/dx = 1/{x+x}^1/2.

dy/dx = 1/2x^(1/2).

Therefore the derivative of x^(1/2) with respect to x by first principle is 1/(2x^(1/2))

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giorgiana1976 | College Teacher | Valedictorian

Posted December 6, 2010 at 4:59 PM (Answer #3)

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The definition of the derivative is to determine the limit of the function for a given point. We'll have in this case x -> 0.

We'll put f(x) = y = sqrt x

f'(x) = lim [f(x) - f(0)]/(x-0), for x->0

f'(x) = lim (sqrtx - sqrt0)/x

We'll substitute x by 0:

lim (sqrtx - sqrt0)/x = (sqrt0 - sqrt0)/0 = 0/0

 Since we have an indetermination case, 0/0, we'll apply L'Hospital rule:

lim f/g = lim f'/g'

lim sqrtx/x = lim (sqrt x)'/x'

lim (sqrt x)'/x' = lim (1/2sqrt x)/1

We'll substitute x by 0 and we'll get:

lim (1/2sqrt x)/1 = (1/2sqrt 0)

f'(x) = 1/2 sqrt x

f'(0) = 1/2

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