Limit definition of derivative problem

physicsernaw
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Homework Statement



Using the definition of derivative find f'(x) for f(x) = x - sqrt(x)

Homework Equations



None.

The Attempt at a Solution



lim h --> 0 : ((x + h) - sqrt(x + h) - x + sqrt(x))/h

1 - (sqrt(x + h) - sqrt(x))/h

Multiply by conjugate..

1 - h/(h*(sqrt(x) + sqrt(x+h)))

1 - 1/(sqrt(x+h) + sqrt(x))

lim as h --> 0 makes it: 1 - 1/2sqrt(x)

-------------------------------------------------------
QUESTION:

My question is, is there a way to solve this problem without multiplying by the conjugate? My friend says there's more ways but I don't see how?

Also, how come using the limit definition of derivative with
(f(x) - f(a)) / (x - a) yields zero?
 
on Phys.org
physicsernaw said:

Homework Statement



Using the definition of derivative find f'(x) for f(x) = x - sqrt(x)

Homework Equations



None.

The Attempt at a Solution



lim h --> 0 : ((x + h) - sqrt(x + h) - x + sqrt(x))/h

1 - (sqrt(x + h) - sqrt(x))/h

Multiply by conjugate..

1 - h/(h*(sqrt(x) + sqrt(x+h)))

1 - 1/(sqrt(x+h) + sqrt(x))

lim as h --> 0 makes it: 1 - 1/2sqrt(x)

-------------------------------------------------------
QUESTION:

My question is, is there a way to solve this problem without multiplying by the conjugate? My friend says there's more ways but I don't see how?

Also, how come using the limit definition of derivative with
(f(x) - f(a)) / (x - a) yields zero?

There are general rules for obtaining the derivative of powers like f(x) = x^k (k = any number, positive or negative).

RE: your second question: the ratio you write does NOT give 0. After all, you just finished finding the result for f(x) = sqrt(x) and a = 0: you did not get zero then, did you?
 
Ray Vickson said:
There are general rules for obtaining the derivative of powers like f(x) = x^k (k = any number, positive or negative).

RE: your second question: the ratio you write does NOT give 0. After all, you just finished finding the result for f(x) = sqrt(x) and a = 0: you did not get zero then, did you?

I got it, was making a silly error :biggrin:
 

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