Using the limit definition, how do you differentiate f(x)=sqrt(x+1)?

2 Answers
May 27, 2016

lim_{Delta->0}((sqrt(x+1+Delta)-sqrt(x+1)))/Delta=1/(2sqrt(x+1))

Explanation:

Let us calculate the limit indirectly.
First we calculate the limit:

l(x)=lim_{Delta->0}(sqrt(x+1+Delta)+sqrt(x+1))times lim_{Delta->0}((sqrt(x+1+Delta)-sqrt(x+1)))/Delta
This limit is equivalent to
lim_{Delta->0}((x+1+Delta)-(x+1))/Delta=1
but
lim_{Delta->0}(sqrt(x+1+Delta)+sqrt(x+1)) = 2sqrt(x+1)
Putting all together
l(x)=2sqrt(x+1) lim_{Delta->0}((sqrt(x+1+Delta)-sqrt(x+1)))/Delta = 1
and finally
lim_{Delta->0}((sqrt(x+1+Delta)-sqrt(x+1)))/Delta=1/(2sqrt(x+1))

May 27, 2016

See explanantion

Explanation:

To solve this you use 2 general principles.

color(blue)("Principle 1:")

Multiply any number by 1 and you do not change its value. However you can change the way it looks. Consider a basic case of 3xx1=3

Multiply 3 by 1 but in the form of 1=sqrt(2)/sqrt(2) -> 3xxsqrt(2)/sqrt(2) =(3sqrt(2))/sqrt(2)

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color(blue)("Principle 2:")

Suppose you had (a+b)

Multiply it by 1 but in the form iof 1=(a-b)/(a-b)

Then (a+b)(a-b)/(a-b) = (a^2-b^2)/(a-b)
color(red)("Your question has square roots and to get rid of this you square it!")
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Given:" "f(x)=sqrt(x+1)

Write as y=sqrt(x+1) ..................................(1)

Increment your values giving

y+delta y =sqrt(x+delta x+1).........................(2)

Equation (2) - Equation (1)

delta y=sqrt(x+delta x + 1) - sqrt(x+1)

Divide throughout by delta x

(delta y)/(delta x) = (sqrt(x+delta x + 1) - sqrt(x+1))/(delta x) .......(3)
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Multiply equation (3) by 1 but where 1=(sqrt(x+delta x + 1) + sqrt(x+1))/(sqrt(x+delta x + 1) + sqrt(x+1))

Now we have our (a-b)(a+b)=a^2-b^2 condition giving:

(delta y)/(delta x) = ((x+delta x + 1) - (x+1))/(deltax(sqrt(x+delta x + 1) + sqrt(x+1)))

(delta y)/(delta x) =( cancel(x)+cancel(delta x) +cancel( 1) -cancel( x)-cancel(1))/(cancel(deltax)(sqrt(x+delta x + 1) + sqrt(x+1)))

lim_(delta x->0) (deltay)/(deltax)= 1/(sqrt(x+1)+sqrt(x+1)) = 1/(2sqrt(x+1))

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Which is of form d/(dx) (x^n) = nx^(n-1)