# Using the definition of derivative find the derivative of y=(2x+4)^1/2

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y= (2x+4)^1/2

Using the definition of the derivative, we need to find f'(x).

We know that:

f'(x) = lim ( f(x+h) - f(x) / h h --> 0.

Let us substitute:

f(x) = (2x+4)^1/2.

f(x+h) = (2(x+h) + 4)^1/2 = (2x+2h+4)^1/2.

==> f'(x) = lim ( 2x+2h+4)^1/2 - (2x+4)^1/2 / h h-->0

Let us simplify.

We will multiply and divide by (2x+2h+4)^1/2 + (2x+4)^1/2.

==> f'(x) = lim (2x+2h+4)^1/2-(2x+4)^1/2]*(2x+2h+4)^1/2 + (2x+4)^1/2 / h*(2x+2h+4)^1/2+ ( 2x+4)^1/2

==> f'(x) = lim ( 2x+2h+4) - ( 2x+4) / h(2x+2h+4)^1/2+ (2x+4)^1/2.

==> f'(x) = lim ( 2/ (2x+2h+4)^1/2 + (2x+4)^1/2)

Now we will substitute with h=0.

==> f'(x) = 2/ (2x+4)^1/2 + ( 2x+4)^1/2

= 2/2(2x+4)^1/2

= 1/(2x+4)^1/2

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

Let y = f(x) = (2x+4)^(1/2)

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

dy/dx = Lt {(2(x+h)+4)^1/2 - (2x+4))/h

We multiply both numerator and denominator by {(2x+h+4)^1/2 +(2x+4)^(1/2)}.

dy/dx = Lt {{(2(x+h)+4)^1/2 - (2x+4)} {(2x+h+4)^1/2 + (2x+4)^(1/2)}/ {(2(x+h)+4)^1/2 +(2x+4)^(1/2)}h.

dy/dx = Lt {(2x+2h+4 -2x-4}/{(2x+h+4)^1/2 +(2x+4)^(1/2)}h.

dy/dx = Lt 2h/{(2x+h+4)^1/2 +(2x+4)^(1/2)}h.

dy/dx= 2/{2(2x+4)^(1/2)}.

dy/dx = 1/(2x+4)^(1/2).

First, we'll express the first principle of finding the derivative of a given function:

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)**