# What is the derivative of y=square root (3x+7)?(use definition of derivative)

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### 2 Answers

To find the derivative of y = (3x+7)^(1/2).

Let y = f(x) = (3+70^(1/2).

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

{f(x+h)-f(x) }/h = {(3(x+h)+7)^(1/2) - (3x+7)^(1/2)}/h.

= {(3(x+h)+7)^(1/2) - (3x+7)^(1/2)}{(3(x+h)+7)^(1/2) + (3x+7)^(1/2)}/h{(3(x+h)+7)^(1/2) - (3x+7)^(1/2)}.

= {3(x+h)+7 - (3x+7)}/h{(3(x+h)+7)^(1/2)+(3x+7)^(1/2)}.

= 3h/h(3x+h+7)(1/2)+(3x+7)(1/2)}

Therefore , Lt x-> 0 { f(x+h)-f(x)}/h = Lt h->0 {3/(3x+h+7)(1/2)+(3x+7)(1/2)}.

dy/dx = (3/2)(3x+7)^(-1/2).

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 [3(x+h)+7] - sqrt(3x+7)}/h

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

lim [sqrt (3x+3h+7) - sqrt(3x+7)]/h

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

lim [sqrt (3x+3h+7) - sqrt(3x+7)][sqrt (3x+3h+7)+sqrt(3x+7)]/h*[sqrt (3x+3h+7)+sqrt(3x+7)]

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

lim [(3x+3h+7) - (3x+7)]/h*[sqrt (3x+3h+7)+sqrt(3x+7)]

We'll eliminate like terms form numerator:

lim 3h/h*[sqrt (3x+3h+7)+sqrt(3x+7)]

We'll simplify and we'll get:

lim 3/[sqrt (3x+3h+7)+sqrt(3x+7)]

We'll substitute h by 0:

lim 3/[sqrt (3x+3h+7)+sqrt(3x+7)] = 3/[sqrt(3x+7)+sqrt(3x+7)]

We'll combine like terms from denominator:

**f'(x)=3/2sqrt(3x+7)**