# If `f(x)= (14x)^(1/2)` find f '(4).

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`f(x)=(14x)^(1/2)`

To take the derivative of f(x), apply the power rule which is `(u^n)'=n*u^(n-1)*u' ` .

`f'(x)=(14x)^(1/2) = 1/2*(14x)^(-1/2)*(14x)' = 1/2(14x)^(-1/2)*14

= 7(14x)^(-1/2)`

Apply the negative exponent rule which is `a^-m=1/a^m` . And then, express the fractional exponent as radical.

`f'(x)=7/(14x)^(1/2) = 7/sqrt(14x)` **Hence, the derivative of the given function is `f'(x)= 7/sqrt(14x)` .**

To solve for f'(4), plug-in x=4 to the derivative.

`f'(4)=7/sqrt(4*14)=7/(2sqrt14)`

Rationalize the denominator to simplify.

`f'(4)=7/(2sqrt14) * sqrt14/sqrt14= (7sqrt14)/(2*14)=sqrt14/4` **Hence, `f'(4)=sqrt14/4` .**

Hi, elpony,

There are several ways we can do this problem. Let's do it like this.

First, you need to be aware of how to "take a derivative". For instance, if I have you:

x^7

You would need to tell me the first derivative is:

7x^6

Then, the second derivative is:

42x^5

The third derivative:

210x^4

The fourth derivative:

840x^3

And, so forth.

In each case, there are only two pieces to the pattern we are doing:

1) we multiply the exponent and the coefficient (number in front of the variable) to get the new coefficient. Like

6*7 = 42

42*5 = 210

210*4 = 840

2) Subtract 1 from the exponent each time.

For your problem, we can use an exponent property from Algebra to get:

(14x)^1/2 = **14^(1/2) * x^(1/2)**

Since **14^1/2** is a constant number, **√14**, this will be just like the 7, 42, and 210 in the example. So, it follows along, just like the 7, 42, and 210 did.

Now, we will take the second part only at first. We take the derivative of:

**x^(1/2)**

by default, "1" is in front of the x:

**1x^(1/2)**

Multiplying 1 and 1/2, we get 1/2, goes in front of the x. Subtract 1 from the exponent, we have

**1/2 x^(-1/2)**

We have to include the √14

So:

** √14 * 1/2 x^(-1/2)**

Which is:

** √14 * 1 x^(-1/2)**

** 2**

Which is:

** √14 * x^(-1/2)**

** 2 **

Some teachers will want this done in different ways. Like, the negative exponent would go to the bottom and changes to positive, as in:

√14 * ** x^(-1/2) = ** √14 *** 1 **

2 2 **x^(1/2)**

I hesitate to go into more detail on that part. For, I believe the main part of your problem would be finding f'(x) and f'(4). We have f'(x). Your teacher may want it in another form.

For f'(4), plug in 4 for x and calculate:

√14 * 1 = √14 * 1 ** = √14 **

2 x^(1/2) 2 √x ** 2√4**

√4 = 2. So:

** √14 ** =

**√14**** 2*2 4**

**So, the answer is:**

** √14/****4**