When asked to differentiate f(x)= squareroot x, what should i do, thanks?

Answer 1

The squareroot of x can also be expressed as x to the power of 1/2.

f(x) = squareroot x
f(x) = x ^ (1/2)

Now, take the usual approach to differentiating a function of x:

df(x)/dx = 1/2 * x ^ (1/2 - 1)
df(x)/dx = 1/2 * x ^ (-1/2)

X to a negative power is the same as 1/(x to the positive power).

df(x)/dx = 1/2 * 1 / (x ^ (1/2))

Replacing x^(1/2) with squareroot x again, you get:

df(x)/dx = 1/(2*squarerootx)

Answer 2

All of the first six answers are correct, except that puggy (who is usually spot on!) let a minus sign creep in that shouldn't be there.

Were you asked to differentiate it from first principles?
If so, the easiest way is to use the definition
f'(x) = limit((f(x) - f(c))/(x - c) as c approaches x. [OK, purists, I know it's usually the other way round, but this way is valid too!]

Now (sqrtx - sqrtc)/(x - c), if we multiply both top and bottom by sqrtx + sqrtc , becomes

(x - c)/(x - c)(sqrtx + sqrtc) using the process that's usually for rationalising the denominator, but in this case it rationalises the numerator, which then cancels with the x - c in the denominator, leaving us with

1/(sqrtx + sqrtc)

As c approaches x, this approaches

1/(sqrtx + sqrtx)

= 1/(2*sqrtx).

Same as by the index method!!! Wonderful.

Answer 3

Write it in exponent form, and use the rule of exponents for differentiating f(x)=x^n kinds of functions.

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

f'(x)= (1/2) x ^(-1/2)

Done!

Answer 4

Write square root of x as a power of x.

Then use the usual rule: multiply by the power, then reduce the power by 1.

Answer 5

sqrt(x) can be expressed as x^(1/2), so you just use the power rule as normal.

f(x) = sqrt(x)
f(x) = x^(1/2)

f'(x) = (1/2)x^(-1/2)

We normally don't want negative exponents; taht's why we have to make x^(-1/2) into a fraction so we'll have a positive exponent.

f'(x) = (1/2) (-1/[x^(1/2)])

Merging into one fraction, we have

f'(x) = -1/[2x^(1/2)]

Answer 6

f(x)= x^(1/2)
I forget how to differentiate would it be
f'(x)=(1/2)x^(-1/2) ?
(hey it was 25 years ago)

Answer 7

(I hope I remember this right...)

That's the same as saying:

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

So, the derivative is (1/2)x^(-1/2)

Answer 8

sqrt(x) or x^(1/2)

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

f'(x) = nx^(n - 1)

f'(x) = (1/2)x^((1/2) - 1)
f'(x) = (1/2)x^(-1/2)

ANS :
f'(x) = 1/(2x^(1/2))
f'(x) = 1/(2sqrt(x))
f'(x) = 1/(sqrt(4x))

Multiply top and bottom by sqrt(4x)

f'(x) = (sqrt(4x))/(4x)
f'(x) = (2sqrt(x))/(4x)
f'(x) = sqrt(x)/(2x)

ANS : f'(x) = sqrt(x)/(2x)