Question

Are there any tricks for finding the derivative of a function containing sqr rt?

Answer 1

http://www.analyzemath.com/calculus/Differentiation/find_derivative_function.html

Answer 2

One of the problems I'm working on is
\[4\sqrt{x}-x\]

Answer 3

can hrlp this to u

Answer 4

indeed there is \[\huge \frac{d}{dx} \left(\sqrt{f(x)}\right) = \frac{f(x)}{2\sqrt{f(x)}}\]

Answer 5

\(\sqrt{x}=x^{(1/2)}\\ then \quad use \quad (d/dx) x^n=nx^{n-1}\)

Answer 6

I would write it out... sqrt(x) is x^(1/2) when you subtract 1 from 1/2 you can that easily... -1/2

Answer 7

for example \[\huge \frac{d}{dx} \left(2\sqrt x\right)\]
here, f(x) = x and f'(x) = 1 so you'll have \[2 \left(\frac{1}{2\sqrt x}\right)\]
and simplifying \[\implies \frac{1}{\sqrt x}\]

Answer 8

So for the problem I'm working on would it be 2x-1?

Answer 9

not exactly

Answer 10

\[\huge \frac{d}{dx} (4\sqrt x) \implies 4\left(\frac{1}{2\sqrt x}\right)\]
do you get what i'm implying?

Answer 11

Yes, but I'm still getting confused. I can't seem to get to the answer.

Answer 12

Move the constant out from the equation and just find the derivative of sqrt(x) which can be written as: x^(1/2)

Bring the 1/2 down in front and then you subtract 1 from 1/2... .5-1=-.5

We know that x^(-1/2) is negative, so it is in the denominator

Answer 13

So \[\frac{ 4 }{ 2\sqrt{x} }-1\] would b the derivative?

Answer 14

Yep!

Answer 15

Yes! Thanks for the help guys!