Question

how do you differentiate the square root of 2x-3?

Answer 1

chain rule.

Answer 2

\[\frac{d}{dx}[\sqrt{f(x)}]=\frac{f'(x)}{2\sqrt{f(x)}}\]

Answer 3

At first, do you know the derivative of a sqrt(x) ?

Answer 4

\(\large\color{black}{ \sqrt{x} }\) is same as \(\large\color{black}{ x^{1/2} }\). So apply the power rule,

\(\large\color{black}{ \frac{d}{dx}~x^n ~=n~\times x^{n-1} }\)

Answer 5

I'll give you examples if you want to.

Answer 6

I keep getting disconnected. god dang it.

Answer 7

1) Find the derivative of \(\large\color{black}{ \sqrt{x^2-4x+5} }\).

First derive the function as if it was just a square root of \(\large\color{black}{ \sqrt{x} }\) , then using a chain rule, mutliply times the derivative of the inner function, of \(\large\color{black}{ x^2-4x+5 }\).

the derivative of \(\large\color{black}{ \sqrt{x} }\) is

\(\large\color{black}{ \frac{d}{dx}~\sqrt{x}=\frac{d}{dx}x^{1/2}=\frac{1}{2}\times x^{1/2~~~-1}=\frac{1}{2}\times x^{-1/2}=\frac{1}{2x^{1/2}}=\frac{1}{2\sqrt{x}}}\)

the derivative of \(\large\color{black}{ x^2-4x+5 }\) (applying power rule to each term)
is: \(\large\color{black}{ 2x-4 }\).

So, we get this:

\(\Large\color{black}{ \frac{d}{dx}(\sqrt{x^2-4x+5}~~)=\frac{1}{2\sqrt{x^2-4x+5}}\times (2x-4) }\)

this simplifies to,

\(\Large\color{blue}{ \frac{d}{dx}(\sqrt{x^2-4x+5}~~)=\frac{(2x-4)}{2\sqrt{x^2-4x+5}} }\)

and to completely simplify:

\(\Large\color{blue}{ \frac{d}{dx}(\sqrt{x^2-4x+5}~~)=2\frac{(x-2)}{2\sqrt{x^2-4x+5}} }\)

\(\Large\color{red}{ \frac{d}{dx}(\sqrt{x^2-4x+5}~~)=\frac{(x-2)}{\sqrt{x^2-4x+5}} }\) and this is the final answer.