Question

What is the derivative of sqrt(9x^2)?

Answer 1

3

Answer 2

thats what i thought but the answer says

Answer 3

arent they the same thing?

Answer 4

d/dx(3 \sqrt(x^2)) | Factor out constants: = | 3 (d/dx(\sqrt(x^2))) | Use the chain rule, d/dx(\sqrt(x^2)) = ( dsqrt(u))/( du) ( du)/( dx), where u = x^2 and ( dsqrt(u))/( du) = 1/(2 \sqrt(u)): = | 3 (d/dx(x^2))/(2 \sqrt(x^2)) |The derivative of x^2 is 2 x: = | (3 (2 x))/(2 \sqrt(x^2))

Answer 5

\[\frac{d}{dx} (\sqrt{9x^2}) = \frac{1}{2\sqrt{9x^2}} \times (18x)\]

Answer 6

:|

Answer 7

therefore it becomes \[\frac{18x}{2\sqrt{9x^2}} = \frac{9x}{\sqrt{9x^2}} = \frac{9x}{3x}\]

Answer 8

:|

Answer 9

\[\frac{3x}{\sqrt{x^2}} = \frac{3x}{x} = 3\]

Answer 10

the sqrt of x^2 is the absolute value of x

so you cannot simplify it unless you assume x to be positive, or unless you take cases

Answer 11

smart @funinabox

Answer 12

:D haha thanks

that one always caught me off guard one tests >_>

Answer 13

on tests*

Answer 14

Be careful you guys: \[\frac{d}{dx}\sqrt{9x^2}=3\cdot \color{red}{\text{sgn}(x)}\]Here, \(\text{sgn}(x)\) is the SIGN or SIGNUM of \(x\)...it's 1 when \(x\) is positive, \(-1\) when \(x\) is negative, and not defined when \(x=0\).

Answer 15

@yakeyglee

via the chain rule the derivative becomes

3x/sqrt(x^2)

at this point, x can still be any real number

however, if you wish to simplify it further, the sqrt(x^2) must be absolute value of x

and it will be 3x/|x| or just 3 if x is an assumed value or you can take cases.

simplifying during the derivative results in losing solutions

Answer 16

I don't see what your correction is? \(\text{sgn}(x)=\dfrac{x}{|x|}\)

Answer 17

thanks guys :D