Question

Hello, stuck on finding the derivative of f(x)=√ x^2-5x , i will write where i stopped working on in a minute @bahrom7893

Answer 1

ok

Answer 2

\[f(x)=\sqrt{x ^{2}-5x}\]

Answer 3

Go for it

Answer 4

this is what I came up with:
\[f \prime(x)= \lim_{h \rightarrow 0} \frac{ f(x+h) - f(x) }{ h }\]
\[=> \lim_{h \rightarrow 0} \frac{ \sqrt{(x+h)^{2} - 5 (x+h)} - \sqrt{x ^{2} - 5x} }{ h }\] we use conjugate to get square roots out
\[==> \lim_{h \rightarrow 0} \frac{ ((x+h)^{2}-(5h - 5x)) - (x ^{2} -5x) }{ h ( \sqrt{(x+h)^{2}} - 5(x+h) - \sqrt{x ^{2}-5x} }\]
\[==> \lim_{h \rightarrow 0} \frac{ 5h ^{3} }{ h ( \sqrt{(x+h)^{2}} - 5(x+h) - \sqrt{x ^{2}-5x} }\]

Answer 5

wait you need to use the limit definition?

Answer 6

yes, sorry forgot to mention that

Answer 7

ok let me look through it

Answer 8

don't you need a plus on the bottom
\[h(\sqrt{(x+h)^2-5(x+h)}+\sqrt{x^2-5x})\]

Answer 9

oh, yes, forgot that i have to change the sign when i use conjugate, can we simplify that before plugging 0?

Answer 10

No, cancel out the h from the denominator, and then plug it in. With all the h = 0, it will be much easier to simplify.

Answer 11

so the answer will be zero? because 5h^2 when plugging 0 will equal 0.

Answer 12

\[h ^{2} + 5h \] is the numerator right? it will equal 0

Answer 13

Sorry I was afk. Gonna take a look.

Answer 14

\[(x+h)^{2} - 5 (x+h) - (x ^{2} - 5x)=x^2+2xh+h^2-5x-5h-x^2+5x=2xh+h^2-5h\]

Answer 15

\[x^2+2xh+h^2-5x-5h-x^2+5x=2xh+h^2-5h\]

Answer 16

how did you take out the square root in the denominator ?

Answer 17

i didn't. I just multiplied by the conjugate.

Answer 18

I still didnt get how you came out with that answer, i thought i should multiply it with conjugate once to get out the square root from the numerator @bahrom7893