Question

FInd the derivative of the function by using the definition of the derivative… please help
f(x)=sqrtx + 1

Answer 1

it is all inside the radical?
\[f(x)=\sqrt{x+1}\]?

Answer 2

noo only x is inside..

Answer 3

@satellite73

Answer 4

\[f(x)=\sqrt{x}+1\]

Answer 5

yes

Answer 6

\[f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}\] so we get
\[\frac{\sqrt{x+h}+1-(\sqrt{x}+1)}{h}\] and now some algebra

Answer 7

evidently the ones up top add up to zero, so we can ignore them and compute
\[\frac{\sqrt{x+h}-\sqrt{x}}{h}\] gimmick is to multiply top and bottom by the conjugate of the numerator, namely \(\sqrt{x+h}+\sqrt{x}\) leave the denominator in factored form you get
\[\frac{(\sqrt{x+h}-\sqrt{x})(\sqrt{x+h}+\sqrt{x})}{h(\sqrt{x+h}+\sqrt{x})}\]
\[=\frac{x+h-x}{h(\sqrt{x+h}+\sqrt{x})}\]

Answer 8

all that is left in the numerator is \(h\) which cancels with the \(h\) in the denominator leaving
\[\frac{1}{\sqrt{x+h}+\sqrt{x}}\] now take the limit by replacing \(h\) by zero and you are done

Answer 9

I got lost on the third part… :/ @satellite73

Answer 10

this one
\[\frac{(\sqrt{x+h}-\sqrt{x})(\sqrt{x+h}+\sqrt{x})}{h(\sqrt{x+h}+\sqrt{x})}\]
??

Answer 11

yes that one,,… but I think I kinda know what you did… thats the product rule right???

Answer 12

ok it is this, you want to get rid of the radical, so you know \((a-b)(a+b)=a^2-b^2\)

Answer 13

which means
\[(\sqrt{h+h}-\sqrt{x})(\sqrt{x+h}+\sqrt{x})=\sqrt{x+h}^2-\sqrt{x}^2\]
\[=x+h-x=h\]

Answer 14

yess...

Answer 15

but isn't that a rule….

Answer 16

and your real job is to figure out how to get rid of the \(h\) in the denominator

Answer 17

so now that you have an \(h\) up top and a factor of \(h\) in the bottom, with
\[\frac{h}{h(\sqrt{x+h}+\sqrt{x})}\] you can cancel the \(h\) giving
\[\frac{1}{\sqrt{x+h}+\sqrt{x}}\] now you can replace \(h\) by 0 without getting a zero in the denominator

Answer 18

ok and to do that you replace it with zero..?

Answer 19

yes, now you can take the limit as \(h\to 0\) by simply erasing the \(h\)

Answer 20

ahhh ok ok… makes sense.

Answer 21

write
\[\lim_{h\to 0}\frac{1}{\sqrt{x+h}+\sqrt{x}}=\frac{1}{\sqrt{x}+\sqrt{x}}\]
\[=\frac{1}{2\sqrt{x}}\]

Answer 22

and dont forget this one
the derivative of
\[\sqrt{x}\] is always \[\frac{1}{2\sqrt{x}}\] so while your classmates are trying to figure it out on a test or quiz using the stupid exponential notation and the power rule, you write the answer instantly. it never changes, you never have to compute it again

Answer 23

omg thankyou so much. we have a quiz on this tomorrow also. thank you . really . Im glad you actually went through and took the time to explain this to me . thank you again.

Answer 24

@satellite73