Question

find the derivative...
f(x) = sqrt(x^2+5)

Answer 1

\[f(x)=\sqrt(x^2+5)\]

Answer 2

actually find the equation of the tangent line of the above function at [2,3]

Answer 3

\[f(x)=(x^2+5)^{1/2}\]
Power rule followed by chain rule

Answer 4

What do you get?

Answer 5

When you use power rule, you should get
\[f(x)=(x^2+5)^{1/2} \\ \\ f'(x)=\frac{1}{2}(x^2+5)^{-1/2}[\frac{d}{dx}(x^2+5)]\]
Where \(\frac{d}{dx}(x^2+5)\) is the result of the chain rule

Answer 6

ok

Answer 7

but I never learned about d/dx

Answer 8

That is just telling you to differentiate x^2+5

Answer 9

all I know is the f(x+h) - f(x) /h

Answer 10

That's another method, but way confusing to solve this, you'll have x and h after differentiating

Answer 11

so the derivative of \[f(x)=\sqrt(x^2-5)\] is just \[\frac{ 1 }{ 2 }(x^2-5)^{1/2}\]

Answer 12

negative 1/2

Answer 13

\[\frac{ 1 }{ 2 }(x^2\color{red}{+}5)^{-1/2}\]

Answer 14

You still need to multiply by the differential of \(x^2+5\) according to chain rule, the final result is
\[f'(x)=\frac{1}{2}(x^2+5)^{-1/2}[\frac{d}{dx}(x^2+5)] \\ \\ f'(x)=\frac{1}{2}(x^2+5)^{-1/2}(2x) \\ \\ f'(x)=x(x^2+5)^{-1/2} \]

Answer 15

ok, i think i got it now, but with other method, I just multiply \[\sqrt{(x+h)^2-5}-\sqrt{x^2-5}\] by the conjugate pair \[\frac{ \sqrt{(x+h)^2-5}+\sqrt{x^2-5} }{\sqrt{(x+h)^2-5}+\sqrt{x^2-5}}\]

Answer 16

Ok did you got it with your method?

Answer 17

\[\lim_{h \rightarrow 0}\]\[\frac{ 1 }{ \sqrt{x^2+5} }\]

Answer 18

?

Answer 19

\[f'(x)=\frac{ x }{ \sqrt{x^2+5} }\]

Answer 20

oh yeah

Answer 21

ok, equation of tangent line is \[y=3-\frac{ 1 }{ 6 }(x-2)\] at the point [2,3]

Answer 22

Gotta go I'll be back later

Answer 23

well im done thanks!