Question

find f'(x) of f(x)=√(x^2 +4)

Answer 1

f'(x)=(1/2)(x^2+4)^(-1/2)*2x=x(x^2+4)^(-1/2)

Answer 2

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

Answer 3

so you can use that here to get
\[\frac{2x}{2\sqrt{x^2+4}}\]

Answer 4

or if you like
\[\frac{x}{\sqrt{x^2+4}}\]

Answer 5

what if the question then ask to find the equation of the tangent line y=f(x)=√(x^2 +4) at the ponint x=1?

Answer 6

well you have the point, namely 1, whatever you get when you make x=1

Answer 7

i get \[(1,\sqrt{5})\]

Answer 8

and you have the slope, namely whatever you get when you put x = 1 into the deivative

Answer 9

make that "derivative"
i get
\[m=\frac{1}{\sqrt{5}}\]

Answer 10

the slope is the f'(x), right?

Answer 11

well
\[f'(x)\] is the formula for the slope. the slope is a number. namely what you get when you replace x by whatever it is. in this case it is 1

Answer 12

so the point is
\[(1,\sqrt{5})\] and the slope is
\[m=\frac{1}{\sqrt{5}}\] so use "point-slope" formula to get the equation for the line

Answer 13

don't forget the slope is a number. the derivative is a formula for the slope, not the slope itself

Answer 14

i get
\[y-y_1=m(x-x_1)\]
\[y-\sqrt{5}=\frac{1}{\sqrt{5}}(x-1)\] you can multiply out if you like

Answer 15

ooo! Thank you! :)