Tsunami Gaming

i'm kind of rusty on the limit definition of the derivative, but so that you can check your answers, they should be

1) 1 - (1/x^2)

2) (1/2)(1/sqrt(x)) one half times 1 over the squareroot of x ......(1 over the quantity 2times squareroot(x))

and for the final one, the derivative gives you the slope at any one point so what you have to do is find a point at which the derivative will yield the m term (slope) of the linear equation

solve the linear equation for y=
3x-y+1=0
3x-y=-1
-y=-1-3x
y=3x+1

so you know the slope needs to be 3 (from y=mx+b form)

set the derivative equal to 3
3=3x^2
1=x^2
x= (+/-) 1

(+/-) denotes the plus or minus operator

that take care of it?

sorry i'm a bit late lol

how wud i find the derivative of these functions, and i have to actually use the definition of a derivative, not the power rule.

f(x)= x + (1/x)

f(x)= √X thats square root of X

and this is what the third question reads:

Find an equation of the line that is tangent to the graph of f AND parallel to the given line.

f(x)= x^3

3x-y +1=0

see now I found the derivative of the function as 3x^2 but how da hell is dat a slope i mean i parallel lines should have the same slope so i have no idea how 3x^2 wud b considered as a slope... i'm mad confused