Question

find the eq. to the tangent line and the normal line to the curve y= x+ square root of x, (1,2)

Answer 1

dy/dx= 1+1/square root of x. at (1,2)
dy/dx=2=slope of tangent. and -1/2=slope of normal.
equation of tangent, y-2=2(x-1).
equation of normal, y-2=-1/2(x-1)

Answer 2

nop

Answer 3

if u change the square into exponential u will have (x)^1/2

Answer 4

sorry dy/ dx= 1+1/2squre root of x at (1,2)
slope of tangent = 3/2..now you can solve ... i just missed 1/2 i above solution

Answer 5

mm
nop

Answer 6

u have dy/dx=1+1/2square root of x^-1/2

Answer 7

that give u that give u 5/4

Answer 8

if you put 1 at the place of x you wil get. 1+1/2 sruare root of 1.
now 1+1/2=3/2

Answer 9

yea
but when u take the derivative of (x)^1/2
you will get 1/2(x)1/2-1
= to 1/2 (x)-1/2... now u plug 1

Answer 10

sorry

Answer 11

1/2(x)^-1/2

Answer 12

now plug 1