Find the point on the curve y = √ x where the tangent line is parallel to the line y = x 8 .

If tangent to the curve y = √x  is parallel to the line  y = 8x,  then this implies that the tangent to  y = √x  has the same slope as the line  y = 8x. In other words, the derivative (slope) function of  y = √x  is equal to the slope of the line  y = 8x, which is m = 8.  Hence y' = 8  once we find  y'

y =  √x = x^(1/2)

Applying the power rule and simplifying, we find that the derivative is

y' = 1/(2√x)

Now remember that  y'  must equal 8

1/(2√x) = 8

Multiplying both sides by  2√x,  we obtain

1 = 16√x

Dividing both sides by  16,  yields

√x = 1/16

But wait a minute, √x = y.  Thus  1/16  must be the y-coordinate of the point at which the tangent to  y = √x  is drawn.

Squaring both sides, yields

x = 1/256 

This is the x-coordinate of the point on the curve where the tangent is drawn.

∴ the required point must be  (1/256, 1/16)

GOOD LUCK!!!

5/ 1/3=15 i think sorry if im wrong

The first one is 2/3the second is 1/160the third is 4/125