Question

Find the equation of the tangent line y=2/(squareroot) x

Answer 1

@ApoorvaAnand Help me again? you explain really well!

Answer 2

y = 2*x^(-1/2)
That is the question

Now slope of tangent can again be written as,
dy/dx = 2*(-1/2)*x^(-3/2)

Since we do not know the points, we can't find the eq of tangent. You need to tell us the point at which the eq of tangent should be formed.

Answer 3

sorry there are points I didnt see them written they are 4 and 2/2

Answer 4

@ApoorvaAnand

Answer 5

dy/dx = 2*(-1/2)*(1/8) = -8 now

Sub in eq and find your answer (like previous question) :)

Answer 6

so then do I do 4/(2/2)? @ApoorvaAnand

Answer 7

No. The eq of tangent is now y-4/x-1 = -8

2/2 = 1

Answer 8

so then if im solving for y? @ApoorvaAnand

Answer 9

Yes. Solve for y for y = mx+c form.

Answer 10

y=-8(4) not sure about the c @ApoorvaAnand

Answer 11

No. You only substitute the values to y1 and x1. Now y and x.
The answer is y-4 = -8x+8 => y = - 8x+12 or the eq of tangent is y + 8x - 12 = 0

Answer 12

ohh so then y=-8x+12? @ApoorvaAnand

Answer 13

Yes. :)

Answer 14

is the given point \[\large (4, \frac{2}{2})\]
hmmm.....

Answer 15

what is the given point?

Answer 16

the way you wrote it is not clear
and apoorva.. used it even though it was not clear

Answer 17

you know post the picture for a better understanding

Answer 18

take a snapshot and post :)

Answer 19

@shaekitchen you should relook at this