Thread: Simple Differentiation

Hello I have the question 'Find the equations of the tangents to these curves at correspnding points '

I'm getting very frustrated becuase I cannot where I'm going wrong . Heres how i've been doing it.

y=2/x² , x=2 hence points (2,1/2)

y=2x-² (Is this where I went wrong?)

y=-4x^-3

dy/dx = 4*1/8
=1/2

y-1/2=4/8(x-2)

y=1/2x-1/2


Yet this isnt the line equation

Hi there,

You know the equation of a line given a gradient and point is

you also know:







Find m by finding the derivitive of when x is 2

you have already found the point

looking at your working it looks correct to me, except for the sign on m

Thats what I did though, and I doing the geometry right?

check the sign on your gradient you should then get the correct line.

Should this come under Pre-Calculus or Calculus?

Originally Posted by alibond07

Hello I have the question 'Find the equations of the tangents to these curves at correspnding points '

I'm getting very frustrated becuase I cannot where I'm going wrong . Heres how i've been doing it.

y=2/x² , x=2 hence points (2,1/2)

y=2x-² (Is this where I went wrong?)

y=-4x^-3

dy/dx = 4*1/8
=1/2

y-1/2=4/8(x-2)

y=1/2x-1/2


Yet this isnt the line equation

Code:

y=2x-²   (Is this where I went wrong?)
 
y=-4x^-3
 
dy/dx = 4*1/8
        =1/2

should be y=2x-²

dy/dx = -4x^-3

dy/dx = - 4*1/8
=- 1/2

y-1/2=-4/8(x-2)


y=-1/2x+3/2