1. ## 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

2. Hi there,

You know the equation of a line given a gradient and point is $y-y_{1}=m(x-x_{1})$

you also know:

$y=c \Rightarrow \frac{dy}{dx}=0$

$y=kx \Rightarrow \frac{dy}{dx}=k$

$y=kx^n \Rightarrow \frac{dy}{dx}=nkx^{n-1}$

Find m by finding the derivitive of $y = \frac{2}{x^2} = 2x^{-2}$ when x is 2

you have already found the point $(2, \frac{1}{2})$

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

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

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

5. Should this come under Pre-Calculus or Calculus?

6. 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

$\rightarrow$y-1/2=-4/8(x-2)

$\rightarrow$y=-1/2x+3/2