Equation of the tangent line using differentiation properties

[hopelynnwelch]

So here is how I am thinking I need to do this:

f '(x)=2x^(-2)
= -4x^(-1)
=-4/x

then if I plug 1 for x I get a y value

-4/1 = -4

y-y1=m(x-x1)

y+4= -4/x(x-1)

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

Is this wrong? I've tried to enter this in like 5 different ways and it keeps telling me I'm wrong. I probably am... lol

 

[hopelynnwelch]

Ohhh I figured this out!

it is y=-4x+6

Wooot woooot!~

 

[HallsofIvy]

View attachment 4928


So here is how I am thinking I need to do this:

f '(x)=2x^(-2)

This is bad notation f(x)= 2x^{-2}, not f'. You have not yet differentiated.

= -4x^(-1)
=-4/x

And here you differentiated incorrectly. The derivative of x^n is x^{n-1}. With n= -2, that would be -2 x^{-2-1}= -2x^{-3}. You added 1 to the derivative when you should have subtracted. Since x= 1, you got the right value even though your derivative was wrong.

then if I plug 1 for x I get a y value

-4/1 = -4

This is not "y", it is "m". The y value when x= 1 is 2/(-1)^2= 2.

y-y1=m(x-x1)

y+4= -4/x(x-1)[/quote]
where did that "/x" come from? You understand that "m" is the derivative evaluated at the given x, don't you?

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

Is this wrong? I've tried to enter this in like 5 different ways and it keeps telling me I'm wrong. I probably am... lol

When x= 1, y= 2 and m= -4. You should have y- 2= -4(x- 1)= -4x+ 4 so y= -4x+ 6.