# Differentiate/Implicit/TangentWatch

#1
Heres the q:

A curve has the equation

a Differentiate by rearranging the equation to the form

b Differentiate the original equation implicitly and then find the equation of the tangent to this curve at the point (1,1)

I dont have answers in the book so im not sure if im right

for (a) I got . Is that right?

for (b) I got to . Is that right/what do i do next? I know the equation of the tangent, but not sure how to get the gradient from that or the other x and y value
0
9 years ago
#2
Factorise dy/dx out.
0
#3
what does that do? I dont see what to do next
0
9 years ago
#4
Well you can rearrange for dy/dx on one side, hence you have gradient function, and you know the x and y co-ords at the point you want to find gradient for (1,1).. Sub them in and you'll get gradient.
0
#5
you mean

for (1,1) it equals -5/3

but how do i get y1 and x1?

for:
0
9 years ago
#6
Erm well i just use y=mx+c, and i would get,

1=-5/3+c
c=1--5/3
c=8/3

Don't know if that's any help, but i've never use the y2-y1=m(x2-x1) so don't know, sorry.
0
#7
what do i need c for/how does that help me get the equation of the tangent?
0
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