Question

tangent line equation of curve y^3 + 7y = x^3 at point (2,1) is ...

Answer 1

to find the slope use implicit differentiation:-

3y^2 . dy/dx + 7 dy/dx = 3x^2

dy/dx = 3x^2 / ( 3y^2 + 7) = slope

Answer 2

plug in x=2 and y=1 to evaluate the slope

Answer 3

then to find the required equation
use
y - y1 = m(x - x1)

where m = slope and (x1,x2) is the point (2,1)

Answer 4

@welshfella
I'm just a beginner in differentiation too

I have an example in my book saying if I have an equation of y = x^3 - 3x^2 + 2x -1
for this, If I were to differentiate I would get

dx/dy = 3x^2 - 6x + 2

but in this question the equation is y^3 + 7y = x^3

how do I differentiate this when there is two different variables?

Answer 5

two different variables as in the y and x

Answer 6

treat y as a function of x and use the chain rule if necessary

so it you want to differentiate y^2
the derivative of y is dy/dx
now find derivative of y^2 like you would x^2 - this gives 2y

so derivative of y^2 with respect to x is 2y. dy/dx

Answer 7

but maybe you haven't done the chain rule yet?

Answer 8

lol yes I was about to say that I didn't really do chain rule
the only thing I know so far is for instance you have

2x^2 the derivative of this will be 4x
things like that..

Answer 9

RIGHT WELL YOU WONT COME TO IMPLICIT DIFFERENTIATION FOR A WHILE YET

Answer 10

is the point really (2,1) ?
it doesn't even lie on the given curve

Answer 11

you use the chain rule for 'a function in a function'.

for example f(x) = (x^2 + 1)^20
the x^2+1 is a function within the function f(x)

so if we let u = x^2 + 1 then f(x) = u^20

then du/dx = 2x and d(fx) du = 20u^19


so d(fx) .dx = d(f(x) / du * du/dx = 2x . 20u^19

= 2x. 20(x^2 + 1)^19
= 40x(x^2 + 1)^19

Answer 12

@ganeshie8 - you are right Might be a typo somewhere

Answer 13

@welshfella Oh! okay
Sorry just a question though...
at

"so d(fx) .dx = d(f(x) / du * du/dx = 2x . 20u^19"

the du are getting cancelled, so I just wanna know where are you getting the 2x in 2x . 20u^19 from, if you are canceling out the du

Answer 14

i'm finding that a bit difficult to explain

the left side contains du/dx which we found to be 2x so we cant eliminate that

we need to find d(f(x)/dx and it equals the same thing as d(fx)/dx * du / dx

Answer 15

* d(fx)/du

Answer 16

oh so I can't treat it like a fraction then...?

Answer 17

yes you can treat it like a fraction

Answer 18

we are treating it like a fraction when we are eliminating du

Answer 19

yes that is exactly what I mean by treating it like a fraction when canceling du

Ok so wait, the way how I understood it.. it's like multiplying

2x with 20u^19
when you said d(fx)/du * du / dx

right?

Answer 20

so thats why you're getting 2x . 20u^19

Answer 21

yes - you got it. Better explanation than mine..

Answer 22

Thank you so much Denis! :D

Answer 23

yw