Question

Find the slope of the tangent line to the curve (a lemniscate) 2(x^2+y^2)^2=25(x^2-y^2) at the point (-3,1)?

Answer 1

could you implicitly differentiate that ?

Answer 2

you will need chain rule here
whats d/dx x^2 =...?
whats d/dx y^2=...?

Answer 3

what ? where does 2*2 come from ?

Answer 4

d/dx[2(x^2+y^2)^2]

= 2*(2 (x^2+y^2)) times d/dx(x^2+y^2)
this is because of chain rule, got this step ?

Answer 5

d/dx (x^2) = 2x
thats why
and the other 2 was already there

Answer 6

ok, so now what about d/dx (x^2+y^2) =...?

Answer 7

2x+2y?

Answer 8

here, chain rule again comes into picture
d/dx f(y) = f'(y) dy/dx

so d/dx (y^2) = 2y dy/dx
got this ?

Answer 9

if you got that we are almost done :)

Answer 10

d/dx[2(x^2+y^2)^2]

= 2*(2 (x^2+y^2)) times d/dx(x^2+y^2)


this part ?

Answer 11

whats d/dx x^2 =.... ?

Answer 12

we don't rewrite anything

Answer 13

ohman :/

Answer 14

derivative of outer times derivative of inner
like for (sin x)^3
we have outer = x^3, derivative = 3x^2
inner = sin x derivative = cos x

so,
3 (sin x)^2 cos x
got this ?

Answer 15

in same way
d/dx [(x^2+y^2)^2] = 2 (x^2+y^2) times d/dx (x^2+y^2)

Answer 16

i get that, but in this problem im confused with why the exponent goes away

Answer 17

ok, you said d/dx (x^2) = 2x
why did the exponent go away here ?

Answer 18

2(x^2+y^2)^2=25(x^2-y^2)

Answer 19

so you're finding the derivative of the inside?

Answer 20

we do find derivative of inside in chain rule, right ?

Answer 21

yup

Answer 22

my professor also taught us a way of outer first then rewrite inner and then multiple by the derivative of the inner.

Answer 23

he must have meant "derivative of outer first ....."

Answer 24

d/dx[f(g(x))]= f' (g(x))*g'(x)

Answer 25

yes,
here f(x) was x2

Answer 26

well isn't it 4?

Answer 27

2(x^2+y^2)^2