Question

i cant solve this--->x^3y^3 - 4=0...find d2y/dx2 by implicit differentiation..please....

Answer 1

You have to use the product rule for the first term.

Answer 2

\[\frac{d}{dx}[x^3y^3] = x^3\frac{d}{dx}[y^3] + y^3\frac{d}{dx}[x^3]\]

Answer 3

Then recall that \[\frac{d}{dx}f(y) = \frac{d}{dy}f(y)\cdot \frac{dy}{dx}\]
by the chain rule.

Answer 4

product rule??already,,,but dy/dx got -y/x,..then dont know..coz the answer 2y/x^2..

Answer 5

You have to take the derivative again to find \[\frac{d^2y}{d^2x^2}\]

Answer 6

yup,,but my answer didnt same like the ans given
i got (-3x^2y^3) / (3x^3y^2)...can it be simplified???

Answer 7

One sec. That doesn't seem right.

Answer 8

:(

Answer 9

\[x^3\frac{d}{dx}[y^3] + y^3\frac{d}{dx}[x^3]\]
\[=2x^3y^2\frac{dy}{dx} + 2x^2y^3 = 0\]\[\implies \frac{dy}{dx} = -\frac{y}{x}\]

Answer 10

With me so far?

Answer 11

yup...so far same..

Answer 12

\[\implies \frac{d^2y}{d^2x^2} = -(\frac{1}{x}\frac{dy}{dx} + y\frac{d}{dx}[\frac{1}{x}])\]
\[=-(\frac{1}{x}\cdot\frac{-y}{x} -\frac{y}{x^2})\] \[=\frac{2y}{x^2}\]

Answer 13

make sense?

Answer 14

cant get it:((i use quotient rule..

Answer 15

What? where?

Answer 16

I used the product rule.

Answer 17

\[\frac{dy}{dx} = -\frac{y}{x} = -yx^{-1}\]
\[\]

Answer 18

\[\implies \frac{d^2y}{d^2x^2} = \frac{d}{dx}[-yx^-1]\]\[=-y\frac{d}{dx}[x^{-1}] + x^{-1}\frac{d}{dx}[-y]\]\[= \frac{y}{x^2} + x^{-1}(-\frac{dy}{dx})\]\[ = \frac{y}{x^2} + x^{-1}(-(-yx^{-1}))\]\[= \frac{y}{x^2} + yx^{-2} \]\[=\frac{y}{x^2} + \frac{y}{x^2} \]\[=\frac{2y}{x^2}\]

Answer 19

is another way if you prefer.

Answer 20

ok2..i got it finally:)
use quotient rule for -y/x
then - ((x(dy/dx)-y) / x^2)
then i subtitute dy/dx with -y/x...so got sme ans with u...

Answer 21

yep. Should be the same either way. I just usually ignore quotient by using power rule with negative exponents and the product rule..

Answer 22

then can u help me with this
d/dx \[d/dx \left( 1/2 \tan^{-1} x + 1/4 \ln \left( x+1 \right)^{2}\div \left( x ^{2}+1 \right) \right)\]