Question

find y'' of xy + tan^-1 (xy) = 4
a. -(2y/x^2)
b. -(2x/y^2)
c. 2y/x^2
d. 2x/y^2

Answer 1

\[y'=-\frac{y}{x}\] is a start

Answer 2

then

Answer 3

i get letter A when i do that but i dont trust always my answer

Answer 4

We have
\[xy+\tan ^{-1} (xy)=4\]
Let's differentiate this
we get
we know
\[\frac{d}{dx} \tan^{-1} x=\frac{1}{1+x^2}\]

\[y+ x\frac{dy}{dx}+ \frac{1}{1+x^2y^2} (y+x\frac{dy}{dx})=0\]
Let's take the common term out
\[(y+x\frac{dy}{dx})(1+\frac{1}{1+x^2y^2})=0\]
we get
\[(y+x\frac{dy}{dx})(1+x^2y^2+1)=0\]
or
\[(y+x\frac{dy}{dx})(2+x^2y^2)=0\]
2+x^y^ can't be zero, so
\[y+x\frac{dy}{dx}=0\]
or
\[ \frac{dy}{dx}=-\frac{y}{x}\]
we have
\[y+x\frac{dy}{dx}=0\]
Let's differentiate this
\[\frac{dy}{dx}+ \frac{dy}{dx}+x\frac{d^2y}{dx^2}=0\]
now we get
\[2\frac{dy}{dx}+x\frac{d^2y}{dx^2}=0\]
or
\[\frac{d^2y}{dx^2}=-2\frac{dy}{xdx}\]
we have \( \frac{dy}{dx}=-\frac{y}{x}\)
so
\[\frac{d^2y}{dx^2}=-2\frac{-y}{x\times x}\]
we get finally
\[\frac{d^2y}{dx^2}=\frac{2y}{x^2}\]

Answer 5

omg thats a big help

Answer 6

:D :D

Answer 7

can i learn this thing in 1 day or not?

Answer 8

You want to learn differentiation?

Answer 9

yes i do but how many days do i need to learn this thing? For example the topic right now on school is about derivative of trigonometric function.

Answer 10

Just go along with the class. Learn the topic the same day when it's taught in School. It'll be a slow process but you'll learn it beautifully:)

Answer 11

maybe i do that after class i will try my best.