Question

derivatives of sin x/y = y/x

Answer 1

Anyone can help me to solve this

Answer 2

Do you have any numbers?

Answer 3

nothing

Answer 4

implicit diff right?

Answer 5

In this problem we'll find the derivative of sin−1(x) (also written arcsin(x)), the inverse of the sine ... (a) If we differentiate the implicit function x = sin(y), we find.

Answer 6

\[\sin \frac{ x }{ y } = \frac{ y }{ x }\]

Answer 7

the equation is
\[\sin(\frac{x}{y})=\frac{y}{x}\]

Answer 8

what is the first step to solve that

Answer 9

think of it as
\[\sin(\frac{x}{f(x)})=\frac{f(x)}{x}\]

Answer 10

the derivative on the right requires the quotient rule
i would be
\[\frac{xf'(x)-f(x)}{x^2}\]

Answer 11

with the \(y\) notation this is
\[\frac{xy'-y}{x^2}\]

Answer 12

the derivative of
\[\sin(\frac{x}{f(x)})\] requires the chain rule
the derivative of sine is cosine, and the derivative of \(\frac{x}{f(x)}\) is
\[\frac{f(x)-xf'(x)}{f^2(x)}\] again by the quotient rule

Answer 13

in the \(y\) notation it is
\[\cos(\frac{x}{y})\left(\frac{y-xy'}{y^2}\right)\]

Answer 14

then whats next

Answer 15

a boatload of algebra is next
set
\[\cos(\frac{x}{y})\left(\frac{y-xy'}{y^2}\right)=\frac{xy'-y}{x^2}\] and solve for \(y'\)

Answer 16

what should i do 1st