Question

we are given that $\frac{dy}{dx}=\frac{\sin(y)}{x}$. find an expression for $\frac{d^{2}y}{dx^{2}}$ in terms of $x$ and $y$. $\frac{d^{2}y}{dx^{2}}=\square$

Explanation:

Step1: Differentiate both sides with respect to \(x\)

Use the quotient rule \((\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}\), where \(u = \sin(y)\) and \(v=x\).
By the chain - rule, \(\frac{d}{dx}(\sin(y))=\cos(y)\frac{dy}{dx}\).
So, \(\frac{d^{2}y}{dx^{2}}=\frac{\cos(y)(\frac{dy}{dx})\cdot x-\sin(y)\cdot1}{x^{2}}\)

Step2: Substitute \(\frac{dy}{dx}=\frac{\sin(y)}{x}\) into the above formula

Substitute \(\frac{dy}{dx}\) into \(\frac{d^{2}y}{dx^{2}}=\frac{\cos(y)(\frac{dy}{dx})\cdot x-\sin(y)\cdot1}{x^{2}}\)
We get \(\frac{d^{2}y}{dx^{2}}=\frac{\cos(y)\cdot\frac{\sin(y)}{x}\cdot x-\sin(y)}{x^{2}}\)

Step3: Simplify the expression

Simplify \(\frac{\cos(y)\cdot\frac{\sin(y)}{x}\cdot x-\sin(y)}{x^{2}}\)
\(\frac{d^{2}y}{dx^{2}}=\frac{\sin(y)\cos(y)-\sin(y)}{x^{2}}=\frac{\sin(y)(\cos(y) - 1)}{x^{2}}\)

Answer:

\(\frac{\sin(y)(\cos(y)-1)}{x^{2}}\)