Question

use implicit differentiation to find \\( \frac { d y } { d x } \\).

\\( x ^ { 4 } + \sin y = x ^ { 3 } y ^ { 5 } \\)

\\( \frac { d y } { d x } = \\)

Explanation:

Step1: Differentiate both sides

Differentiate \(x^{4}+\sin y\) and \(x^{3}y^{5}\) with respect to \(x\).
For \(x^{4}\), using the power rule \(\frac{d}{dx}(x^{n})=nx^{n - 1}\), we have \(\frac{d}{dx}(x^{4}) = 4x^{3}\).
For \(\sin y\), using the chain - rule \(\frac{d}{dx}(f(g(x)))=f^{\prime}(g(x))\cdot g^{\prime}(x)\), where \(f(u)=\sin u\) and \(u = y\), we get \(\cos y\frac{dy}{dx}\).
For \(x^{3}y^{5}\), using the product rule \(\frac{d}{dx}(uv)=u^{\prime}v + uv^{\prime}\) (where \(u=x^{3}\), \(u^{\prime}=3x^{2}\) and \(v = y^{5}\), \(v^{\prime}=5y^{4}\frac{dy}{dx}\)), we have \(3x^{2}y^{5}+5x^{3}y^{4}\frac{dy}{dx}\).
So, \(4x^{3}+\cos y\frac{dy}{dx}=3x^{2}y^{5}+5x^{3}y^{4}\frac{dy}{dx}\).

Step2: Solve for \(\frac{dy}{dx}\)

Rearrange the terms:
\(\cos y\frac{dy}{dx}-5x^{3}y^{4}\frac{dy}{dx}=3x^{2}y^{5}-4x^{3}\).
Factor out \(\frac{dy}{dx}\):
\(\frac{dy}{dx}(\cos y - 5x^{3}y^{4})=3x^{2}y^{5}-4x^{3}\).
Then \(\frac{dy}{dx}=\frac{3x^{2}y^{5}-4x^{3}}{\cos y - 5x^{3}y^{4}}\).

Answer:

\(\frac{3x^{2}y^{5}-4x^{3}}{\cos y - 5x^{3}y^{4}}\)