Question

1. use implicit differentiation to find \\( \frac { d y } { d x } \\) where \\( x ^ { 2 } - x \cos y + y ^ { 3 } = 0 \\).
2. find \\( \frac { d y } { d x } \\) where \\( y = \frac { e ^ { x ^ { 2 } } } { \cos ^ { 2 } x } + \frac { 1 } { x ^ { 2 } + 1 } + e ^ { x } \csc x \\).
3. find the equation of the line tangent to the graph of \\( y = \ln \frac { x } { 3 } \\) at \\( x = e \\).

Explanation:

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

Differentiate \(x^{2}-x\cos y + y^{3}=0\) term - by - term.
For the \(x^{2}\) term: \(\frac{d}{dx}(x^{2}) = 2x\).
For the \(-x\cos y\) term: Use the product rule \((uv)^\prime=u^\prime v+uv^\prime\), where \(u = - x\) and \(v=\cos y\). So \(\frac{d}{dx}(-x\cos y)=-\cos y + x\sin y\frac{dy}{dx}\).
For the \(y^{3}\) term: Use the chain rule \(\frac{d}{dx}(y^{3}) = 3y^{2}\frac{dy}{dx}\).
So, \(2x-\cos y+x\sin y\frac{dy}{dx}+3y^{2}\frac{dy}{dx}=0\).

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

Group the terms with \(\frac{dy}{dx}\) together: \((x\sin y + 3y^{2})\frac{dy}{dx}=\cos y - 2x\).
Then \(\frac{dy}{dx}=\frac{\cos y - 2x}{x\sin y+3y^{2}}\).

Answer:

\(\frac{dy}{dx}=\frac{\cos y - 2x}{x\sin y + 3y^{2}}\)