Explain why the function is differentiable at the given point. Then find the linearization L(x,y) of the function at that point. f(x,y)=y+sin(x/y), (0,3)

2021-10-07

Explain why the function is differentiable at the given point. Then find the linearization L(x,y) of the function at  that point.

\(f(x,y)=y+\sin(\frac{x}{y}), (0,3)\)

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Answered 2021-10-21 Author has 2252 answers

\(f(x,y)=y+\sin(\frac{x}{y})\)

By the Sum Rule, the derivative of \(y+\sin(\frac{x}{y})\)  with respect to \(y\) is \(\frac{d}{dy}[y]+\frac{d}{dy}[\sin(\frac{x}{y})]\)

Differentiate using the Power Rule which states that \(\frac{d}{dy}[y^n]\) is \(ny^{-1}\) where \(n=1\)

\(1+\frac{d}{dy}[\sin(\frac{x}{y})]\)

Evaluate \(\frac{d}{dy}[\sin(\frac{x}{y})]\)

Differentiate using the chain rule, which states that \(\frac{d}{dy}[f(g(y)) ]\) is \(f'(g(y))g'(y)\) where \(f(y)=\sin(y)\) and \(g(y)=\frac{x}{y}\)

\(1+\cos(\frac{x}{y}) \frac{d}{dy}[\frac{x}{y}]\)

Since x is constant with respect to y, the derivative of \(\frac{x}{y}\) with respect to y is \(x\frac{d}{dy}[\frac{1}{y}]\)

\(1+\cos(\frac{x}{y})(x\frac{d}{dy}[\frac{1}{y}])\)

Rewrite \(\frac{1}{y}\) as \(y^{-1}\)

\(1+\cos(\frac{x}{y})(x\frac{d}{dy}[y^{-1}])\)

 Differentiate using the Power Rule which states that \(\frac{d}{dy}[y^n]\) is \(ny^{-1}\) where \(n=-1\)

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

Simplify.

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

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