 # 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.

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

• Questions are typically answered in as fast as 30 minutes

### Plainmath recommends

• Get a detailed answer even on the hardest topics.
• Ask an expert for a step-by-step guidance to learn to do it yourself. content_user

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

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.

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