Q. 71 - Calculus

Question

Let $$f(x, y)$$ be a function of two variables. Prove that if $$f$$ is differentiable at the point $$(a, b)$$, then there is a unit vector, $$u$$, such that $$D_{u} f(a, b) \neq 1$$. (Hint: See Exercise 70.)

Step-by-Step Solution

Answer

It is proved that if $$f$$ is differentiable at the point $$(a, b)$$, then there is a unit vector, $$u$$, such that $$D_{u} f(a, b) \neq 1$$.

Step 1. Given Information

• Let $$f(x, y)$$ be a function of two variables.  
• The function $$f$$ is differentiable at the point $$(a, b)$$.

Step 2. Explanation

From Exercise 70, we know that, if $$f$$ is differentiable at the point $$(a,b)$$ and $$D_{u} f(a, b) = k$$, where $$k$$ is a constant for every unit vector $$u\epsilon R^{2}$$, then $$k = 0$$.

Here, by definition, we have, $$k=0$$.

$$\implies D_{u} f(a, b) \neq 1$$$$

Now, we find that $$f$$ is not differentiable at the point $$(a,b)$$ when $$D_{u} f(a, b)=1$$ for every unit vector $$u$$.

So, we can understand that function $$f$$ is differentiable at the point $$(a,b)$$ only when $$D_{u} f(a, b) \neq 1$$ for every unit vector $$u$$.

Hence, it is proved that if $$f$$ is differentiable at the point $$(a, b)$$, then there is a unit vector, $$u$$, such that $$D_{u} f(a, b) \neq 1$$.