Find the directional derivative of the function at the given point in the direction of the vector v. f(x,y)=xx2+y2,(1,2),u=&lt;3,5&gt;

Question

Find the directional derivative of the function at the given point in the direction of the vector v. f(x,y)=xx2+y2,(1,2),u=&amp;lt;3,5&amp;gt;

Brittany Patton · Accepted Answer

Knowning that the directional derivative is Du→f(x,y)=▽f(x,y)⋅u→ Where ▽f(x,y)=&amp;lt;fx(x,y),fy(x,y)&amp;gt; and u→ is a unit vector. f(x,y)=xx2+y2, (1,2), v→=&amp;lt;3,5&amp;gt; Using the quotient rule to find the partial derivative with respect to x fx=(1)(x2+y2)−(x)(2x)(x2+y2)2 fx=y2−x2(x2+y2)2 Rewriting f(x,y) and using the chain rule to acquire the partial derivative with respect to y. f(x,y)=x(x2+y2)−1 fy=(−1)(x)(x2+y2)−2(2y) fy=−2xy(x2+y2)2 Plugging in the given point into the partial derivatives gives us ▽f(1,2)=&amp;lt;325,−425&amp;gt; fx=(2)2−(1)2((1)2+(2)2)2=325 fy=−2(1)(2)((1)2+(2)2)=−425 Need a unit vector, so have to divide the components of the given vector by its length |v→|=32+52=34 So u→=v→|v→|=&amp;lt;334,534&amp;gt; Dotting the gradient vector with our newly found unit vector

Find the directional derivative of the function at the given point in the direct

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