Question

Find Duf at P

Answer 1

\(\large f(x_0,y_0)=(1+xy)^{\frac{3}{2}}\)

\(\large P(3,1)\)

\(\large u=\frac{1}{\sqrt{2}}i+\frac{1}{\sqrt{2}}j\)

Answer 2

So, i used this formula
\(\large D_uf(x_0,y_0)=\frac{d}{ds}[f(x_0+su_1,y_0+su_2)]\)

Answer 3

and got
\(\large D_uf(3,1)=\frac{d}{ds}[f(3+\frac{s}{\sqrt{2}},1+\frac{s}{\sqrt{2}})]\)

Answer 4

But since
\(\large f(3+\frac{s}{\sqrt{2}},1+\frac{s}{\sqrt{2}})=(1+(3+\frac{s}{\sqrt{2}})(1+\frac{s}{\sqrt{2}}))^{\frac{3}{2}}\)

Answer 5

Then we will have

\(\large D_uf(3,1)=\frac{d}{ds}[(1+(3+\frac{s}{\sqrt{2}})(1+\frac{s}{\sqrt{2}}))^{\frac{3}{2}}]=\frac{d}{ds}[\Large \frac{(s^2+4\sqrt{2}s+8)^{3/2}}{2\sqrt{2}}]\)

Answer 6

@SolomonZelman

Answer 7

Is this right?

Answer 8

@jim_thompson5910

Answer 9

Taking the derivative and plug s=0
\(\large D_uf(3,1)=\frac{d}{ds}[\Large \frac{(s^2+4\sqrt{2}s+8)^{3/2}}{2\sqrt{2}}]_{s=0}=\frac{3\sqrt{2}\sqrt{\sqrt{2}(0)+8}+2(0)}{\sqrt{2}}\)
\(=\large 6\sqrt{2}\)

Answer 10

NVM i got it!