Question

Find:

will post equation below

Answer 1

\[f _{x}(1,0) for f(x,y)=\frac{ xe ^{\sin(x ^{2}y)} }{ (x ^{2}+y ^{2})^{3/2} }\]

Answer 2

that looks messy. I'd rather quit ^^

Answer 3

haha me too

the answer is -2 lol i just need to know how to do it! stupid final

Answer 4

So we have to start by finding the partial of f with respect to x?
Hmmm ok.

Answer 5

The numerator AND denominator contain x's, so we need to apply the quotient rule to start, yes?

Answer 6

im guna go with yes

Answer 7

lol! XD

Answer 8

\[\Large \bf f(x,y)=\frac{ xe ^{\sin(x ^{2}y)} }{ (x ^{2}+y ^{2})^{3/2} }\]

\[\large\bf f_x=\frac{\color{#DD4747 }{\left[x e^{\sin(x^2y)}\right]_x}(x^2+y^2)^{3/2}-(x e^{\sin(x^2y)})\color{#DD4747 }{\left[(x^2+y^2)^{3/2}\right]_x}}{\left[(x^2+y^2)^{3/2}\right]^2}\]

Answer 9

So here is our quotient rule setup.
We need to take the derivative of the pink portions ( partial derivatives ).

Answer 10

For the first pink part, it looks like we have to apply the product rule.

Answer 11

make sense...still sucks lol

Answer 12

\[\Large \left[x e^{\sin(x^2y)}\right]_x\quad=\quad (x)_xe^{\sin(x^2y)}+x \left[e^{\sin(x^2y)}\right]_x\]So we have the product rule within our quotient.
Hmm that second term is going to be a little tricky.

Answer 13

Treat it as a normal derivative until you get to the innermost function.
So you have to take the derivative of the exponential, and then apply the chain rule a couple times, yes?

Answer 14

where is the small x coming from?

Answer 15

and i have no idea what u just did

Answer 16

I'm using the small x as the derivative operator.
Similar to how you might see primes when you setup product rule normally.\[\Large (x e^x)'\quad=\quad (x)'e^x+x(e^x)'\]If it was a partial derivative, we could write it like this.\[\Large (x e^x)_x\quad=\quad (x)_xe^x+x(e^x)_x\]

Answer 17

When we are taking partial derivative with respect to x we treat y as a constant.
And they want you to evaluate the partial derivative at (1,0)
Since y is a constant how about putting y = 0 and simplifying the expression BEFORE taking the partial derivative?

Answer 18

Oh that's a nifty idea ^^ hehe

Answer 19

im super confused

Answer 20

Treat y as a constant and set y = 0 in the function.
Then find the derivative with respect to x and evaluate it at x = 1

Answer 21

im guna need someone to write it out so i can see it..im a visual learner and im studying for my final so just seeing the steps is going to be the best thing

Answer 22

\[\Large \bf f(x,y)=\frac{ xe ^{\sin(x ^{2}y)} }{ (x ^{2}+y ^{2})^{3/2} }\]
\[\Large \bf f(x,0)=\frac{ xe ^{\sin(x ^{2}\cdot0)} }{ (x ^{2}+0 ^{2})^{3/2} }\]
Hmm that'll simplify down really nicely :O

Answer 23

See how we plugged in y=0?
Can you see how it will simplify?

Answer 24

yup thank you

Answer 25

f(x,0) = x * e^(sin(0)) / x^3
= x * e^(0) / x^3
= x / x^3
= x^(-2)

f'(x,0) = -2x^(-3)
f'(1,0) = -2(1)^(-3) = -2

Answer 26

beautiful thank you both that was very helpful

Answer 27

can you do that in every equation like that?

Answer 28

You are welcome.
And thank you @zepdrix.