Question

partial derivative with respect to y- fy(x,y)= sqrt 4x^2 - 5xy^3 + 12y^2x^2??

Answer 1

$$f(x,y)=\sqrt{4x^2-5xy^3-12y^2x^2}$$
consider x as a constant and do the normal differentiation wrt y

Answer 2

i haven't done this kind of math in a long time... can u show me the steps??

Answer 3

Are you familiar with single variable differentiation?

Answer 4

uh sorta... it's f(x+h) - f(x)/ h right?

Answer 5

that's the basis,
I meant the proven things that we normally apply directly for differentiation.
such as
$$\frac{d x^n}{dx}=nx^{n-1}$$
and some other rules such as chain rule
How about them?

Answer 6

You should be quite comfortable with single variable calculus to handle multi-variate differentiation. Other wise I recommend that you should first remind the things in single variable calculus

Answer 7

ok thank you i'll go from there!

Answer 8

Since you've asked, I'll show the steps for your question
since the differentiation is respect to y we can consider x as a constant thus the expression is like,
$$f(y)=\sqrt{a + by^3 + cy^2}$$
where a,b,c are considered constants
(a=4x^2, b=-5x, c=-12x^2)

if we take $$v=a+by^3+cy^2$$
$$f(v)=\sqrt{v}$$
Now we have to do the differentiation. For that we use chain rule i.e.
$$\frac{du}{dy}=\frac{du}{dv}\frac{dv}{dy}$$

here we substitute u with f,
$$\frac{df}{dy}=\frac{df}{dv}\frac{dv}{dy}$$
This means that we need to find df/dv and dv/dy to find df/dy
$$\frac{df}{dv}=\frac{d\sqrt{v}}{dv}=\frac{1}{2\sqrt{v}}$$
$$\frac{dv}{dy}=\frac{d(a+by^3+cy^2)}{dy}=3by^2+2cy$$
so,
$$\frac{df}{dy}=\frac{1}{2\sqrt{v}}(3by^2+2cy)=\frac{3by^2+2cy}{2(a+by^3+cy^2)}=\frac{3(-5x)y^2+2(-12x^2)y}{2(4x^2-5xy^3-12y^2x^2)}$$

Answer 9

the last line should be changed to,
$$\frac{df}{dy}=\frac{3by^2+2cy}{2\sqrt{a+by^3+cy^2}}=\frac{3(-5x)y^2+2(-13x^2)y}{2\sqrt{4x^2-5xy^3-12x^2y^2}}$$
I forgot the square root :(