Differentiation and integration?

Question

1.x^2+xy^2-5=0
Find the value of dy/dx at the point (1,2)

2. Also i had trouble integrating (x+3)/x^2

thanks very much for any help.

Answer 1

1. we have to differentiate each term and then solve for dy/dx

i'll differentiate each term seperatley for you:
x^2 --> 2x

xy^2 --> we have to do product rule which states derivative of first times 2nd + first times derivative of 2nd.

(1)(y^2) + (x)(2y)(dy/dx)

5 --> 0 derivative of a number is always zero.

so we know have

2x + y^2 + 2xy(dy/dx) = 0 subtract the 2x and the y^2
2xy(dy/dx) = -2x - y^2 divide both sides by 2xy
dy/dx = (-2x - y^2)/(2xy)

Now plug in x = 1, and y = 2
dy/dx = (-2 - 4)/(2*1*2)
dy/dx = -6/4 = -3/2

2. split the fraction into the following:

Integral(x/x^2) + Integral(3/x^2) Simplifying
Integral(1/x) + Integral(3x^-2)
Integral(1/x) = ln (x) put x in absolute value
Integral(3x^(-2)) add one to the exponent and then divide the leading coefficient by that term --> -3x^-1 = -3/x

So the answer becomes: ln|x| - 3/x + C

Answer 2

x² +xy² - 5 = 0

2x +2xy dydx + y² = 0

2 + 4dy/dx + 4 = 0

dy/dx = -3/2

2. Integral of (x + 3)/x² =

(x + 3)x^-2 = x^-1 + 3x^-2

= 1/x + 3x^-2

Integral = ln x - 3x^-1 + C

Answer 3

#1: First, differentiate:

d/dx (x² + xy² - 5) = d/dx 0 = 0
2x + d/dx (xy²) = 0

Use the product and chain rules:

2x + y² + 2xy dy/dx = 0

Let x = 1 and y = 2:

2 + 4 + 4 dy/dx = 0

Solve for dy/dx:

4 dy/dx = -6
dy/dx = -3/2

#2: Just do it term by term:

∫(x+3)/x² dx
∫1/x + 3/x² dx
ln |x| - 3/x + C