Trying to find second derivative in respect to y?

Question

I have the problem pretty much down but this last part is messing with me.

Taking the second derivative from e^x - x(1/y)

I'm getting -xy^-2, but the book is saying the answer is xy^-2. I don't understand how the negative sign in front of the x can just disappear. Am I missing something or does my book have a typo?

Answer 1

f (x , y) = e^x - x y^(-1)
f(x,y) = e^x + (-1) x y^(-1)
df /dy = 0 + (-1) (-1) x y^(-2)
df /dy = 1 x y^(-2)
df /dy = x / y ²

Answer 2

Well first off the first derivative of y^-1 is -y^-2 so i believe u mean take the first derivative. The part you are missing is that the intial negative infront of x(1/y) if this was a plus sign then yes your answer would be write. However, you know that a negative times a negative is positive so the answer is positive.

Answer 3

u = e^x - x/y
= e^x - x * y^(-1)

du/dy = 0 -x*(-1)y^(-2)
= xy^-2

You get a second negative sign from the exponent.
y = x^n
dy/dx = n*x^(n-1)
If n = -1 as is the case in this question, the minus sign would be multiplied by the minus sign that is already there. The product of two negative numbers is a positive number.

Answer 4

f(x,y)=xln(xy) here (x,y) is unmarried variable so differentiating with appreciate to (x,y) df(x,y)/d(x,y) = d/d(x,y) (xln(xy) (the place x is a few consistent) = x(a million)/(xy) . (a million) now lower back diff w.r.t (x,y) f ' (x,y) =x (-a million)/(xy)^2 .(a million) =-x/(xy)^2

Answer 5

z = e^x - x(1/y)
z= e^x - x(y^-1)
dz/dy = 0 -(-1)xy^-2
dz/dy = xy^-2
d^2z/dy^2 = -2xy^-3