What are the first and second derivatives of y = (1/x^2)-(1/(x-2)^2)?
I will do the first...
y'= -2/x^3 - dy/dx (1/(x-2)^2)=
= -2/x^3 +2/(x-1)^3
check that.
Yes, that is what I got as well. So would we have to go further, or does that suffice? Thank you.
But isn't it supposed to be 2/x^3 - 2/(x-2)^3?
To find the first and second derivatives of the given function, y = (1/x^2) - (1/(x-2)^2), we can apply the quotient rule and power rule. Let's start by finding the first derivative:
Step 1: Rewrite the function using negative exponents:
y = x^(-2) - (x-2)^(-2)
Step 2: Apply the power rule:
dy/dx = -2x^(-3) + 2(x-2)^(-3)
Step 3: Simplify the expression:
dy/dx = -2/x^3 + 2/(x-2)^3
Now, let's find the second derivative:
Step 1: Differentiate the first derivative with respect to x:
d^2y/dx^2 = d/dx (-2/x^3 + 2/(x-2)^3)
Step 2: Apply the power rule and chain rule:
d^2y/dx^2 = 6/x^4 - 6(x-2)^(-4)
Therefore, the first derivative is dy/dx = -2/x^3 + 2/(x-2)^3, and the second derivative is d^2y/dx^2 = 6/x^4 - 6(x-2)^(-4).