What is the derivative of x - 4/x square.
Just the denominator is squared.
The derivative is
1+ 8/x^3
Just add the derivatives of the two separate terms: x and -4/x^2.
Wait I think I got it
The equation can be written as x-4x^-2
so d/dx(x)=1
and d/dx(-4x^-2)= -2(-4)(x^-2-1)
which is 8x^-3
d/dx(x-4/x^2)= 1+8x^-3
correct?
To find the derivative of the function, we can use the quotient rule. The quotient rule states that if we have a function in the form f(x)/g(x), the derivative is given by:
(f'(x)g(x) - f(x)g'(x)) / (g(x))^2
In this case, our function is (x - 4)/(x^2). To apply the quotient rule, we need to find the derivatives of both the numerator and denominator.
First, find the derivative of the numerator:
f'(x) = d/dx(x - 4) = 1
Next, find the derivative of the denominator:
g'(x) = d/dx(x^2) = 2x
Now, we can apply the quotient rule:
(f'(x)g(x) - f(x)g'(x)) / (g(x))^2
= (1 * x^2 - (x - 4) * 2x) / (x^2)^2
= (x^2 - 2x(x - 4)) / x^4
= (x^2 - 2x^2 + 8x) / x^4
= (-x^2 + 8x) / x^4
So, the derivative of (x - 4)/(x^2) with respect to x is (-x^2 + 8x) / x^4.