find the 1st derivative of f(x) = x/(x-1)2
Is the 1st derivative f'(x)= -x-1/(x-1)^3
yes , but you need brackets like this
(-1-x)/(x-1)^3
To find the derivative of a function, you can use the power rule and quotient rule. Let's find the derivative of the given function step by step:
Step 1: Rewrite the function using exponent notation:
f(x) = x/(x-1)^2
Step 2: Apply the quotient rule to find the derivative:
The quotient rule states that for a function h(x) = f(x)/g(x), the derivative is given by:
h'(x) = (f'(x) * g(x) - f(x) * g'(x))/(g(x))^2
In our case, f(x) = x and g(x) = (x-1)^2. Let's find the derivatives of f(x) and g(x):
f'(x) = 1 (the derivative of x is 1)
g'(x) = 2(x-1) * 1 (using the power rule, derivative of (x-1)^2 is 2(x-1))
Now, plug these derivatives into the quotient rule formula:
h'(x) = (1 * (x-1)^2 - x * (2(x-1) * 1))/((x-1)^2)^2
Step 3: Simplify the expression:
Expand the numerator and denominator:
h'(x) = ((x-1)^2 - 2x(x-1))/((x-1)^4)
Simplify further:
h'(x) = ((x^2 - 2x + 1) - (2x^2 - 2x))/((x-1)^4)
= (-x^2 + 1)/((x-1)^4)
Therefore, the first derivative of f(x) = x/(x-1)^2 is f'(x) = (-x^2 + 1)/((x-1)^4).