Let f(x)=(x^(2)-2)/(x-1). f^(')(x)=

Question

Let  f(x)=(x^(2)-2)/(x-1).  f^(')(x)=

Bytelearn · Accepted Answer

Apply Quotient Rule: To find the derivative of the function f(x) = (x^2 - 2) / (x - 1), we will use the quotient rule. The quotient rule states that if we have a function that is the quotient of two functions, u(x) / v(x), then its derivative f'(x) is given by (v(x) * u'(x) - u(x) * v'(x)) / (v(x))^2.Identify u(x) and v(x): Let's identify u(x) and v(x) for our function. Here, u(x) = x^2 - 2 and v(x) = x - 1. We will need to find the derivatives of both u(x) and v(x), which are u'(x) and v'(x), respectively.Find u'(x) and v'(x): The derivative of u(x) = x^2 - 2 with respect to x is u'(x) = 2x, since the derivative of x^2 is 2x and the derivative of a constant is 0.Derivative of u(x): The derivative of v(x) = x - 1 with respect to x is v'(x) = 1, since the derivative of x is 1 and the derivative of a constant is 0.Derivative of v(x): Now we apply the quotient rule. We have u'(x) = 2x and v'(x) = 1, so we plug these into the quotient rule formula to get f'(x) = ((x - 1) * (2x) - (x^2 - 2) * (1)) / (x - 1)^2.Plug into Quotient Rule: Let's simplify the numerator of the derivative. We have f'(x) = (2x^2 - 2x - (x^2 - 2)) / (x - 1)^2 = (2x^2 - 2x - x^2 + 2) / (x - 1)^2.Simplify Numerator: Further simplifying the numerator, we combine like terms to get f'(x) = (x^2 - 2x + 2) / (x - 1)^2.Combine Like Terms: The derivative f'(x) is now in its simplest form, so we have completed the problem. The derivative of f(x) = (x^2 - 2) / (x - 1) is f'(x) = (x^2 - 2x + 2) / (x - 1)^2.

Let $f(x)=\frac{x^{2}-2}{x-1}$ . $\newline$ $f^{\prime}(x)=$

Full solution

More problems from Find the roots of factored polynomials

Related topics

Let f(x)=x2−2x−1 f(x)=\frac{x^{2}-2}{x-1} f(x)=x−1x2−2​.\newlinef′(x)= f^{\prime}(x)= f′(x)=

Let $f(x)=\frac{x^{2}-2}{x-1}$ . $\newline$ $f^{\prime}(x)=$