Final answer:
To find the derivative of the function f(x) = 2 / sqrt(x) - 3 / x^3, we can use the quotient rule. The derivative is (-x^3 + 12 * sqrt(x)) / (2 * x^8 * sqrt(x)). The equation of the tangent line at x = 1 can be found using the point-slope form of a line.
Step-by-step explanation:
To find the derivative of the function f(x) = 2 / sqrt(x) - 3 / x^3, we can use the quotient rule. The quotient rule states that if we have a function in the form f(x) = g(x) / h(x), then the derivative f'(x) can be calculated as (g'(x) * h(x) - g(x) * h'(x)) / (h(x)^2). In this case, g(x) = 2 / sqrt(x) and h(x) = x^3. Taking the derivatives of g(x) and h(x), we get g'(x) = -1 / (2 * sqrt(x)^3) and h'(x) = -3 / x^4. Plugging these values into the quotient rule formula, we get f'(x) = (-1 / (2 * sqrt(x)^3) * x^3 - 2 / sqrt(x) * (-3 / x^4)) / (x^3)^2.
Simplifying this expression, we get f'(x) = (-x^3 / (2 * sqrt(x)^3) + 6 / (x^2 * sqrt(x))) / x^6 = (-x^3 + 12 * sqrt(x)) / (2 * x^8 * sqrt(x)).
To find the equation of the tangent line to the graph of f(x) at x = 1, we can use the point-slope form of a line. The point-slope form states that if we have a point (x₁, y₁) on the line and the slope m, then the equation of the line is given by y - y₁ = m(x - x₁). In this case, we have the point (1, f(1)) on the line and the slope f'(1) at x = 1. Plugging these values into the point-slope form, we get y - f(1) = f'(1)(x - 1). Simplifying this equation will give us the equation of the tangent line.