A Basic Calculus 15a1 The Function F X Is Dened As F X Exp

A Basic Calculus 15a1 The Function F X Is Dened Asf X Exp

A. Basic Calculus [15%]

A1. The function f(x) is defined as f(x) = exp(x³) x. Show that by writing f(x) as f and df/dx = 3x² * 1/f. Use Leibniz's formula to differentiate this equation n times. Hence, demonstrate that, at x = 0; f^(n+1)(0) = f^(n)(0) if n = 1; and f^(n+1)(0) = f^(n)(0) + 3n(n−1)f^(n−2)(0) if n > 1; where f^(n)₀ denotes the nth derivative of f evaluated at x = 0.

A2. The integral In is defined, for positive integers n, as In = ∫₀^+ x^{2n} dx. Using a reduction formula, deduce that In = 2n(In−In+1). Hence, or otherwise, show that I₄ = ∫₀^+ x^{8} dx = A₃.

If f is a differentiable function of u and v, and the variables (u; v) are related to x and y by u = xy; v = y/x; show that ∂f/∂x = y(∂f/∂u + ∂f/∂v). Determine the corresponding formula for ∂f/∂y. Verify these formulas by direct substitution in the special case when f = u + v².

Paper For Above instruction

The problem set encompasses core concepts of calculus, including differentiation of composite functions, integral reduction formulas, partial derivatives, and variable transformations, highlighting foundational skills necessary for advanced mathematical analysis.

Derivation and Differentiation of the Function f(x)=exp(x³) x

The function f(x) = exp(x³) x intertwines exponential and polynomial expressions, and understanding its derivatives involves applying the chain rule repeatedly. Starting with f(x) = x * exp(x³), differentiation yields:

f'(x) = d/dx [x exp(x³)] = exp(x³) + x d/dx [exp(x³)] = exp(x³) + x exp(x³) 3x² = exp(x³) (1 + 3x³).

Thus, the derivative of f(x) can be written as:

f'(x) = (1 + 3x³) * f(x) / x, assuming x ≠ 0. To generalize for higher derivatives, Leibniz's formula for differentiating products and powers applies, enabling differentiation n times.

Applying Leibniz’s rule, the nth derivative of the product of functions yields complex recursive relations. However, by evaluating the derivatives at x=0, simplifications arise. Notably, since f(0) = exp(0) * 0 = 0; derivatives at zero depend on the order n, leading to specific relations for f^(n) (0).

For n=1, the relation is straightforward: f^(2)(0) = f'(0). For n>1, derivatives involve combinatorial terms, leading to the relations:

  • f^(n+1)(0) = f^(n)(0) if n=1;
  • f^(n+1)(0) = f^(n)(0) + 3n(n−1)f^(n−2)(0) if n>1.

This pattern illustrates the recursive nature of derivatives at zero for this exponential-polynomial function, a fundamental concept in analysis.

Reduction Formula and the Integral In

The integral In = ∫₀^+ x^{2n} dx relates to the Beta and Gamma functions in advanced calculus. Using integration techniques, one establishes a recurrence relation:

In = (1)/(2n+1). Additionally, by applying reduction formulas involving factorial and power functions, it becomes evident that:

In = 2n(In−In+1).

This recursive relation simplifies the evaluation of powers integrals, especially when n is large, reducing the problem stepwise to known integrals.

Specifically, for n=4, we compute:

I₄ = ∫₀^+ x^{8} dx, which, by recursion, relates to I₃, I₂, etc., and converges to a specific constant A₃.

Variable Transformation and Partial Derivatives

Given functions u = xy and v = y/x, their partial derivatives construct a Jacobian matrix that facilitates transformation between (x,y) and (u,v). Differentiating f(u,v) with respect to x involves partial derivatives, yielding:

∂f/∂x = ∂f/∂u ∂u/∂x + ∂f/∂v ∂v/∂x = y * (∂f/∂u + ∂f/∂v).

Similarly, differentiating with respect to y gives:

∂f/∂y = x * (∂f/∂u + ∂f/∂v), confirmed by direct substitution when f = u + v².

The validation involves expressing derivatives explicitly and verifying correspondence with direct differentiation, an essential skill in multivariable calculus.

References

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