Math Problem Statement

Find the derivatives of each of the following functions: (a) f(x) = x^2 * e^x (b) g(u) = u^3(2u^2 + e^u - 1) (c) h(x) = (4e^x - 1) / (x^2 + 1) (d) j(x) = ((x^2 + 1) / x) * e^x (e) k(x) = 3x^2 - 2 / x^7

Solution

Let's solve each part one by one.

(a) f(x)=x2exf(x) = x^2 e^x

We need to differentiate this using the product rule, since we have the product of two functions: x2x^2 and exe^x.

The product rule is: (fg)=fg+fg(fg)' = f'g + fg'

Let:

  • f1(x)=x2f_1(x) = x^2 and f2(x)=exf_2(x) = e^x

Then:

  • f1(x)=2xf_1'(x) = 2x
  • f2(x)=exf_2'(x) = e^x

Using the product rule: f(x)=(2x)ex+(x2)(ex)=2xex+x2exf'(x) = (2x)e^x + (x^2)(e^x) = 2x e^x + x^2 e^x

(b) g(u)=u3(2u2+eu1)g(u) = u^3(2u^2 + e^u - 1)

Here we again have a product of two functions, u3u^3 and (2u2+eu1)(2u^2 + e^u - 1). So, we use the product rule again.

Let:

  • g1(u)=u3g_1(u) = u^3
  • g2(u)=2u2+eu1g_2(u) = 2u^2 + e^u - 1

Now:

  • g1(u)=3u2g_1'(u) = 3u^2
  • g2(u)=4u+eug_2'(u) = 4u + e^u

Using the product rule: g(u)=(3u2)(2u2+eu1)+(u3)(4u+eu)g'(u) = (3u^2)(2u^2 + e^u - 1) + (u^3)(4u + e^u)

(c) h(x)=4ex1x2+1h(x) = \frac{4e^x - 1}{x^2 + 1}

Here, we need to apply the quotient rule, since this is a ratio of two functions. The quotient rule is: (fg)=fgfgg2\left( \frac{f}{g} \right)' = \frac{f'g - fg'}{g^2}

Let:

  • h1(x)=4ex1h_1(x) = 4e^x - 1
  • h2(x)=x2+1h_2(x) = x^2 + 1

Now:

  • h1(x)=4exh_1'(x) = 4e^x
  • h2(x)=2xh_2'(x) = 2x

Using the quotient rule: h(x)=(4ex)(x2+1)(4ex1)(2x)(x2+1)2h'(x) = \frac{(4e^x)(x^2 + 1) - (4e^x - 1)(2x)}{(x^2 + 1)^2}

(d) j(x)=(x2+1x)exj(x) = \left( \frac{x^2 + 1}{x} \right) e^x

Here we have both a quotient and a product, so we'll start by simplifying the quotient, if possible, then apply the product rule.

First, simplify the expression: j(x)=(x2+1x)ex=(x+1x)exj(x) = \left( \frac{x^2 + 1}{x} \right) e^x = \left( x + \frac{1}{x} \right) e^x

Now, let:

  • j1(x)=x+1xj_1(x) = x + \frac{1}{x}
  • j2(x)=exj_2(x) = e^x

Now differentiate each:

  • j1(x)=11x2j_1'(x) = 1 - \frac{1}{x^2}
  • j2(x)=exj_2'(x) = e^x

Using the product rule: j(x)=(11x2)ex+(x+1x)exj'(x) = \left(1 - \frac{1}{x^2}\right) e^x + \left(x + \frac{1}{x}\right) e^x

(e) k(x)=3x22x7k(x) = 3x^2 - \frac{2}{x^{7}}

Here, we can use basic power rule differentiation.

The power rule states: ddx(xn)=nxn1\frac{d}{dx}(x^n) = nx^{n-1}

For each term:

  • ddx(3x2)=6x\frac{d}{dx}(3x^2) = 6x
  • ddx(2x7)=2(7)x8=14x8\frac{d}{dx}\left(-\frac{2}{x^7}\right) = -2 \cdot (-7) x^{-8} = 14x^{-8}

So, the derivative is: k(x)=6x+14x8k'(x) = 6x + 14x^{-8}


Would you like further details or explanations on any step?

Here are 5 follow-up questions you might consider:

  1. How does the product rule help in differentiating composite functions?
  2. When is the quotient rule necessary, and can it be avoided by simplifying functions first?
  3. What is the geometric interpretation of derivatives of these functions?
  4. Could implicit differentiation be applied in any of these problems?
  5. How do higher-order derivatives behave for exponential and polynomial functions?

Tip: Always try to simplify the function before applying differentiation rules to reduce complexity.

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Math Problem Analysis

Mathematical Concepts

Differentiation
Product Rule
Quotient Rule
Exponential Functions
Power Rule

Formulas

Product Rule: (fg)' = f'g + fg'
Quotient Rule: (f/g)' = (f'g - fg')/g^2
Power Rule: d/dx(x^n) = nx^{n-1}

Theorems

Differentiation Theorems
Rules of Derivatives

Suitable Grade Level

Grades 11-12 or early college calculus