Math Problem Statement

Differentiate the given mathematical functions.

Solution

The image contains 8 mathematical expressions where each f(x)f(x) appears to be a function involving different algebraic, exponential, logarithmic, or trigonometric forms. I'll provide solutions by differentiating each function f(x)f(x).

(1) f(x)=73x3x2+3xf(x) = 7^{3x^3 - x^2 + 3x}

To differentiate this, let's denote: y=73x3x2+3xy = 7^{3x^3 - x^2 + 3x} Taking the natural logarithm on both sides, we get: lny=(3x3x2+3x)ln7\ln y = (3x^3 - x^2 + 3x) \ln 7 Now, differentiate implicitly: 1ydydx=(9x22x+3)ln7\frac{1}{y} \frac{dy}{dx} = (9x^2 - 2x + 3) \ln 7 Thus: f(x)=y(9x22x+3)ln7=73x3x2+3x(9x22x+3)ln7f'(x) = y(9x^2 - 2x + 3) \ln 7 = 7^{3x^3 - x^2 + 3x}(9x^2 - 2x + 3) \ln 7

(2) f(x)=e5x33x2f(x) = e^{5x^3 - 3x^2}

To differentiate: f(x)=ddx(e5x33x2)=e5x33x2(15x26x)f'(x) = \frac{d}{dx} \left( e^{5x^3 - 3x^2} \right) = e^{5x^3 - 3x^2} \cdot (15x^2 - 6x) f(x)=(15x26x)e5x33x2f'(x) = (15x^2 - 6x) e^{5x^3 - 3x^2}

(3) f(x)=e2x+1f(x) = e^{\sqrt{2x + 1}}

Let u=2x+1u = \sqrt{2x + 1}, then: f(x)=e2x+1ddx(2x+1)f'(x) = e^{\sqrt{2x + 1}} \cdot \frac{d}{dx}(\sqrt{2x + 1}) =e2x+1122x+12=e2x+12x+1= e^{\sqrt{2x + 1}} \cdot \frac{1}{2\sqrt{2x + 1}} \cdot 2 = \frac{e^{\sqrt{2x + 1}}}{\sqrt{2x + 1}}

(4) f(x)=4log(2x53x3+x)f(x) = 4 \log(2x^5 - 3x^3 + x)

Using the chain rule: f(x)=412x53x3+x(10x49x2+1)f'(x) = 4 \cdot \frac{1}{2x^5 - 3x^3 + x} \cdot (10x^4 - 9x^2 + 1) f(x)=4(10x49x2+1)2x53x3+xf'(x) = \frac{4(10x^4 - 9x^2 + 1)}{2x^5 - 3x^3 + x}

(5) f(x)=(5x2+3x)log(7x+3)f(x) = (5x^2 + 3x) \log(7x + 3)

Let u=5x2+3xu = 5x^2 + 3x and v=log(7x+3)v = \log(7x + 3). Using the product rule: f(x)=uv+uv=(10x+3)log(7x+3)+(5x2+3x)77x+3f'(x) = u'v + uv' = (10x + 3) \log(7x + 3) + (5x^2 + 3x) \frac{7}{7x + 3}

(6) f(x)=(2x34x)e2x5f(x) = (2x^3 - 4x) e^{2x^5}

Using the product rule: f(x)=(6x24)e2x5+(2x34x)e2x510x4f'(x) = (6x^2 - 4) e^{2x^5} + (2x^3 - 4x) \cdot e^{2x^5} \cdot 10x^4

(7) f(x)=sin(4x32x)f(x) = \sin(4x^3 - 2x)

f(x)=cos(4x32x)(12x22)f'(x) = \cos(4x^3 - 2x) \cdot (12x^2 - 2)

(8) f(x)=ln(1x2+3x5)f(x) = \ln\left(\frac{1}{\sqrt[5]{x^2 + 3x}}\right)

Simplify first: f(x)=15ln(x2+3x)f(x) = -\frac{1}{5} \ln(x^2 + 3x) Differentiate: f(x)=152x+3x2+3xf'(x) = -\frac{1}{5} \cdot \frac{2x + 3}{x^2 + 3x}

These are the derivatives for each problem. Would you like further details or explanations on any specific function?

Follow-up Questions:

  1. Can you simplify f(x)f'(x) further in example (6)?
  2. How would you find the second derivative of f(x)f(x) in problem (7)?
  3. What is the interpretation of the derivative result in problem (5)?
  4. Could f(x)f'(x) be zero for any xx in the given functions?
  5. What would happen if you changed the base of the logarithmic functions in problems (4) and (5)?

Tip:

Always simplify the function as much as possible before differentiating; this can often make the process easier and reduce errors.

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Math Problem Analysis

Mathematical Concepts

Differentiation
Exponential functions
Logarithmic functions
Trigonometric functions
Product rule
Chain rule

Formulas

d/dx (e^x) = e^x
d/dx (ln(x)) = 1/x
Product rule: (uv)' = u'v + uv'
Chain rule: (f(g(x)))' = f'(g(x))g'(x)

Theorems

Fundamental Theorem of Calculus
Exponential Differentiation Rule
Logarithmic Differentiation Rule
Product Rule of Differentiation
Chain Rule of Differentiation

Suitable Grade Level

Grade 11-12 (Advanced Calculus)