Math Problem Statement

Solve

Solution

The image contains a mathematics question asking to find the derivatives of the following functions:

  1. y=(2x3+4x)5(3x2)y = \frac{(2x^3 + 4x)^5}{(3x^2)}
  2. y=2x3xy = 2x^{3x}
  3. y=sin(4x)x2y = \frac{\sin(4x)}{x^2}
  4. y=2x3+2x+5y = \sqrt{2x^3 + \sqrt{2x + 5}}

Let's solve each derivative step by step:

1. y=(2x3+4x)53x2y = \frac{(2x^3 + 4x)^5}{3x^2}

To differentiate this, we can use the quotient rule. The quotient rule is given by:

ddx(uv)=vuuvv2\frac{d}{dx} \left(\frac{u}{v}\right) = \frac{v \cdot u' - u \cdot v'}{v^2}

Where u=(2x3+4x)5u = (2x^3 + 4x)^5 and v=3x2v = 3x^2.

First, differentiate u=(2x3+4x)5u = (2x^3 + 4x)^5 using the chain rule:

u=5(2x3+4x)4(6x2+4)u' = 5(2x^3 + 4x)^4 \cdot (6x^2 + 4)

Now, differentiate v=3x2v = 3x^2:

v=6xv' = 6x

Applying the quotient rule:

dydx=3x2[5(2x3+4x)4(6x2+4)](2x3+4x)56x(3x2)2\frac{dy}{dx} = \frac{3x^2 \cdot [5(2x^3 + 4x)^4 \cdot (6x^2 + 4)] - (2x^3 + 4x)^5 \cdot 6x}{(3x^2)^2}

Simplify the expression.

2. y=2x3xy = 2x^{3x}

To differentiate this, it's easier to first take the natural logarithm of both sides to deal with the exponent:

lny=ln(2x3x)=ln2+ln(x3x)=ln2+3xlnx\ln y = \ln(2x^{3x}) = \ln 2 + \ln(x^{3x}) = \ln 2 + 3x \ln x

Now differentiate both sides with respect to xx:

1ydydx=3lnx+3\frac{1}{y} \cdot \frac{dy}{dx} = 3 \ln x + 3

Thus:

dydx=y(3lnx+3)=2x3x(3lnx+3)\frac{dy}{dx} = y \cdot (3 \ln x + 3) = 2x^{3x} \cdot (3 \ln x + 3)

3. y=sin(4x)x2y = \frac{\sin(4x)}{x^2}

This is another quotient rule problem where u=sin(4x)u = \sin(4x) and v=x2v = x^2.

Differentiate uu and vv:

u=4cos(4x)u' = 4\cos(4x) v=2xv' = 2x

Applying the quotient rule:

dydx=x24cos(4x)sin(4x)2xx4\frac{dy}{dx} = \frac{x^2 \cdot 4\cos(4x) - \sin(4x) \cdot 2x}{x^4}

Simplify the expression.

4. y=2x3+2x+5y = \sqrt{2x^3 + \sqrt{2x + 5}}

Use the chain rule to differentiate this function.

Let u=2x3+2x+5u = 2x^3 + \sqrt{2x + 5}, then y=u1/2y = u^{1/2}.

First, differentiate uu:

u=6x2+122x+52u' = 6x^2 + \frac{1}{2\sqrt{2x + 5}} \cdot 2

Now, differentiate yy:

dydx=12uu\frac{dy}{dx} = \frac{1}{2\sqrt{u}} \cdot u'

Substitute uu back in:

dydx=122x3+2x+5(6x2+12x+5)\frac{dy}{dx} = \frac{1}{2\sqrt{2x^3 + \sqrt{2x + 5}}} \cdot \left(6x^2 + \frac{1}{\sqrt{2x + 5}}\right)

These are the steps to find the derivatives for each given function. Would you like any further details or explanations?

Here are some related questions:

  1. How does the chain rule work for composite functions?
  2. What is the importance of the quotient rule in differentiation?
  3. Can the natural logarithm method be used for other exponential functions?
  4. How do trigonometric derivatives interact with the chain rule?
  5. Why do we simplify derivatives, and what are the benefits?

Tip: When solving derivative problems, always start by identifying which rules apply (product, quotient, chain, etc.), which makes the process more straightforward.

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Math Problem Analysis

Mathematical Concepts

Differentiation
Quotient Rule
Chain Rule
Exponential Functions
Trigonometric Functions

Formulas

Quotient Rule
Chain Rule

Theorems

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Suitable Grade Level

Advanced High School