Math Problem Statement

a) Find the integral of (10x - 20) / (x^2 - 4x + 5) dx. b) If x^2y^2 + xy^2 = 4x, evaluate dy/dx when x = 1 and y = 1. c) i) Given that y = x^(3x), find dy/dx. ii) Given that y = tan^4(x), find dy/dx.

Solution

Let's solve the questions shown in the image step by step:

Question 2 (a)

We are asked to find the integral of:

10x20x24x+5dx\int \frac{10x - 20}{x^2 - 4x + 5} \, dx

Step-by-step solution:

  1. First, notice that the numerator 10x2010x - 20 is the derivative of the denominator x24x+5x^2 - 4x + 5, which suggests that a substitution method will work here.

  2. Let: u=x24x+5u = x^2 - 4x + 5 Then, differentiate uu with respect to xx: dudx=2x4\frac{du}{dx} = 2x - 4

    Now, rewrite the integral: 10x20x24x+5dx=10(x2)udx\int \frac{10x - 20}{x^2 - 4x + 5} \, dx = \int \frac{10(x - 2)}{u} \, dx

    But since 10(x2)10(x - 2) is a constant multiple of dudx\frac{du}{dx}, the integral simplifies: duu=lnu+C=lnx24x+5+C\int \frac{du}{u} = \ln|u| + C = \ln|x^2 - 4x + 5| + C So, the final result is: 10x20x24x+5dx=lnx24x+5+C\int \frac{10x - 20}{x^2 - 4x + 5} \, dx = \ln|x^2 - 4x + 5| + C


Question 2 (b)

We are given: x2y2+xy2=4xx^2y^2 + xy^2 = 4x and are asked to find dydx\frac{dy}{dx} when x=1x = 1 and y=1y = 1.

Step-by-step solution:

  1. Differentiate both sides of the equation implicitly with respect to xx: ddx(x2y2+xy2)=ddx(4x)\frac{d}{dx}(x^2 y^2 + xy^2) = \frac{d}{dx}(4x)

  2. Use the product rule and chain rule: ddx(x2y2)=2xy2+x2(2ydydx)\frac{d}{dx}(x^2 y^2) = 2x y^2 + x^2(2y \frac{dy}{dx}) and ddx(xy2)=y2+x(2ydydx)\frac{d}{dx}(xy^2) = y^2 + x(2y \frac{dy}{dx})

    So, the derivative becomes: 2xy2+2x2ydydx+y2+2xydydx=42x y^2 + 2x^2 y \frac{dy}{dx} + y^2 + 2xy \frac{dy}{dx} = 4

  3. Substitute x=1x = 1 and y=1y = 1 into the equation: 2(1)(1)2+2(1)2(1)dydx+(1)2+2(1)(1)dydx=42(1)(1)^2 + 2(1)^2(1) \frac{dy}{dx} + (1)^2 + 2(1)(1) \frac{dy}{dx} = 4 Simplifying: 2+2dydx+1+2dydx=42 + 2 \frac{dy}{dx} + 1 + 2 \frac{dy}{dx} = 4 3+4dydx=43 + 4 \frac{dy}{dx} = 4 4dydx=14 \frac{dy}{dx} = 1 dydx=14\frac{dy}{dx} = \frac{1}{4}


Question 2 (c)

Part (i)

We are given y=x3xy = x^{3x} and need to find dydx\frac{dy}{dx}.

  1. Take the natural logarithm of both sides to simplify: lny=ln(x3x)=3xlnx\ln y = \ln(x^{3x}) = 3x \ln x

  2. Differentiate both sides using implicit differentiation: 1ydydx=3lnx+3\frac{1}{y} \frac{dy}{dx} = 3 \ln x + 3

  3. Multiply through by y=x3xy = x^{3x}: dydx=x3x(3lnx+3)\frac{dy}{dx} = x^{3x} (3 \ln x + 3)


Part (ii)

We are given y=tan4(x)y = \tan^4(x) and need to find dydx\frac{dy}{dx}.

  1. Use the chain rule. First, rewrite the function: y=(tanx)4y = (\tan x)^4

  2. Differentiate using the chain rule: dydx=4(tanx)3sec2x\frac{dy}{dx} = 4(\tan x)^3 \cdot \sec^2 x


Do you want any further details or have any questions on any of the steps?

Here are 5 related questions to expand your understanding:

  1. How do you identify when substitution is useful in integration?
  2. What are some common applications of implicit differentiation in physics or engineering?
  3. Can you generalize how to find the derivative of exponential functions like y=xnxy = x^{nx}?
  4. What is the difference between differentiating algebraic functions and trigonometric functions?
  5. How can you apply the chain rule when differentiating composite functions?

Tip: In implicit differentiation, always apply the chain rule to terms involving yy, as they are treated as functions of xx.

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Math Problem Analysis

Mathematical Concepts

Integration
Implicit Differentiation
Differentiation of Exponential Functions
Trigonometric Differentiation

Formulas

∫ (f'(x)/f(x)) dx = ln|f(x)| + C
Product rule: d(uv)/dx = u'v + uv'
Chain rule for differentiation
Derivative of a^b where a and b are functions of x

Theorems

Fundamental Theorem of Calculus
Implicit Differentiation
Chain Rule in Differentiation

Suitable Grade Level

Grades 11-12