Math Problem Statement

Differentiate the following expressions: (a) (i) x^2 * e^(3x + 2), (ii) cos(2x^3) / (3x), (b) Given x = 4 * sin(2y + 6), find dy/dx in terms of x.

Solution

Let's work through the given problems:

(a) Differentiate with respect to xx

(i) x2e3x+2x^2 e^{3x + 2}

To differentiate this, we will use the product rule and chain rule.

The product rule states: ddx(uv)=uv+uv\frac{d}{dx}(uv) = u'v + uv' where u=x2u = x^2 and v=e3x+2v = e^{3x+2}.

  • u=x2u = x^2, so u=2xu' = 2x.
  • v=e3x+2v = e^{3x+2}, and by the chain rule, v=e3x+23=3e3x+2v' = e^{3x+2} \cdot 3 = 3e^{3x+2}.

Using the product rule: ddx(x2e3x+2)=(2x)(e3x+2)+(x2)(3e3x+2)=2xe3x+2+3x2e3x+2\frac{d}{dx}(x^2 e^{3x+2}) = (2x)(e^{3x+2}) + (x^2)(3e^{3x+2}) = 2x e^{3x+2} + 3x^2 e^{3x+2} Thus, the derivative is: 2xe3x+2+3x2e3x+2\boxed{2x e^{3x+2} + 3x^2 e^{3x+2}}

(ii) cos(2x3)3x\frac{\cos(2x^3)}{3x}

This requires the quotient rule and chain rule.

The quotient rule is: ddx(uv)=vuuvv2\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v u' - u v'}{v^2} where u=cos(2x3)u = \cos(2x^3) and v=3xv = 3x.

  • u=cos(2x3)u = \cos(2x^3), and by the chain rule, u=sin(2x3)(6x2)=6x2sin(2x3)u' = -\sin(2x^3) \cdot (6x^2) = -6x^2 \sin(2x^3).
  • v=3xv = 3x, so v=3v' = 3.

Now applying the quotient rule: ddx(cos(2x3)3x)=(3x)(6x2sin(2x3))(cos(2x3))(3)(3x)2\frac{d}{dx} \left( \frac{\cos(2x^3)}{3x} \right) = \frac{(3x)(-6x^2 \sin(2x^3)) - (\cos(2x^3))(3)}{(3x)^2} Simplifying: =18x3sin(2x3)3cos(2x3)9x2= \frac{-18x^3 \sin(2x^3) - 3\cos(2x^3)}{9x^2} =6x3sin(2x3)cos(2x3)3x2= \frac{-6x^3 \sin(2x^3) - \cos(2x^3)}{3x^2} Thus, the derivative is: 6x3sin(2x3)cos(2x3)3x2\boxed{\frac{-6x^3 \sin(2x^3) - \cos(2x^3)}{3x^2}}

(b) Given x=4sin(2y+6)x = 4\sin(2y + 6), find dydx\frac{dy}{dx} in terms of xx.

We will use implicit differentiation. Differentiate both sides with respect to xx:

ddx(x)=ddx(4sin(2y+6))\frac{d}{dx}(x) = \frac{d}{dx}(4 \sin(2y + 6)) The left side is: ddx(x)=1\frac{d}{dx}(x) = 1

The right side requires the chain rule: ddx(4sin(2y+6))=4cos(2y+6)ddx(2y+6)=8cos(2y+6)dydx\frac{d}{dx}(4\sin(2y+6)) = 4\cos(2y+6) \cdot \frac{d}{dx}(2y+6) = 8\cos(2y+6) \cdot \frac{dy}{dx}

Thus, we have: 1=8cos(2y+6)dydx1 = 8\cos(2y+6) \cdot \frac{dy}{dx}

Solving for dydx\frac{dy}{dx}: dydx=18cos(2y+6)\frac{dy}{dx} = \frac{1}{8\cos(2y+6)}

To express dydx\frac{dy}{dx} in terms of xx, recall that x=4sin(2y+6)x = 4\sin(2y+6), so: sin(2y+6)=x4\sin(2y+6) = \frac{x}{4}

Using the identity sin2θ+cos2θ=1\sin^2 \theta + \cos^2 \theta = 1, we can solve for cos(2y+6)\cos(2y+6): cos(2y+6)=1sin2(2y+6)=1(x4)2=16x216=16x24\cos(2y+6) = \sqrt{1 - \sin^2(2y+6)} = \sqrt{1 - \left( \frac{x}{4} \right)^2} = \sqrt{\frac{16 - x^2}{16}} = \frac{\sqrt{16 - x^2}}{4}

Substitute into the equation for dydx\frac{dy}{dx}: dydx=1816x24=1216x2\frac{dy}{dx} = \frac{1}{8 \cdot \frac{\sqrt{16 - x^2}}{4}} = \frac{1}{2\sqrt{16 - x^2}}

Thus, the final result is: dydx=1216x2\boxed{\frac{dy}{dx} = \frac{1}{2\sqrt{16 - x^2}}}


Would you like any additional details or explanations? Here are some follow-up questions you might find helpful:

  1. How does the product rule work when differentiating products of functions?
  2. What is the chain rule, and how does it apply in these problems?
  3. Can you explain the quotient rule in further detail with an example?
  4. How does implicit differentiation differ from regular differentiation?
  5. What are the main trigonometric identities used in calculus?

Tip: When differentiating complex functions, always break them down step by step using the product, quotient, or chain rule to avoid mistakes!

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Math Problem Analysis

Mathematical Concepts

Differentiation
Product Rule
Quotient Rule
Chain Rule
Implicit Differentiation
Trigonometry

Formulas

Product Rule: (d/dx)(uv) = u'v + uv'
Quotient Rule: (d/dx)(u/v) = (v * u' - u * v') / v^2
Chain Rule: d/dx(f(g(x))) = f'(g(x)) * g'(x)
Implicit Differentiation: d/dx(f(x, y)) using dy/dx

Theorems

Chain Rule
Product Rule
Quotient Rule
Trigonometric Identities

Suitable Grade Level

Grades 11-12 or College Level