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Solve the following differentiation problems using the chain rule.

Mathematics

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Here are the solutions using the chain rule:

Basic Polynomial Composition:

  1. To find dydx\frac{dy}{dx} for y=(3x+2)5y = (3x + 2)^5:

    dydx=5(3x+2)5−1⋅ddx(3x+2)\frac{dy}{dx} = 5(3x + 2)^{5-1} \cdot \frac{d}{dx}(3x + 2) dydx=5(3x+2)4⋅3\frac{dy}{dx} = 5(3x + 2)^4 \cdot 3 15(3x+2)4\boxed{15(3x + 2)^4}
  2. To differentiate f(x)=(2x2−5x+1)4f(x) = (2x^2 - 5x + 1)^4:

    f′(x)=4(2x2−5x+1)4−1⋅ddx(2x2−5x+1)f'(x) = 4(2x^2 - 5x + 1)^{4-1} \cdot \frac{d}{dx}(2x^2 - 5x + 1) f′(x)=4(2x2−5x+1)3⋅(4x−5)f'(x) = 4(2x^2 - 5x + 1)^3 \cdot (4x - 5) 4(4x−5)(2x2−5x+1)3\boxed{4(4x - 5)(2x^2 - 5x + 1)^3}
  3. To find f′(x)f'(x) if f(x)=(4−x2)3f(x) = (4 - x^2)^3:

    f′(x)=3(4−x2)3−1⋅ddx(4−x2)f'(x) = 3(4 - x^2)^{3-1} \cdot \frac{d}{dx}(4 - x^2) f′(x)=3(4−x2)2⋅(−2x)f'(x) = 3(4 - x^2)^2 \cdot (-2x) −6x(4−x2)2\boxed{-6x(4 - x^2)^2}

Square Root and Fractional Powers:

  1. To differentiate y=5x2+3x−1=(5x2+3x−1)1/2y = \sqrt{5x^2 + 3x - 1} = (5x^2 + 3x - 1)^{1/2}:

    dydx=12(5x2+3x−1)12−1⋅ddx(5x2+3x−1)\frac{dy}{dx} = \frac{1}{2}(5x^2 + 3x - 1)^{\frac{1}{2}-1} \cdot \frac{d}{dx}(5x^2 + 3x - 1) dydx=12(5x2+3x−1)−1/2⋅(10x+3)\frac{dy}{dx} = \frac{1}{2}(5x^2 + 3x - 1)^{-1/2} \cdot (10x + 3) 10x+325x2+3x−1\boxed{\frac{10x + 3}{2\sqrt{5x^2 + 3x - 1}}}
  2. To find dydx\frac{dy}{dx} if y=(x3+2x)1/3y = (x^3 + 2x)^{1/3}:

    dydx=13(x3+2x)13−1⋅ddx(x3+2x)\frac{dy}{dx} = \frac{1}{3}(x^3 + 2x)^{\frac{1}{3}-1} \cdot \frac{d}{dx}(x^3 + 2x) dydx=13(x3+2x)−2/3⋅(3x2+2)\frac{dy}{dx} = \frac{1}{3}(x^3 + 2x)^{-2/3} \cdot (3x^2 + 2) 3x2+23(x3+2x)2/3\boxed{\frac{3x^2 + 2}{3(x^3 + 2x)^{2/3}}}
  3. To find the derivative of f(x)=(1−2x)−2f(x) = (1 - 2x)^{-2}:

    f′(x)=−2(1−2x)−2−1⋅ddx(1−2x)f'(x) = -2(1 - 2x)^{-2-1} \cdot \frac{d}{dx}(1 - 2x) f′(x)=−2(1−2x)−3⋅(−2)f'(x) = -2(1 - 2x)^{-3} \cdot (-2) f′(x)=4(1−2x)−3f'(x) = 4(1 - 2x)^{-3} 4(1−2x)3\boxed{\frac{4}{(1 - 2x)^3}}

Trigonometric Functions:

  1. To differentiate y=sin⁡(4x2+3x)y = \sin(4x^2 + 3x):

    dydx=cos⁡(4x2+3x)⋅ddx(4x2+3x)\frac{dy}{dx} = \cos(4x^2 + 3x) \cdot \frac{d}{dx}(4x^2 + 3x) dydx=cos⁡(4x2+3x)⋅(8x+3)\frac{dy}{dx} = \cos(4x^2 + 3x) \cdot (8x + 3) (8x+3)cos⁡(4x2+3x)\boxed{(8x + 3)\cos(4x^2 + 3x)}
  2. To find f′(x)f'(x) if f(x)=cos⁡3(2x)=(cos⁡(2x))3f(x) = \cos^3(2x) = (\cos(2x))^3:

    f′(x)=3(cos⁡(2x))3−1⋅ddx(cos⁡(2x))f'(x) = 3(\cos(2x))^{3-1} \cdot \frac{d}{dx}(\cos(2x)) f′(x)=3cos⁡2(2x)⋅(−sin⁡(2x))⋅ddx(2x)f'(x) = 3\cos^2(2x) \cdot (-\sin(2x)) \cdot \frac{d}{dx}(2x) f′(x)=3cos⁡2(2x)⋅(−sin⁡(2x))⋅2f'(x) = 3\cos^2(2x) \cdot (-\sin(2x)) \cdot 2 −6cos⁡2(2x)sin⁡(2x)\boxed{-6\cos^2(2x)\sin(2x)}
  3. To find dydx\frac{dy}{dx} if y=tan⁡(x2−5x)y = \tan(x^2 - 5x):

    dydx=sec⁡2(x2−5x)⋅ddx(x2−5x)\frac{dy}{dx} = \sec^2(x^2 - 5x) \cdot \frac{d}{dx}(x^2 - 5x) dydx=sec⁡2(x2−5x)⋅(2x−5)\frac{dy}{dx} = \sec^2(x^2 - 5x) \cdot (2x - 5) (2x−5)sec⁡2(x2−5x)\boxed{(2x - 5)\sec^2(x^2 - 5x)}
  4. To differentiate y=sin⁡2(3x+π/4)=(sin⁡(3x+π/4))2y = \sin^2(3x + \pi/4) = (\sin(3x + \pi/4))^2:

    dydx=2sin⁡(3x+π/4)⋅ddx(sin⁡(3x+π/4))\frac{dy}{dx} = 2\sin(3x + \pi/4) \cdot \frac{d}{dx}(\sin(3x + \pi/4)) dydx=2sin⁡(3x+π/4)⋅cos⁡(3x+π/4)⋅ddx(3x+π/4)\frac{dy}{dx} = 2\sin(3x + \pi/4) \cdot \cos(3x + \pi/4) \cdot \frac{d}{dx}(3x + \pi/4) dydx=2sin⁡(3x+π/4)⋅cos⁡(3x+π/4)⋅3\frac{dy}{dx} = 2\sin(3x + \pi/4) \cdot \cos(3x + \pi/4) \cdot 3 6sin⁡(3x+π/4)cos⁡(3x+π/4)\boxed{6\sin(3x + \pi/4)\cos(3x + \pi/4)}

Exponential Functions:

  1. To find dydx\frac{dy}{dx} if y=e3x2+2xy = e^{3x^2 + 2x}: dydx=e3x2+2x⋅ddx(3x2+2x)\frac{dy}{dx} = e^{3x^2 + 2x} \cdot \frac{d}{dx}(3x^2 + 2x) dydx=e3x2+2x⋅(6x+2)\frac{dy}{dx} = e^{3x^2 + 2x} \cdot (6x + 2)
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