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Differentiate the following functions using the power rule: i) y = 3x7 + 2x3 − 3x ii) f(x) = 4x9 − 5x10 + 17 iii) g(x) = x3 − 4x5 + 16x − 3

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3 parts
i

dydx=21x6+6x2−3\frac{dy}{dx} = 21x^6 + 6x^2 - 3

ii

f′(x)=36x8−50x9f'(x) = 36x^8 - 50x^9

iii

g′(x)=3x2−20x4+16g'(x) = 3x^2 - 20x^4 + 16

Step-by-step solution

9 steps

Here are the differentiations for each function:

To differentiate, we use the power rule: ddx(axn)=anxn−1\frac{d}{dx}(ax^n) = anx^{n-1} and the rule for constants: ddx(c)=0\frac{d}{dx}(c) = 0.

Part i· 3 steps

y=3x7+2x3−3xy = 3x^7 + 2x^3 - 3x

  1. Differentiate each term with respect to xx.

    dydx=ddx(3x7)+ddx(2x3)−ddx(3x)\frac{dy}{dx} = \frac{d}{dx}(3x^7) + \frac{d}{dx}(2x^3) - \frac{d}{dx}(3x)

  2. Apply the power rule to each term.

    dydx=(3⋅7x7−1)+(2⋅3x3−1)−(3⋅1x1−1)\frac{dy}{dx} = (3 \cdot 7x^{7-1}) + (2 \cdot 3x^{3-1}) - (3 \cdot 1x^{1-1}) dydx=21x6+6x2−3x0\frac{dy}{dx} = 21x^6 + 6x^2 - 3x^0

  3. Simplify the expression (x0=1x^0 = 1).

    dydx=21x6+6x2−3\boxed{\frac{dy}{dx} = 21x^6 + 6x^2 - 3}

Part ii

f′(x)=36x8−50x9f'(x) = 36x^8 - 50x^9

f(x)=4x9−5x10+17f(x) = 4x^9 - 5x^{10} + 17

  1. Differentiate each term with respect to xx.

    f′(x)=ddx(4x9)−ddx(5x10)+ddx(17)f'(x) = \frac{d}{dx}(4x^9) - \frac{d}{dx}(5x^{10}) + \frac{d}{dx}(17)

  2. Apply the power rule and the constant rule.

    f′(x)=(4⋅9x9−1)−(5⋅10x10−1)+0f'(x) = (4 \cdot 9x^{9-1}) - (5 \cdot 10x^{10-1}) + 0 f′(x)=36x8−50x9f'(x) = 36x^8 - 50x^9

  3. Simplify the expression.

    f′(x)=36x8−50x9\boxed{f'(x) = 36x^8 - 50x^9}

Part iii

g′(x)=3x2−20x4+16g'(x) = 3x^2 - 20x^4 + 16

g(x)=x3−4x5+16x−3g(x) = x^3 - 4x^5 + 16x - 3

  1. Differentiate each term with respect to xx.

    g′(x)=ddx(x3)−ddx(4x5)+ddx(16x)−ddx(3)g'(x) = \frac{d}{dx}(x^3) - \frac{d}{dx}(4x^5) + \frac{d}{dx}(16x) - \frac{d}{dx}(3)

  2. Apply the power rule and the constant rule.

    g′(x)=(1⋅3x3−1)−(4⋅5x5−1)+(16⋅1x1−1)−0g'(x) = (1 \cdot 3x^{3-1}) - (4 \cdot 5x^{5-1}) + (16 \cdot 1x^{1-1}) - 0 g′(x)=3x2−20x4+16x0g'(x) = 3x^2 - 20x^4 + 16x^0

  3. Simplify the expression (x0=1x^0 = 1).

    g′(x)=3x2−20x4+16\boxed{g'(x) = 3x^2 - 20x^4 + 16}
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