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Hint

It might help to write f'(x) = 2x as f'(x) = 2x¹.

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The coefficient of the derivative is the exponent of the original function, and the exponent of the derivative is one less than the exponent of the original function. If

f(x) = x⁴

then

f'(x) = 4x^{(4 - 1)} = 4x³

The derivative of the function

f(x) = xⁿ

is

f'(x) = nx^(n-1).

We take the exponent n from the original function. This n is the coefficient for the derivative. The value n-1 is the exponent for the derivative.

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We use the rule for finding the derivatives of power functions.

f(x) = x¹⁰

f'(x) = 10x^10–1 = 10x⁹

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f(x) = x⁸⁵

f'(x) = 85x ^85–1 = 85x⁸⁴

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k(x) = x^3.5

k'(x) = 3.5x^3.5–1 = 3.5x^2.5

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k(x) = x^–6

k'(x) = (–6)x^–6–1 = –6x^-7

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This function can be rewritten as

r(x) = x^1/2,

therefore

r'(x) = 1/2x^{1/2 - 1} = 1/2x^-1/2.

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h(x) = x^{π + e}

The constant (π + e) is kind of unpleasant-looking, but it's still a constant.

h'(x) = (π + e)x^{(π + e) – 1} = (π + e)x^{π + e – 1}.

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This can be rewritten as

g(x) = x^–2,

which looks a lot more like a power function.

g'(x) = (–2)x^–2–1 = –2x^–3.

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Again,  is an ugly constant, but it is a constant.

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k(x) = x⁰

This is the power function. The function

k(x) = x⁰ = 1

is a constant function, therefore

k'(x) = 0.

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This function can be rewritten as

therefore

r'(x) = ( –1)x ^–1–1 = –x^–2.

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This derivative came from a power function, the only thing we need to figure out is the exponent of the original power function.

f'(x) = 3x² = 3x^3–1

the original function could have been

f(x) = x³.

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f'(x) = 8x⁷ = 8x ^8–1

the original function could have been

f(x) = x⁸.

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f'(x) = –3x^–4 = –3x^–3–1

the original function could have been

f(x) = x^–3

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g'(x) = –9x^–10 = –9x^–9–1

the original function could have been

g(x) = x^–9.

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the original function could have been

We're saying the original function "could have been" instead of "must have been" because, as we'll see later, for each derivative there are infinitely many functions with that same derivative.