Differentiate and Skate

Calculus Level 4

Consider the function $f: \mathbb{R}^2 \to \mathbb{R}$ such that $f(x)f'(x) = 2017.$

Knowing that $f(1) = 64$ and $f'(3907)$ can be expressed as $\dfrac ab$ , where $a$ and $b$ are coprime positive integers.

Evaluate $a+b$ .

Notation: $\mathbb R$ denotes the set of real numbers .

The answer is 5987.

1 solution

Guilherme Dela Corte
Jan 5, 2017

Let $y = f(x)$ . First we must solve the differential equation $yy' = 2017$ .

See that $y \dfrac{dy}{dx} = 2017 \Leftrightarrow ydy = 2017dx$ . Integrating each side, we get $\dfrac{y^2}{2} + c_1 = 2017x + c_2 \Leftrightarrow y = \pm \sqrt{c+4034x}$ .

Since $x = 1$ yields $y = 64$ , we get that $64 = \sqrt{c+4034} \Leftrightarrow c = 62$ .

Substituting in the initial differential equation, we get $f'(x) = \dfrac{2017}{\sqrt{62+4034x}}$ . This way, plugging in $x = 3907$ we get that $f'(3907) = \dfrac{2017}{3970}$ , and thus $a + b = \boxed{5987.}$ .

Differentiate and Skate

The answer is 5987.

1 solution

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