A calculus problem by Rahil Sehgal

Calculus Level 3

d d x x ( cot ⁡ ( ln ⁡ ( x + 10 ) ) + 6 ) = − x csc ⁡ 2 ( ln ⁡ ( x + a ) ) x + 10 + cot ⁡ ( ln ⁡ ( x + 10 ) ) + b \frac{d}{dx} x(\cot (\ln (x+10) ) +6) = - \dfrac{x \csc^2 (\ln (x+a) )}{x+10} + \cot ( \ln (x+10) ) +b

If the equation above holds true for positive real number a a and b b , find a − b + 6 a -b+6 .


The answer is 10.

Rahil Sehgal by Rahil Sehgal , 20, India

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1 solution

d d x x ( cot ⁡ ( ln ⁡ x + 10 ) + 6 ) = d d x ( x cot ⁡ u + 6 x ) Let u = ln ⁡ x + 6 = − x csc ⁡ 2 u ⋅ d u d x + cot ⁡ u + 6 = − x csc ⁡ 2 u ⋅ 1 x + 10 + cot ⁡ u + 6 = − x csc ⁡ 2 ( ln ⁡ x + 10 ) x + 10 + cot ⁡ ( ln ⁡ x + 10 ) + 6 \begin{aligned} \frac d{dx} x(\cot({\color{#3D99F6}\ln x + 10})+6) & = \frac d{dx} (x\cot {\color{#3D99F6}u} +6x) & \small \color{#3D99F6} \text{Let } u = \ln x + 6 \\ & = - x \csc^2 u \cdot \frac {du}{dx} + \cot u + 6 \\ & =- x \csc^2 u \cdot \frac 1{x + 10} + \cot u + 6 \\ & = - \frac {x \csc^2 (\ln x + 10)}{x + 10} + \cot (\ln x + 10) + 6 \end{aligned}

⟹ a − b + 6 = 10 − 6 + 6 = 10 \implies a - b + 6 = 10-6+6 = \boxed{10}

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