The rate of change of x 2 + 16 with respect to x − 1 x at x = 3 is

u = x 2 + 16 , d x d u = 2 x 2 + 16 2 x = x 2 + 16 x ; v = x − 1 x ⇒ d x d v = ( x − 1 ) 2 − 1 d v d u = d v / d x d u / d x = − x 2 + 16 x ( x − 1 ) 2 ; at x = 3 : d v d u = − 5 3 × 4 = 5 − 12

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Mathematics - Application of Derivative Question with Solution | TestHub

MathematicsApplication of DerivativeRate MeasureMedium2 minQB

MathematicsMediumsingle choice

The rate of change of $x^{2} + 16$ with respect to $\frac{x}{x - 1}$ at $x = 3$ is

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Solution:

$u = x^{2} + 16, \frac{d u}{d x} = \frac{2 x}{2 x ^{2} + 16} = \frac{x}{x ^{2} + 16}; v = \frac{x}{x - 1} \Rightarrow \frac{d v}{d x} = \frac{- 1}{( x - 1 ) ^{2}}$ $\frac{d u}{d v} = \frac{d u / d x}{d v / d x} = - \frac{x ( x - 1 ) ^{2}}{x ^{2} + 16}; at x = 3 : \frac{d u}{d v} = - \frac{3 \times 4}{5} = \frac{- 12}{5}$

Stream:JEESubject:MathematicsTopic:Application of DerivativeSubtopic:Rate Measure

⏱ 2mℹ️ Source: QB

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