MHT CET · Maths · Differentiation
The rate of change of \(\sqrt{x^2+16}\) with respect to \(\frac{x}{x-1}\) at \(x=5\) is
- A \(\frac{-80}{\sqrt{41}}\)
- B \(\frac{80}{\sqrt{41}}\)
- C \(\frac {12}{5}\)
- D \(\frac {-12}{5}\)
Answer & Solution
Correct Answer
(A) \(\frac{-80}{\sqrt{41}}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & y=\sqrt{x^2+16} \\ & \therefore \quad \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{2 x}{2 \sqrt{x^2+16}} \\ & \therefore \quad \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{x}{\sqrt{x^2+16}}... (i) \\ & \text { Let } \mathrm{z}=\frac{x}{x-1} \\ & \therefore \quad \frac{\mathrm{dz}}{\mathrm{d} x}=\frac{(x-1)-x}{(x-1)^2} \\ & \therefore \quad \frac{\mathrm{dz}}{\mathrm{d} x}=\frac{-1}{(x-1)^2}... (ii) \\ & \therefore \quad\left(\frac{d y}{d z}\right)_{x=5}=\frac{\frac{x}{\sqrt{x^2+16}}}{\frac{-1}{(x-1)^2}} \\ & =\frac{-5}{\sqrt{25+16}} \times 16=\frac{-80}{\sqrt{41}} \\ & \end{aligned}\)
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