Question:
If \(y = x\cdot 7^x\), then the value of \(\frac{dx}{dy}\) when \(x = 1\) is
If \(y = x\cdot 7^x\), then the value of \(\frac{dx}{dy}\) when \(x = 1\) is
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Find \(\frac{dy}{dx}\) by the product rule and take the reciprocal.
Updated On: Oct 1, 2026
- \(7(log7+1)\)
- \(log7+1\)
- \(\frac{1}{7(log7+1)}\)
- \(\frac{1}{log7+1}\)
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept
\(\dfrac{dx}{dy}=\dfrac{1}{dy/dx}\) whenever \(dy/dx\neq0\).
Step 2: Key Formula or Approach
\[ \frac{dy}{dx}=7^x+x\,7^x\log7 \]
Step 3: Detailed Explanation
At \(x=1\): \(\dfrac{dy}{dx}=7+7\log7=7(1+\log7)\).
\[ \frac{dx}{dy}=\frac{1}{7(\log7+1)} \]
Final Answer:
The value is \(\dfrac{1}{7(\log7+1)}\), option (C). \[ \boxed{\dfrac{1}{7(\log7+1)}\ \text{(C)}} \]
\(\dfrac{dx}{dy}=\dfrac{1}{dy/dx}\) whenever \(dy/dx\neq0\).
Step 2: Key Formula or Approach
\[ \frac{dy}{dx}=7^x+x\,7^x\log7 \]
Step 3: Detailed Explanation
At \(x=1\): \(\dfrac{dy}{dx}=7+7\log7=7(1+\log7)\).
\[ \frac{dx}{dy}=\frac{1}{7(\log7+1)} \]
Final Answer:
The value is \(\dfrac{1}{7(\log7+1)}\), option (C). \[ \boxed{\dfrac{1}{7(\log7+1)}\ \text{(C)}} \]
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