Question:
If $(3 + 5x)e^{\frac{y}{x}} = x$, then $\dfrac{dy}{dx}$ is equal to:
If $(3 + 5x)e^{\frac{y}{x}} = x$, then $\dfrac{dy}{dx}$ is equal to:
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Use logarithms when variable appears in exponent.
Updated On: Apr 24, 2026
- $\log\left|\frac{x}{3+5x}\right| + \frac{3}{3+5x}$
- $\log\left|\frac{x}{3+5x}\right| + \frac{5x-3}{3+5x}$
- $\log\left|\frac{x}{3+5x}\right| + \frac{5x-2}{3+5x}$
- $\log\left|\frac{3x}{3+5x}\right| + \frac{10x+3}{3+5x}$
- $\log\left|\frac{x}{3+5x}\right| + \frac{3-10x}{3+5x}$
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The Correct Option is A
Solution and Explanation
Concept:
• Use logarithmic differentiation
Step 1: Take log on both sides
\[ (3+5x)e^{y/x} = x \] \[ \ln(3+5x) + \frac{y}{x} = \ln x \]
Step 2: Differentiate
\[ \frac{5}{3+5x} + \frac{x\frac{dy}{dx} - y}{x^2} = \frac{1}{x} \]
Step 3: Simplify
Multiply by $x$: \[ \frac{5x}{3+5x} + \frac{dy}{dx} - \frac{y}{x} = 1 \]
Step 4: Substitute $\frac{y}{x}$
From original: \[ \frac{y}{x} = \ln\left(\frac{x}{3+5x}\right) \]
Step 5: Solve
\[ \frac{dy}{dx} = 1 - \frac{5x}{3+5x} + \ln\left(\frac{x}{3+5x}\right) \] \[ = \frac{3}{3+5x} + \ln\left(\frac{x}{3+5x}\right) \] Final Conclusion:
Option (A)
• Use logarithmic differentiation
Step 1: Take log on both sides
\[ (3+5x)e^{y/x} = x \] \[ \ln(3+5x) + \frac{y}{x} = \ln x \]
Step 2: Differentiate
\[ \frac{5}{3+5x} + \frac{x\frac{dy}{dx} - y}{x^2} = \frac{1}{x} \]
Step 3: Simplify
Multiply by $x$: \[ \frac{5x}{3+5x} + \frac{dy}{dx} - \frac{y}{x} = 1 \]
Step 4: Substitute $\frac{y}{x}$
From original: \[ \frac{y}{x} = \ln\left(\frac{x}{3+5x}\right) \]
Step 5: Solve
\[ \frac{dy}{dx} = 1 - \frac{5x}{3+5x} + \ln\left(\frac{x}{3+5x}\right) \] \[ = \frac{3}{3+5x} + \ln\left(\frac{x}{3+5x}\right) \] Final Conclusion:
Option (A)
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