MHT CET · Maths · Differentiation
If \(y=x^{x e^{x}}, \frac{d y}{d x}=y \cdot g(x)\), then \(g(x)=\)
- A \(\left[e^{x}+e^{x}(x+1) \log x\right]\)
- B \(\left[e^{x}-e^{x} \cdot x \cdot(1+\log x)\right]\)
- C \(\left[e^{x}+e^{x} \cdot x \cdot(1+\log x)\right]\)
- D \(\left[e^{x}(x+1) \log x\right]\)
Answer & Solution
Correct Answer
(A) \(\left[e^{x}+e^{x}(x+1) \log x\right]\)
Step-by-step Solution
Detailed explanation
We have \(y=x^{x e^{x}}\)
\(\therefore \log y=x e^{x} \log x\)
\(\therefore \frac{1}{y} \frac{d y}{d x}=e^{x} \log x+\frac{x e^{x}}{x}+x e^{x} \log x\)
\(=e^{x} \log x+e^{x}+x e^{x} \log x\)
\(\therefore \frac{d y}{d x}=y\left[e^{x}+e^{x} \log x(1+x)\right]\)
\(\therefore \log y=x e^{x} \log x\)
\(\therefore \frac{1}{y} \frac{d y}{d x}=e^{x} \log x+\frac{x e^{x}}{x}+x e^{x} \log x\)
\(=e^{x} \log x+e^{x}+x e^{x} \log x\)
\(\therefore \frac{d y}{d x}=y\left[e^{x}+e^{x} \log x(1+x)\right]\)
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