Question

If \(y=x^{e^x}+x^{e} for x>0\),the  \( \dfrac{dy}{dx}\) is equal to ?

• A. \(x^{e^x}[1/x+lnx]+e^x\)
• B. \(=e^{x}.x^{e^x-1}+e.x^{e-1}\)
• C. \(e^x.x^{e^x-1}+ex^e\)
• D. \(x^{e^x }e^-x[1/x-lnx]+ex^e-1\)
• E. \(x^{e^x} e^x[1/x-lnx]+ex^e-1\)
• F. \(x^{e^x} e^x[1/x+lnx]+ex^e-1\)

Answer 1

Answer: (F)

Step 1: Differentiate the function \( y = xe^{ex} + x^e \) with respect to \( x \).

Step 2: Differentiate the first term \( xe^{ex} \) using the product rule: \[ \frac{d}{dx}(xe^{ex}) = e^{ex} + x \cdot e^{ex} \cdot e = e^{ex}(1 + ex) \] This can be rewritten as: \[ xe^{ex}\left(\frac{1}{x} + e\right) \]

Step 3: Differentiate the second term \( x^e \): \[ \frac{d}{dx}(x^e) = ex^{e-1} \]

Step 4: Combine the derivatives: \[ \frac{dy}{dx} = e^{ex}(1 + ex) + ex^{e-1} \] Or equivalently: \[ \frac{dy}{dx} = xe^{ex}\left(\frac{1}{x} + e\right) + ex^{e-1} \]

Conclusion: The correct derivative matches option (E): \[ \boxed{E} \quad \left( xe^{ex}\left[\frac{1}{x} + e\right] + ex^{e-1} \right) \]

Answer 2

1. Differentiating \( x^{e^x} \):

Let \( u = x^{e^x} \). Then \( \ln(u) = e^x \ln(x) \).

Differentiating both sides with respect to \( x \):

\[ \frac{1}{u} \cdot \frac{du}{dx} = e^x \ln(x) + \frac{e^x}{x} \] \[ \frac{du}{dx} = u \cdot \left[e^x \ln(x) + \frac{e^x}{x}\right] = x^{e^x} \left[e^x \ln(x) + \frac{e^x}{x}\right] \]

2. Differentiating \( x^e \):

This is a simple power rule differentiation:

\[ \frac{d}{dx} (x^e) = e x^{e-1} \]

3. Combining the derivatives:

\[ \frac{dy}{dx} = \frac{du}{dx} + \frac{d}{dx}(x^e) = x^{e^x} \left[e^x \ln(x) + \frac{e^x}{x}\right] + e x^{e-1} \]

Therefore, the derivative \( \frac{dy}{dx} \) is:

\[ x^{e^x} \left[e^x \ln(x) + \frac{e^x}{x}\right] + e x^{e-1} \]