If \(y = 3^x+e^x+x^x+x^3\), then the value of \(\displaystyle\frac{dy}{dx}\) at \(x = 3\) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(e^3+54\log_e 3+54\)
The term \(x^x\) has a variable base and a variable power, so write it as \(x^x = e^{x\log_e x}\):
\(\displaystyle \frac{d}{dx}\left(x^x\right) = e^{x\log_e x}\cdot\frac{d}{dx}\left(x\log_e x\right)\)
\(\displaystyle = x^x\left(\log_e x + x\cdot\frac{1}{x}\right) = x^x\left(\log_e x + 1\right)\)
The other terms are standard: the derivative of \(3^x\) is \(3^x\log_e 3\), of \(e^x\) is \(e^x\), and of \(x^3\) is \(3x^2\). So
\(\displaystyle \frac{dy}{dx} = 3^x\log_e 3 + e^x + x^x(\log_e x + 1) + 3x^2\)
Putting x = 3:
\(\displaystyle \frac{dy}{dx} = 27\log_e 3 + e^3 + 27(\log_e 3 + 1) + 27\)
\(= e^3 + 54\log_e 3 + 54\)
Why the other options are incorrect
Option (1) \(e^3 +27\log_e 3+27\) — Incorrect. This leaves out the derivative of \(x^x\), which is \(27(\log_e 3 + 1)\) at x = 3.
Option (3) \(e^3+54\log_e 3+27\) — Incorrect. This takes the derivative of \(x^x\) as \(x^x\log_e x\) only (the rule for a constant base), so the term \(x^x \cdot 1 = 27\) is missing.
Option (4) \(e^3 +27\log_e 3+54\) — Incorrect. This takes the derivative of \(x^x\) as \(x \cdot x^{x-1}\) (the power rule, for a constant power), which gives 27 and misses \(27\log_e 3\).