nple 4 . If \( y=e^{3 x} x^{3} \), then find \[ \frac{d y}{d x} \text { at } x=\frac{1}{3} \text {. } \]

Monjur Alahi

nple 4 . If \( y=e^{3 x} x^{3} \), then find \[ \frac{d y}{d x} \text { at } x=\frac{1}{3} \text {. } \]


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Salim Ahmed Kawsar

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y = e3x x3 \(\frac{dy}{dx}\) at x = \(\frac{1}{3}\)

\(\frac{dy}{dx}\) = \(\left[x^3 . \frac{d}{dx}^{(e^{3x})} + e^{3x} . \frac{d}{dx}^{(x^3)}\right]\) [By product rule]

= x3 . e3x . 3 + e3x . 3x2 [By chain rule]

= 3e3x x3 + e3x . 3x2 

= 3x2 [e3x . x + e3x]

Put x = \(\frac{1}{3}\)

= 3 (\(\frac{1}{3}\))2 \(\left[e^{3{(1/3)}}.\frac{1}{3} + e^{3{(1/3)}}\right]\)

= 3 × \(\frac{1}{9}\)[\(\frac{e}{3}\) + e]

= \(\frac{1}{3}\)\(\left[\frac{e + 3e}{3}\right]\)

= \(\frac{4e}{9}\)

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