A calculus problem by Nazmus sakib

Calculus Level 2

d 2 d x 2 ( x 2 ) e 2 = ? \large\dfrac{d^2}{dx^2}(x^2)^{e^{2}}=?

2 e 2 ( 2 e 2 − 1 ) ( x 2 ) e 2 − 1 2e^2(2e^{2}-1)(x^{2})^{e^{2}-1} 2 e 3 ( x 2 ) e 3 x \dfrac{2 e^3 (x^2)^{e^3}}{x} 1 1 2 e 2 ( x 2 ) e 2 x \dfrac{2 e^2 (x^2)^{e^2}}{x}

Nazmus Sakib by Nazmus sakib , 24, Bangladesh

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1 solution

Chew-Seong Cheong
Nov 19, 2017

y = x 2 e 2 = x a Let y = x 2 e 2 and a = 2 e 2 (a constant) d y d x = a x a − 1 d 2 y d x 2 = a ( a − 1 ) x a − 2 Put back y = x 2 e 2 and a = 2 e 2 d 2 x 2 e 2 d x 2 = 2 e 2 ( 2 e 2 − 1 ) x 2 e 2 − 2 = 2 e 2 ( 2 e 2 − 1 ) ( x 2 ) e 2 − 1 \begin{aligned} \color{#3D99F6}y & = x^{\color{#D61F06}2e^2} = x^{\color{#D61F06}a} & \small \text{Let }{\color{#3D99F6} y = x^{2e^2}} \text{ and }\color{#D61F06} a = 2e^2 \text{ (a constant)} \\ \frac {dy}{dx} & = ax^{a-1} \\ \frac {d^2\color{#3D99F6}y}{dx^2} & = {\color{#D61F06}a}({\color{#D61F06}a}-1)x^{{\color{#D61F06}a}-2} & \small \text{Put back }{\color{#3D99F6} y = x^{2\color{#3D99F6}e^2}} \text{ and }\color{#D61F06} a = 2e^2 \\ \frac {d^2\color{#3D99F6}x^{2e^2}}{dx^2} & = {\color{#D61F06}2e^2}({\color{#D61F06}2e^2}-1)x^{{\color{#D61F06}2e^2}-2} \\ & = \boxed{2e^2(2e^2-1)\left(x^2\right)^{e^2-1}} \end{aligned}

4 Helpful 1 Interesting 2 Brilliant 0 Confused

@Chew-Seong Cheong Thanks for the solution,It is quite clear ⌣ ¨ \huge\ddot\smile .

Nazmus sakib - 3 years, 6 months ago

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