Transparent deviation

Calculus Level 2

$\displaystyle \large \dfrac{d}{dx} \left(e^{3x-2}\right) = 3$

The real value of $x$ can be written in the form $\dfrac{a}{b}$ , where $a$ and $b$ are coprime positive integers. Find the value of $a+b$ .

Hint: Use chain rule .

The answer is 5.

2 solutions

Viki Zeta
Sep 28, 2016

$D = \dfrac{d}{dx} (e^{3x-2}) \\ \text{Let, } f(x) = e^x; g(x) = 3x-2, \text{ such that : } f(g(x)) = e^{3x-2} \\ D = \dfrac{d}{dx} f(g(x)) = \dfrac{d}{dx} f'(g(x)) \cdot \dfrac{d}{dx} g(x) \\ = (\dfrac{d}{dx} e^x)(3x-2) \cdot \dfrac{d}{dx} (3x-2) \\ = e^{3x-2} \cdot 3 \\ = 3e^{3x-2}\\ \text{Given, }D = 3 \\ 3e^{3x-2} = 3 \\ e^{3x-2} = 1 \\ e^{3x-2} = e^0 \\ 3x- 2 = 0 \\ 3x = 2 \\ x = \dfrac{2}{3} = \dfrac{a}{b} \\ \boxed{\therefore a + b = 2 + 3 = 5}$

There's probably a typo error in the question. It asked us to calculate a-b instead of a+b.

Noah Hunter - 4 years, 8 months ago

Thanks. I think I wrote $a+b$ , maybe some staff edited with a typo.

Viki Zeta - 4 years, 8 months ago

My records indicate that you were the one that made the $a-b \rightarrow a+b$ change, about 6 hours after you posted the problem.

Calvin Lin Staff - 4 years, 8 months ago

@Calvin Lin – I don't remember, anyway it's now resolved

Viki Zeta - 4 years, 8 months ago

Chew-Seong Cheong
Sep 28, 2016

$\begin{aligned} \frac d{dx} e^{3x-2} & = 3 \\ e^{-2} \frac d{dx} e^{3x} & = 3 \\ \frac d{dx} e^{3x} & = 3e^2 \\ 3e^{3x} & = 3e^2 \\ e^{3x} & = e^2 \\ \implies 3x&=2\\ x&= \frac 23 \end{aligned}$

$\implies a+b=2+3=\boxed {5}$

Transparent deviation

The answer is 5.

2 solutions

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