Question #ea87b

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Question #ea87b

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Answer 1 · 2016-03-26T11:55:35

$\frac{d}{d x} (e^{k tan \sqrt{3 x}}) = \frac{k \cdot \sqrt{3 x}}{2 x} \cdot ({sec}^{2} \sqrt{3 x}) \cdot e^{k tan (\sqrt{3 x})}$ $d/dx(e^(ktan sqrt(3x)))=(k*sqrt(3x))/(2x)*(sec^2 sqrt(3x))*e^(k tan(sqrt(3x))$

Explanation:

Let $y = e^{k tan \sqrt{3 x}}$ $y=e^(ktan sqrt(3x)$

$\frac{d}{d x} (y) = \frac{d}{d x} (e^{k tan \sqrt{3 x}})$ $d/dx (y)=d/dx (e^(ktan sqrt(3x)))$
$\frac{d}{d x} (y) = e^{k tan \sqrt{3 x}} \cdot \frac{d}{d x} (k tan \sqrt{3 x})$ $d/dx (y)=e^(ktan sqrt(3x))*d/dx (ktan sqrt(3x))$
$\frac{d}{d x} (y) = e^{k tan \sqrt{3 x}} (k {sec}^{2} \sqrt{3 x}) \cdot \frac{d}{d x} (\sqrt{3 x})$ $d/dx (y)=e^(ktan sqrt(3x))(k sec^2 sqrt(3x))*d/dx (sqrt(3x))$

$\frac{d}{d x} (y) = e^{k tan \sqrt{3 x}} (k {sec}^{2} \sqrt{3 x}) \cdot (\frac{1}{\sqrt{3 x}}) \cdot \frac{d}{d x} (3 x)$ $d/dx (y)=e^(ktan sqrt(3x))(k sec^2 sqrt(3x))* (1/sqrt(3x))*d/dx (3x)$

$\frac{d}{d x} (y) = e^{k tan \sqrt{3 x}} (k {sec}^{2} \sqrt{3 x}) \cdot (\frac{1}{\sqrt{3 x}}) \cdot 3 \cdot \frac{d}{d x} (x)$ $d/dx (y)=e^(ktan sqrt(3x))(k sec^2 sqrt(3x))* (1/sqrt(3x))*3*d/dx (x)$

$\frac{d}{d x} (y) = e^{k tan \sqrt{3 x}} (k {sec}^{2} \sqrt{3 x}) \cdot (\frac{1}{\sqrt{3 x}}) \cdot 3 (1)$ $d/dx (y)=e^(ktan sqrt(3x))(k sec^2 sqrt(3x))* (1/sqrt(3x))*3(1)$

$\frac{d}{d x} (y) = e^{k tan \sqrt{3 x}} (k) ({sec}^{2} \sqrt{3 x}) \cdot (\frac{1}{\sqrt{3 x}}) \cdot 3 (1)$ $d/dx (y)=e^(ktan sqrt(3x))(k )(sec^2 sqrt(3x))* (1/sqrt(3x))*3(1)$

Simplification

$\frac{d}{d x} (y) = \frac{3 k}{2 \sqrt{3 x}} ({sec}^{2} \sqrt{3 x}) \cdot e^{k tan \sqrt{3 x}}$ $d/dx (y)=(3k)/(2sqrt(3x)) (sec^2 sqrt(3x))*e^(ktan sqrt(3x) )$

The final answer after some rationalization of radicals

$\frac{d}{d x} (y) = \frac{k \sqrt{3 x}}{2 x} \cdot ({sec}^{2} \sqrt{3 x}) \cdot e^{k tan \sqrt{3 x}}$ $d/dx (y)=(k sqrt(3x))/(2x) *(sec^2 sqrt(3x))*e^(ktan sqrt(3x) )$

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