Question #ea87b

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Answer 1 · 2016-03-26T11:55:35

$d/dx(e^(ktan sqrt(3x)))=(k*sqrt(3x))/(2x)*(sec^2 sqrt(3x))*e^(k tan(sqrt(3x))$

Let $y=e^(ktan sqrt(3x)$

$d/dx (y)=d/dx (e^(ktan sqrt(3x)))$
$d/dx (y)=e^(ktan sqrt(3x))*d/dx (ktan sqrt(3x))$
$d/dx (y)=e^(ktan sqrt(3x))(k sec^2 sqrt(3x))*d/dx (sqrt(3x))$

$d/dx (y)=e^(ktan sqrt(3x))(k sec^2 sqrt(3x))* (1/sqrt(3x))*d/dx (3x)$

$d/dx (y)=e^(ktan sqrt(3x))(k sec^2 sqrt(3x))* (1/sqrt(3x))*3*d/dx (x)$

$d/dx (y)=e^(ktan sqrt(3x))(k sec^2 sqrt(3x))* (1/sqrt(3x))*3(1)$

$d/dx (y)=e^(ktan sqrt(3x))(k )(sec^2 sqrt(3x))* (1/sqrt(3x))*3(1)$

Simplification

$d/dx (y)=(3k)/(2sqrt(3x)) (sec^2 sqrt(3x))*e^(ktan sqrt(3x) )$

The final answer after some rationalization of radicals

$d/dx (y)=(k sqrt(3x))/(2x) *(sec^2 sqrt(3x))*e^(ktan sqrt(3x) )$

God bless...I hope the explanation is useful.