Question

Question #ea87b

Answer 1

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

Explanation:

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.