a clearer picture of y

in Calculus Answers by Level 2 User (1.4k points)

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Let y=tan(u), dy/du=sec²(u), where u=eᵛ, du/dv=eᵛ, where v=√(3x)=(√3)(x^½), dv/dx=(√3)(½x^-½)=√3/(2√x).

du/dx=(du/dv)(dv/dx)=eᵛ(√3/(2√x))=(e^√(3x))(√3/(2√x)).

dy/dx=(dy/du)(du/dx)=sec²(u)(e^√(3x))(√3/(2√x))=

sec²(e^√(3x))(e^√(3x))(√3/(2√x)).

by Top Rated User (1.0m points)

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