Answered

Tuklasin ang maliwanag na mga sagot sa iyong mga tanong sa IDNStudy.com. Ang aming komunidad ay handang magbigay ng malalim at praktikal na mga solusyon sa lahat ng iyong mga katanungan.

what is the derivative of y=ln(cosh2x) :)

Sagot :

[tex]\large \bold {SOLUTION}[/tex]

[tex]\large\sf{y = ln( \cosh(2x) ) }[/tex]

[tex]\small\textsf{By the Chain Rule of differentiation, let u = cosh (2x)}[/tex]

[tex]\small\sf{(f[g(x)])' = f'[g(x)] \: \: • \: \: g'(x)}[/tex]

[tex]\small\sf{y' = \dfrac{d}{du} \: ln(u) \: \:• \: \: \dfrac{d}{dx} \: \cosh(2x) }[/tex]

[tex]\small\textsf{Set aside the first term and differentiate the second term}[/tex]

[tex]\small\textsf{By the Chain Rule of differentiation, let u = 2x}[/tex]

[tex]\small\sf{ \dfrac{d}{du} \: \cosh(u) \: \: • \: \: \dfrac{d}{dx} \: 2x }[/tex]

[tex]\small\sf{y' = \dfrac{d}{du} \: ln(u) \: \: • \: \: \sinh(u) \: \: • \: \: 2 }[/tex]

[tex]\small\textsf{Return u = 2x as the substitution}[/tex]

[tex]\small\sf{y' = \dfrac{d}{du} \: ln(u) \: \: • \: \: \sinh(2x) \: \: • \: \: 2 }[/tex]

[tex]\small\sf{y' = \dfrac{1}{u} \: \: • \: \: \sinh(2x) \: \: • \: \: 2 }[/tex]

[tex]\small\textsf{Return the main u-substitution}[/tex]

[tex]\small\sf{y' = \dfrac{1}{ \cosh(2x) } \: \: • \: \: \sinh(2x) \: \: • \: \: 2 }[/tex]

[tex]\small\sf{y' = \dfrac{1}{ \cosh(2x) } \: \: • \: \: 2\sinh(2x) }[/tex]

[tex]\therefore\small\sf{y' = ln( \cosh(2x) ) \implies\small\boxed{\green{\sf{ \frac{2 \sinh(2x) }{ \cosh(2x) } }}}}[/tex]

[tex]\small\textsf{\#AlwaysBeTheGreat}[/tex]