What is the derivative of ${(tan 2 x + sec 2 x)}^{2}$ $(tan2x+sec2x)^2$ ?

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Answer 1 · 2017-10-11T13:07:05

$\frac{d y}{d x} = 4 sec 2 x {(tan 2 x + sec 2 x)}^{2}$ $(dy)/(dx)=4sec2x(tan2x+sec2x)^2$

we will use teh chain rule

$\frac{d y}{d x} = \frac{d y}{d u} \frac{d u}{d x}$ $(dy)/(dx)=(dy)/(du)(du)/(dx)$

$y = {(tan 2 x + sec 2 x)}^{2}$ $y=(tan2x+sec2x)^2$

$u = tan 2 x + sec 2 x \Rightarrow \frac{d u}{d x} = 2 {sec}^{2} 2 x + 2 sec 2 x tan 2 x$ $u=tan2x+sec2x=>(du)/(dx)=2sec^2 2x+2sec2xtan2x$

$y = u^{2} \Rightarrow \frac{d y}{d u} = 2 u$ $y=u^2=>(dy)/(du)=2u$

$:.(dy)/(dx)=2uxx(2sec^2 2x+2sec2xtan2x)$

substitute back and tidy up

$(dy)/(dx)=2(tan2x+sec2x)(2sec^2 2x+2sec2xtan2x)$

$(dy)/(dx)=4sec2x(tan2x+sec2x)(sec2x+tan2x)$

$(dy)/(dx)=4sec2x(tan2x+sec2x)^2$

What is the derivative of (tan2x+sec2x)^2(tan2x+sec2x)2(tan2x+sec2x)^2?