MCQ
if $\text{y}=\text{e}^{{\tan}\text{x}},$ then $(\cos^2\text{x})\text{y}_2=$
  • A
    $(1-\sin2\text{x})\text{y}_1$
  • B
    $-(1+\sin2\text{x})\text{y}_1$
  • $(1+\sin2\text{x})\text{y}_1$
  • D
    $\text{None of these}$

Answer

Correct option: C.
$(1+\sin2\text{x})\text{y}_1$
$\text{y}=\text{e}^{{\tan}\text{x}},$

$\text{y}_1=\text{sec}^2\text{xe}^{\tan\text{x}}$

$\Rightarrow\cos^2\text{xy}_1=\text{e}^{\tan\text{x}}$

again differentiating w.r.t.x, we get

$\cos^2\text{xy}_2-2\cos\text{x}\sin\text{xy}_1=\sec^2\text{xe}^{\tan\text{x}}$

$\Rightarrow\cos^2\text{xy}_2=\text{y}_1\sin2\text{x}+\text{y}_1$

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