Question

if y=sin(sin x); prove that y2+tanx y1+y cos²x=0​

Answer 1

$\huge\boxed{\underline{\underline{\red{Solution}}}}$

$y=sin(sin x)$

Differentiating both sides with respect to x we get

$\displaystyle \: y_1 \: = \frac{d}{dx} \{sin(sinx) \}$

$\implies \displaystyle \: y_1 \: = cos(sinx)\frac{d}{dx} (sinx)$

$\implies \displaystyle \: y_1 \: = cosx \: \: \times \: cos(sinx)$

Again Differentiating both sides with respect to x

$\displaystyle \: y_2 \: = \frac{d}{dx} \{cosx \: \times cos(sinx) \}$

$\implies \displaystyle \: y_2 \: = - sinx \: \times cos(sinx) - {cos}^{2} x \: \times sin(sinx)$

Now

LHS

$\displaystyle \: y_2 + tanx \times \: y_1 + y {cos}^{2} x$

$= \displaystyle \: - sinx \: \times cos(sinx) - {cos}^{2} x \: \times sin(sinx) \: + tanx \: \times cosx \: \times cos(sinx) + {cos}^{2} x \times sin(sinx)$

$= \displaystyle \: - sinx \: \times cos(sinx) - {cos}^{2} x \: \times sin(sinx) \: +sinx \: \times cos(sinx) + {cos}^{2} x \times sin(sinx)$

$= 0$

Hence Proved