If $y=1/2 sec^2x$ how we can prove that $(d^2y)/dx^2=4y(3y-1)$ ?

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Answer 1 · 2018-02-19T20:37:20

See below

$y = 1/2sec^2x$

Use the chain rule to find $dy/dx$

$d/dx[f(x)]^n=n[f(x)]^(n-1)f'(x)$

$d/dx(1/2sec^2x)=1/2 *2secx*secxtanx=sec^2xtanx$

Now use the chain and product rules to find $(d^2y)/dx^2$

$d/dxf(x)g(x)=g(x)f'(x)+f(x)g'(x)$

$d/dxsec^2xtanx=(d^2y)/dx^2 1/2sec^2x=2sec^2xtanxtanx+sec^2xsec^2x=2sec^2xtan^2x+sec^4x=sec^2x(2tan^2x+sec^2x)$

Now use $tan^2x=sec^2x-1$ to yield the required result

$sec^2x(2tan^2x+sec^2x)=sec^2x(2(sec^2x-1)+sec^2x)=sec^2x(2sec^2x-2+sec^2x)=sec^2x(3sec^2x-2)$

Now substitute $sec^2x=2y$

$sec^2x(3sec^2x-2)=2y(6y-2)$
$=4y(3y-1)$ , as required. $square$

If y=1/2 sec^2xy=1/2 sec^2x how we can prove that (d^2y)/dx^2=4y(3y-1)(d^2y)/dx^2=4y(3y-1) ?