If $y=arcsin(4x^2)$ , then what is $dy/dx$ ?

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Answer 1 · 2016-12-19T14:19:51

The answer is $=(8x)/sqrt(1-16x^4)$

By definition

If $y=arcsinx$ , then, $siny=x$

Here, we have

$y=arcsin4x^2$

So,

$siny=4x^2$

Differentiating

$cosydy/dx=8x$

$dy/dx=(8x)/cosy$

$sin^2y+cos^2y=1$

$cos^2y=1-sin^2y=1-(4x^2)^2$

$=1-16x^4$

$cosy=sqrt(1-16x^4)$

$dy/dx=(8x)/sqrt(1-16x^4)$

If y=arcsin(4x^2)y=arcsin(4x^2), then what is dy/dxdy/dx?