How do you differentiate $y=csc^-1(4x^2)$ ?

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Answer 1 · 2016-11-12T00:24:50

$dy/dx=(-2)/(xsqrt(16x^4 - 1)) (=(-2)/(xsqrt((4x^2+1)(4x^2-1))))$

$y=csc^-1(4x^2) <=> cscy=4x^2$

Differentiating implicitly we have:
$cscy=4x^2$
$-cscycotydy/dx=8x$
$-(4x^2)cotydy/dx=8x$
$cotydy/dx=-2/x$ ..... [1]

Now $1 + cot^2A -= csc^2A$
$:. 1 + cot^2y = 16x^4$
$:. cot^2y = 16x^4 - 1$
$:. coty = sqrt(16x^4 - 1)$

Substituting into [1] we get:
$sqrt(16x^4 - 1)dy/dx=-2/x$

$:. dy/dx=(-2/x)/sqrt(16x^4 - 1)$
$:. dy/dx=(-2)/(xsqrt(16x^4 - 1)) (=(-2)/(xsqrt((4x^2+1)(4x^2-1))))$

How do you differentiate y=csc^-1(4x^2)y=csc^-1(4x^2)?