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

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

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Answer 1 · 2016-11-12T00:24:50

$\frac{d y}{d x} = \frac{- 2}{x \sqrt{16 x^{4} - 1}} (= \frac{- 2}{x \sqrt{(4 x^{2} + 1) (4 x^{2} - 1)}})$ $dy/dx=(-2)/(xsqrt(16x^4 - 1)) (=(-2)/(xsqrt((4x^2+1)(4x^2-1))))$

Explanation:

$y = {csc}^{- 1} (4 x^{2}) \Leftrightarrow csc y = 4 x^{2}$ $y=csc^-1(4x^2) <=> cscy=4x^2$

Differentiating implicitly we have:
$csc y = 4 x^{2}$ $cscy=4x^2$
$- csc y cot y \frac{d y}{d x} = 8 x$ $-cscycotydy/dx=8x$
$- (4 x^{2}) cot y \frac{d y}{d x} = 8 x$ $-(4x^2)cotydy/dx=8x$
$cot y \frac{d y}{d x} = - \frac{2}{x}$ $cotydy/dx=-2/x$ ..... [1]

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

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

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

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

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How do you differentiate y=csc^-1(4x^2)y=csc−1(4x2)y=csc^-1(4x^2)?

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Explanation:

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