What is the derivative of this function $y = {({cos}^{- 1} (4 x^{2}))}^{2}$ $y=(cos^-1(4x^2))^2$ ?

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What is the derivative of this function $y = {({cos}^{- 1} (4 x^{2}))}^{2}$ $y=(cos^-1(4x^2))^2$ ?

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Answer 1 · 2017-08-04T18:10:15

$\frac{d}{d x} (arccos (4 x^{2})) = - \frac{16 x \cdot arccos (4 x^{2})}{\sqrt{1 - 16 x^{4}}}$ $("d")/("d"x) ("arccos"(4x^2)) = -(16x*"arccos"(4x^2))/(sqrt(1-16x^4))$ .

Explanation:

Note that $\frac{d}{d x} (arccos (x)) = - \frac{1}{\sqrt{1 - x^{2}}}$ $"d"/("d"x) ("arccos"(x))=-1/sqrt(1-x^2)$ .

Then, by the chain rule
$\frac{d}{d x} (arccos (4 x^{2})) = 2 \cdot (arccos (4 x^{2})) \cdot \frac{d}{d x} (arccos (4 x^{2}))$ $("d")/("d"x) ("arccos"(4x^2)) = 2*("arccos"(4x^2))*"d"/("d"x) ("arccos"(4x^2))$ .

Then, by the chain rule again,
$\frac{d}{d x} (arccos (4 x^{2})) = - \frac{1}{\sqrt{1 - 16 x^{4}}} \cdot \frac{d}{d x} (4 x^{2})$ $"d"/("d"x) ("arccos"(4x^2))=-1/sqrt(1-16x^4) * "d"/("d"x) (4x^2)$ ,
$\frac{d}{d x} (arccos (4 x^{2})) = - \frac{8 x}{\sqrt{1 - 16 x^{4}}}$ $"d"/("d"x) ("arccos"(4x^2))=-(8x)/sqrt(1-16x^4)$ .

Substituting,

$\frac{d}{d x} (arccos (4 x^{2})) = - \frac{16 x \cdot arccos (4 x^{2})}{\sqrt{1 - 16 x^{4}}}$ $("d")/("d"x) ("arccos"(4x^2)) = -(16x*"arccos"(4x^2))/(sqrt(1-16x^4))$ .

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What is the derivative of this function $y = {({cos}^{- 1} (4 x^{2}))}^{2}$ $y=(cos^-1(4x^2))^2$ ?

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What is the derivative of this function $y = {({cos}^{- 1} (4 x^{2}))}^{2}$ $y=(cos^-1(4x^2))^2$ ?