Inverse Trigonometric Differentiation Rules

Step 1: Apply the Constant Multiple Rule.

$\frac{d}{dx}\left[cf\left(x\right)\right]=c\frac{d}{dx}f\left(x\right)$

$2\frac{d}{dx}{\cos }^{-1}x$ Constant Mul.

Step 2: Take the derivative of cos^-1x.

$2 ·\frac{1}{\sqrt{1-x^2}}$    Arccos Rule

$\frac{2}{\sqrt{1-x^2}}$

Step 1: Apply the chain rule.

$\left(f\circ g\right)\left(x\right)=f^{\prime }\left(g\left(x\right)\right)·g\prime \left(x\right)$

g = sin^-1 x

u = sin^-1 x

f = u³

Step 2: Take the derivative of both functions.

Derivative of f = u³

$\frac{d}{dx}{u}^3$       Original

3u²          Power

$3u^2$

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Derivative of g = sin^-1 x

$\frac{d}{dx}{\sin }^{-1}x$ Original

$\frac{1}{\sqrt{1-x^2}}$      Arcsin Rule

$\frac{1}{\sqrt{1-x^2}}$

Step 3: Substitute the derivatives and the original expression for the variable u into the Chain Rule and simplify.

$\left(f\circ g\right)\left(x\right)=f^{\prime }\left(g\left(x\right)\right)·g\prime \left(x\right)$

$3u^2\left(\frac{1}{\sqrt{1-x^2 }}\right)$ Chain Rule

$3{\left({\sin }^{-1}x\right)}^2\left(\frac{1}{\sqrt{1-x^2 }}\right)$  Sub for u

$\frac{3{\left(si{n}^{-1}x\right)}^2}{\sqrt{1-x^2}}$

Step 1: Apply the quotient rule.

$\frac{d}{dx}\left[\frac{f\left(x\right)}{g\left(x\right)}\right]=\frac{g\left(x\right)\frac{d}{dx}\left[f\left(x\right)\right]-f\left(x\right)\frac{d}{dx}\left[g\left(x\right)\right]}{{\left[g\left(x\right)\right]}^2}$

$\frac{d}{dx}\left[\frac{5ta{n}^{-1}x}{1+x^2}\right]$

$\frac{\left[\left(1+x^2\right)\frac{d}{dx}5{\tan }^{-1}x]-[5{\tan }^{-1} x\frac{d}{dx} \left(1+x^2\right)\right]}{{\left(1+x^2\right)}^2}$

Step 2: Take the derivative of each part.

Apply the appropriate trigonometric differentiation rule.

$\frac{d}{dx}5{\tan }^{-1}x$ Original

$5\frac{d}{dx}{\tan }^{-1}x$ Constant Multiple Rule

$\frac{5}{1+x^2}$       Arctan Rule

$\frac{5}{1+x^2}$

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$\frac{d}{dx}1+x^2$ Original

$\frac{d}{dx}1+\frac{d}{dx}{x}^2$       Sum Rule

0 + 2x  Constant/Power

$2x$

Step 3: Substitute the derivatives & simplify.

$\frac{\left[\left(1+x^2\right)\left(\frac{5}{1+x^2}\right)\right]-\left[\left(5{\tan }^{-1}x\left)\right(2x\right)\right]}{{\left(1+x^2\right)}^2}$

$\frac{5-10x ta{n}^{-1}x}{{\left(1+x^2\right)}^2}$