Inverse Trigonometric Differentiation Rules

A derivative of a function is the rate of change of the function or the slope of the line at a given point. The derivative of f(a) is notated as ${f}^{\prime }\left(a\right)$ or $\frac{d}{dx}f\left(a\right)$.

This discussion will focus on the basic Inverse Trigonometric Differentiation Rules. There are two different inverse function notations for trigonometric functions. The inverse function for sinx can be written as sin-1x or arcsin x.

DERIVATIVES OF INVERSE TRIGONOMETRIC FUNCTIONS:

 FUNCTION DERIVATIVE FUNCTION DERIVATIVE $\frac{d}{dx}{\mathrm{sin}}^{-1}x$ $\frac{1}{\sqrt{1-{x}^{2}}}$ $\frac{d}{dx}{\mathrm{csc}}^{-1}x$ $-\frac{1}{x\sqrt{{x}^{2}-1}}$ $\frac{d}{dx}{\mathrm{cos}}^{-1}x$ $-\frac{1}{\sqrt{1-{x}^{2}}}$ $\frac{d}{dx}{\mathrm{sec}}^{-1}x$ $\frac{1}{x\sqrt{{x}^{2}-1}}$ $\frac{d}{dx}{\mathrm{tan}}^{-1}x$ $\frac{1}{1+{x}^{2}}$ $\frac{d}{dx}{\mathrm{cot}}^{-1}x$ $-\frac{1}{1+{x}^{2}}$

Let's look at some examples:

To work these examples requires the use of various differentiation rules. If you are not familiar with a rule go to the associated topic for a review.

2cos-1 x
 Step 1: Apply the Constant Multiple Rule. $\frac{d}{dx}\left[cf\left(x\right)\right]=c\frac{d}{dx}f\left(x\right)$ $2\frac{d}{dx}{\mathrm{cos}}^{-1}x$ Constant Mul. Step 2: Take the derivative of cos-1x. Arccos Rule $\frac{2}{\sqrt{1-{x}^{2}}}$
Example 1:      (sin-1 x)3
 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 $3{u}^{2}$ __________________________ Derivative of g = sin-1 x $\frac{d}{dx}{\mathrm{sin}}^{-1}x$ 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. $\left(f\circ g\right)\left(x\right)={f}^{\prime }\left(g\left(x\right)\right)·g\prime \left(x\right)$ Chain Rule  Sub for u $\frac{3{\left(si{n}^{-1}x\right)}^{2}}{\sqrt{1-{x}^{2}}}$
Example 2:      $\frac{5ta{n}^{-1}x}{1+{x}^{2}}$
 Step 1: Apply the quotient rule. $\frac{d}{dx}\left[\frac{f\left(x\right)}{g\left(x\right)}\right]=\frac{g\left(x\right)\frac{d}{dx}\left[f\left(x\right)\right]-f\left(x\right)\frac{d}{dx}\left[g\left(x\right)\right]}{{\left[g\left(x\right)\right]}^{2}}$ $\frac{d}{dx}\left[\frac{5ta{n}^{-1}x}{1+{x}^{2}}\right]$ Step 2: Take the derivative of each part. Apply the appropriate trigonometric differentiation rule. $\frac{d}{dx}5{\mathrm{tan}}^{-1}x$ Original $5\frac{d}{dx}{\mathrm{tan}}^{-1}x$ Constant Multiple Rule $\frac{5}{1+{x}^{2}}$       Arctan Rule $\frac{5}{1+{x}^{2}}$ __________________________ $\frac{d}{dx}1+{x}^{2}$ Original $\frac{d}{dx}1+\frac{d}{dx}{x}^{2}$       Sum Rule 0 + 2x  Constant/Power $2x$ Step 3: Substitute the derivatives & simplify. $\frac{\left[\left(1+{x}^{2}\right)\left(\frac{5}{1+{x}^{2}}\right)\right]-\left[\left(5{\mathrm{tan}}^{-1}x\right)\left({2}{x}\right)\right]}{{\left(1+{x}^{2}\right)}^{2}}$

 Related Links: Math algebra Logarithmic Differentiation Implicit Differentiation Calculus Topics

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