Derivatives of Inverse Trigonometric Functions

When finding derivatives of inverse trig functions, we use specific formulas that might look complicated at first, but follow a clear pattern. The three main formulas are:

$\frac{d}{dx} \arcsin u = \frac{\frac{du}{dx}}{\sqrt{1-u^2}}$

$\frac{d}{dx} \arctan u = \frac{\frac{du}{dx}}{1+ u^2}$

$\frac{d}{dx} \arcsec u = \frac{\frac{du}{dx}}{u\sqrt{u^2-1}}$

Let's see how to apply these with an example: For $y = \arcsin 4x$, we identify $u = 4x$ and $\frac{du}{dx} = 4$. Plugging these into the formula, we get $\frac{dy}{dx} = \frac{4}{\sqrt{1- (4x)^2}} = \frac{4}{\sqrt{1- 16x^2}}$.

Pro Tip: Always start by clearly identifying what your "u" is and finding its derivative. This makes applying the formula much more straightforward!

Similarly for $y = \arctan 3x$, with $u = 3x$ and $\frac{du}{dx} = 3$, the derivative is $\frac{dy}{dx} = \frac{3}{1+9x^2}$.

More Complex Inverse Trig Derivatives

When working with fractions inside inverse trig functions, apply the formula carefully. For example, with $y = \arctan \frac{x}{a}$ where $a$ is constant, we identify $u = \frac{x}{a}$ and find $\frac{du}{dx} = \frac{1}{a}$.

Substituting into the formula: $\frac{dy}{dx} = \frac{\frac{1}{a}}{1 + (\frac{x}{a})^2}$. After simplifying, we get the elegant result: $\frac{dy}{dx} = \frac{a}{a^2 + x^2}$.

For problems involving products and compositions, like $x = (\arctan t)^2$, use the chain rule. Since this is in the form $u^n$ where $u = \arctan t$ and $n = 2$, the derivative is $\frac{dx}{dt} = \frac{2 \arctan t}{1+t^2}$.

Remember: When solving problems with products like $y = x^2 \arcsin x$, use the product rule: find the derivative of each part and combine them properly.

Products use the formula $(uv)' = u'v + uv'$, giving us $\frac{dy}{dx} = \frac{x^2}{\sqrt{1-x^2}} + 2x \arcsin x$ for the example above.

Complex Examples with Multiple Rules

More challenging problems combine multiple derivative rules. For example, with $y=(x-1) \sqrt{2x-x²} + \arcsin(x - 1)$, we need both the product rule and the inverse trig formula.

For the first term, we use the product rule where $u=(x-1)$ and $v=\sqrt{2x-x²}$. We find $u'=1$ and calculate $v'$ using the chain rule.

The second term uses the inverse sine formula with $u=x-1$, giving us $\frac{1}{\sqrt{1-(x-1)^2}}$.

Strategy Alert: Break complex expressions into manageable pieces. Solve each piece separately using the appropriate rules, then combine your results.

After careful algebraic manipulation and combining terms, the final derivative simplifies to $\frac{2x(2-x)}{\sqrt{2x-x²}}$, which is much cleaner than the original expression. This demonstrates how derivatives can sometimes be surprisingly simplified.

Final Examples and Practice

The last example showcases how derivatives can help us discover interesting relationships. When finding the derivative of $y = \frac{x}{\sqrt{a^2-x^2}} - \arcsin \frac{x}{a}$, we need both the quotient rule and the inverse trig formula.

For the first term, use the quotient rule with $u=x$ and $v=\sqrt{a^2-x^2}$. Then calculate the derivative of $\arcsin \frac{x}{a}$ using our formula with $u=\frac{x}{a}$.

When the terms are combined and simplified, we get the elegant result: $\frac{dy}{dx} = \frac{x^2}{(a^2-x^2)^{3/2}}$, which is surprisingly compact compared to the original expression.

You've got this! As you practice more problems, you'll develop an intuition for which techniques to apply and how terms might simplify.

The pattern of these problems shows that even complicated expressions can lead to relatively simple derivatives if you apply the rules carefully and look for opportunities to simplify.