Inverse Trig And Log Derivatives

Derivatives of Inverse Trig and Log Functions

These four transcendental function derivatives appear constantly in calculus and are worth memorizing:

$\frac{d}{dx}[\ln x] = \frac{1}{x}, \quad \frac{d}{dx}[e^x] = e^x$

$\frac{d}{dx}[\arcsin x] = \frac{1}{\sqrt{1-x^2}}, \quad \frac{d}{dx}[\arctan x] = \frac{1}{1+x^2}$

The exponential function is remarkable: $e^x$ is its own derivative. The inverse trig derivatives produce algebraic expressions, not trigonometric ones—a common source of confusion.

Chain Rule Application

In practice, these rules typically form the outer function of a composite. For $\ln(g(x))$, $e^{g(x)}$, $\arcsin(g(x))$, or $\arctan(g(x))$, differentiate the outer function and multiply by the inner derivative $g'(x)$:

$\frac{d}{dx}[\ln(g(x))] = \frac{g'(x)}{g(x)}$

Examples

Logarithmic: For $f(x) = \ln(3x^2 + 1)$, treat $3x^2+1$ as the inner function with derivative $6x$: $f'(x) = \frac{1}{3x^2+1} \cdot 6x = \frac{6x}{3x^2+1}$

Inverse trig: For $h(x) = \arctan(5x)$ with inner function $5x$ and derivative $5$: $h'(x) = \frac{1}{1+(5x)^2} \cdot 5 = \frac{5}{1+25x^2}$

For $k(x) = \arcsin(x^2)$ with inner derivative $2x$: $k'(x) = \frac{1}{\sqrt{1-x^4}} \cdot 2x = \frac{2x}{\sqrt{1-x^4}}$

Geometric Insight

$\text{Slope of } e^x \text{ at any point} = \text{height at that point}$

For $\ln x$, the slope $\frac{1}{x}$ decreases as $x$ grows, reflecting the curve's visible flattening.

Common Pitfall

Don't forget the chain rule. The formulas apply only when the argument is exactly $x$. If you see $\ln(3x^2+1)$ or $e^{x^3}$, you must multiply by the inner derivative—these are not $\frac{1}{3x^2+1}$ or $e^{x^3}$ alone.