Differentiation of Exponential Functions (HSC SSCE Mathematics Advanced): Flashcards

$e^x$

$ae^{ax+b}$

When exponent is more complex than linear expression

$\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$

$e^u\frac{du}{dx}$

No, $e^x$ is never zero

$\frac{dy}{dx} = v\frac{du}{dx} + u\frac{dv}{dx}$

$\frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$

Numerator equals zero (denominator non-zero)

To help identify stationary points