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10 cards from this deck
exe^xex
aeax+bae^{ax+b}aeax+b
When exponent is more complex than linear expression
dydx=dydu×dudx\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}dxdy=dudy×dxdu
eududxe^u\frac{du}{dx}eudxdu
No, exe^xex is never zero
dydx=vdudx+udvdx\frac{dy}{dx} = v\frac{du}{dx} + u\frac{dv}{dx}dxdy=vdxdu+udxdv
dydx=vdudx−udvdxv2\frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}dxdy=v2vdxdu−udxdv
Numerator equals zero (denominator non-zero)
To help identify stationary points
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