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Mathematics · Ch 18 — Differentiation

Theorem 2 — Derivative of Difference of Functions

18.2.2

Theorem 2 — Derivative of Difference of Functions

Theorem 2. If uu and vv are differentiable functions of xx such that y=u−vy=u-v, then

dydx=dudx−dvdx\dfrac{dy}{dx}=\dfrac{du}{dx}-\dfrac{dv}{dx}

The book leaves this proof as an exercise, since it is identical in structure to the sum-rule proof of Theorem 1: give xx an increment δx\delta x, note δy=δu−δv\delta y=\delta u-\delta v, divide by δx\delta x and take the limit as δx→0\delta x\to0, using the differentiability of uu and vv exactly as before.

Corollary (linear combination rule). If u,v,w,…u,v,w,\ldots are any finite number of differentiable functions of xx, and k1,k2,k3,…k_1,k_2,k_3,\ldots are constants, then for

y=k1u±k2v±k3w±⋯y=k_1u\pm k_2v\pm k_3w\pm\cdots

dydx=k1dudx±k2dvdx±k3dwdx±⋯\dfrac{dy}{dx}=k_1\dfrac{du}{dx}\pm k_2\dfrac{dv}{dx}\pm k_3\dfrac{dw}{dx}\pm\cdots …

Misc 1Corollary — derivative of a linear combination

Worked out. If u,v,w,…u,v,w,\ldots are any finite number of differentiable functions of xx and k1,k2,k3,…k_1,k_2,k_3,\ldots are constants, then y=k1u±k2v±k3w±⋯y=k_1u\pm k_2v\pm k_3w\pm\cdots is differentiable and dydx=k1dudx±k2dvdx±k3dwdx±⋯\dfrac{dy}{dx}=k_1\dfrac{du}{dx}\pm k_2\dfrac{dv}{dx}\pm k_3\dfrac{dw}{dx}\pm\cdots. This single corollary is what actually gets used in every 'differentiate the following' problem with several algebraic/trig/log/exponential terms added or subtracted together with constant coefficients — it says each term can be differentiated completely independently of the others and the results simply combined with the same …