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Question Number 215918 by issac last updated on 21/Jan/25

$$\mathrm{why}\frac{\mathrm{d}}{\mathrm{d}z}{e}^z=e^z??$$$$\mathrm{why}\mathrm{dose}\mathrm{not}\frac{\mathrm{d}{e}^z}{\mathrm{d}z}=\frac{\cancel{\mathrm{d}}{e}^z}{\cancel{\mathrm{d}}z}=\frac{e^z}{z}???$$

Answered by Frix last updated on 21/Jan/25

$$\mathrm{You}\mathrm{mean}\mathrm{generally}\frac{d}{dx}f\left(x\right)=\frac{f\left(x\right)}{x}?$$

Answered by mahdipoor last updated on 21/Jan/25

$$$$$$\mathrm{The}\mathrm{symbol}d\mathrm{here}\mathrm{is}\mathrm{an}\mathrm{operator}:$$$$\frac{\mathrm{d}}{\mathrm{d}z}{e}^z\equiv \frac{e^{z+\mathrm{\Delta }z}-e^z}{\mathrm{\Delta }z}when(\mathrm{\Delta }z\to 0)$$$$\mathrm{whose}\mathrm{name}\mathrm{is}\mathrm{derivative}$$$$notmean:$$$$\frac{d}{dz}{e}^z=\frac{d\times e^z}{d\times z}$$

Answered by MathematicalUser2357 last updated on 29/Mar/25

$$\mathrm{You}\mathrm{maybe}\mathrm{forgot}\mathrm{the}\mathrm{notation}\mathrm{of}\mathrm{derivatives}.$$$$\mathrm{The}\mathrm{derivative}\mathrm{notations}\mathrm{are}:\frac{dy}{dx},\frac{d}{dx}y,y^{\prime },\mathrm{etc}.\left(\mathrm{Derivative}\mathrm{notation}\mathrm{is}\mathrm{marked}\mathrm{blue}\right)$$$$\mathrm{Solving}\mathrm{it}\mathrm{with}\mathrm{your}\mathrm{derivative}\mathrm{skills}\mathrm{with}\mathrm{that}\mathrm{notation}\mathrm{you}\mathrm{forgot}:$$$$\frac{d}{dz}{e}^z=e^z.$$$$\mathrm{If}\mathrm{you}{\mathrm{didn}}^{\prime }\mathrm{t}\mathrm{forgot}\mathrm{the}\mathrm{notation}\mathrm{of}\mathrm{derivatives},$$$$\mathrm{You}\mathrm{maybe}\mathrm{forgot}\mathrm{the}\mathrm{first}\mathrm{principle}\mathrm{of}\mathrm{derivatives}.$$$$\mathrm{If}\mathrm{you}{\mathrm{didn}}^{\prime }\mathrm{t}\mathrm{forgot}\mathrm{that},\mathrm{you}\mathrm{thinked}\frac{d}{dz}{e}^z\mathrm{as}\mathrm{an}\mathrm{Algebra}\mathrm{question}.$$$$\mathrm{Please},\mathrm{let}\mathrm{that}\mathrm{notation}\mathrm{go}\mathrm{to}\mathrm{your}\mathrm{mind}.$$

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