What is the derivative of y=10000+30xe^{x/(800)} ?

The derivative of y=10000+30xe^{x/(800)} is 30((e^{x/(800)}x)/(800)+e^{x/(800)})

What is the first derivative of y=10000+30xe^{x/(800)} ?

The first derivative of y=10000+30xe^{x/(800)} is 30((e^{x/(800)}x)/(800)+e^{x/(800)})

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derivative of

y = 10000 + 30 x e^{\frac{x}{800}}

30 (\frac{e ^{\frac{x}{800}} x}{800} + e^{\frac{x}{800}})

Solution steps

\frac{d}{d x} (10000 + 30 x e^{\frac{x}{800}})

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d x} (10000) + \frac{d}{d x} (30 x e^{\frac{x}{800}})

\frac{d}{d x} (10000) = 0

\frac{d}{d x} (30 x e^{\frac{x}{800}}) = 30 (e^{\frac{x}{800}} + \frac{e ^{\frac{x}{800}} x}{800})

= 0 + 30 (e^{\frac{x}{800}} + \frac{e ^{\frac{x}{800}} x}{800})

Simplify

= 30 (\frac{e ^{\frac{x}{800}} x}{800} + e^{\frac{x}{800}})

y^{^{'}} - \frac{4}{x} y = \frac{y ^{4}}{x ^{8}}

\int_{0}^{y + z} \frac{1}{( z + y + x ) ^{3}} d x

y^{^{'}} = \frac{1}{x} y - (\frac{x ^{2}}{3}) y^{4}

x + 2 y = 2, y - x = 1, 2 x + y = 7

\frac{d}{d x} (\frac{π}{x})

What is the derivative of y=10000+30xe^{x/(800)} ?
The derivative of y=10000+30xe^{x/(800)} is 30((e^{x/(800)}x)/(800)+e^{x/(800)})
What is the first derivative of y=10000+30xe^{x/(800)} ?
The first derivative of y=10000+30xe^{x/(800)} is 30((e^{x/(800)}x)/(800)+e^{x/(800)})