What is derivative of f(x)=e^r+r^e ?

The answer to derivative of f(x)=e^r+r^e is (dr)/(dx)=0

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derivative of f(x)=e^r+r^e

derivative of

f (x) = e^{r} + r^{e}

\frac{d r}{d x} = 0

Solution steps

f (x) = e^{r} + r^{e}

Treat

r

r (x)

Differentiate both sides:

0 = e^{r} \frac{d r}{d x} + e r^{e - 1} \frac{d r}{d x}

Isolate

\frac{d r}{d x} : \frac{d r}{d x} = 0

\frac{d r}{d x} = 0

\int \frac{3 x}{x ^{2}} d x

f^{^{'}} (x) = 3 sec (x)

\int_{- 1}^{4} \frac{x}{5 + x} d x

x \to 3 - lim (\frac{∣ x - 3 ∣}{x - 3})

\frac{d}{d x} (arcsin (1 - x))

What is derivative of f(x)=e^r+r^e ?
The answer to derivative of f(x)=e^r+r^e is (dr)/(dx)=0