What is derivative of f(y)= A/(y^9)+Be^y ?

The answer to derivative of f(y)= A/(y^9)+Be^y is (dy)/(dx)=0

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derivative of f(y)= A/(y^9)+Be^y

derivative of

f (y) = \frac{A}{y ^{9}} + B e^{y}

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

Solution steps

f (y) = \frac{A}{y ^{9}} + B e^{y}

Treat

y

y (x)

Differentiate both sides:

0 = - \frac{9 A \frac{d y}{d x}}{y ^{10}} + e^{y} \frac{d y}{d x}

Isolate

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

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

x \to 2 - lim (\frac{3}{x - 2})

\frac{d}{d x} ((cos (x))^{2} + sin (x^{2}))

\int \frac{x}{( x ^{2} + R ^{2} ) ^{\frac{3}{2}}} d x

\int (x^{2} + 3 x) 2^{x} d x

\int \frac{x ^{3} + 4 x - 13}{x ^{2} + 4} d x

What is derivative of f(y)= A/(y^9)+Be^y ?
The answer to derivative of f(y)= A/(y^9)+Be^y is (dy)/(dx)=0