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

The derivative of A/(y^9)+Be^y is -(9A)/(y^{10)}+Be^y

What is the first derivative of A/(y^9)+Be^y ?

The first derivative of A/(y^9)+Be^y is -(9A)/(y^{10)}+Be^y

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derivative of

\frac{A}{y ^{9}} + B e^{y}

- \frac{9 A}{y ^{10}} + B e^{y}

Solution steps

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

Apply the Sum/Difference Rule:

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

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

\frac{d}{d y} (\frac{A}{y ^{9}}) = - \frac{9 A}{y ^{10}}

\frac{d}{d y} (B e^{y}) = B e^{y}

= - \frac{9 A}{y ^{10}} + B e^{y}

What is the derivative of A/(y^9)+Be^y ?
The derivative of A/(y^9)+Be^y is -(9A)/(y^{10)}+Be^y
What is the first derivative of A/(y^9)+Be^y ?
The first derivative of A/(y^9)+Be^y is -(9A)/(y^{10)}+Be^y