What is the d/(dy)(A/(y^7)+Be^y) ?

The d/(dy)(A/(y^7)+Be^y) is -(7A)/(y^8)+Be^y

What is the first d/(dy)(A/(y^7)+Be^y) ?

The first d/(dy)(A/(y^7)+Be^y) is -(7A)/(y^8)+Be^y

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\frac{d}{d y} (\frac{A}{y ^{7}} + B e^{y})

- \frac{7 A}{y ^{8}} + B e^{y}

Solution steps

\frac{d}{d y} (\frac{A}{y ^{7}} + B e^{y})

Treat

A, B

as constants

Apply the Sum/Difference Rule:

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

= \frac{d}{d y} (\frac{A}{y ^{7}}) + \frac{d}{d y} (B e^{y})

\frac{d}{d y} (\frac{A}{y ^{7}}) = - \frac{7 A}{y ^{8}}

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

= - \frac{7 A}{y ^{8}} + B e^{y}

\int 6 x d x

f (x) = \frac{x ^{3} + 2 x ^{2} - 4}{3}

\int (x^{2} + 1) e^{x^{3} + 3 x} d x

\int_{0}^{1} \frac{1}{x ^{0.4}} d x

\frac{1}{2} (8 t + 3)^{- \frac{1}{2}}

What is the d/(dy)(A/(y^7)+Be^y) ?
The d/(dy)(A/(y^7)+Be^y) is -(7A)/(y^8)+Be^y
What is the first d/(dy)(A/(y^7)+Be^y) ?
The first d/(dy)(A/(y^7)+Be^y) is -(7A)/(y^8)+Be^y