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

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

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

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

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

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

Solution steps

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

Treat

A, B

as constants

Apply the Sum/Difference Rule:

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

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

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

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

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

(4 + x) y^{^{'}} = 8 y

\int (x^{2} - 7 x + 2) d x

y = e^{5 x}, at x = \frac{1}{5} ln (5)

\int 5 sin^{3} (x co) s^{7} x d x

\frac{d}{d x} (arccsc (4 x^{2} + 1))

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