What is the derivative of f(x)=(log_{9}(x))/(7+x^9) ?

The derivative of f(x)=(log_{9}(x))/(7+x^9) is (7+x^9-18ln(3)x^9log_{9}(x))/(2ln(3)x(7+x^9)^2)

What is the first derivative of f(x)=(log_{9}(x))/(7+x^9) ?

The first derivative of f(x)=(log_{9}(x))/(7+x^9) is (7+x^9-18ln(3)x^9log_{9}(x))/(2ln(3)x(7+x^9)^2)

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derivative of

f (x) = \frac{lo g _{9} ( x )}{7 + x ^{9}}

\frac{7 + x ^{9} - 18 ln ( 3 ) x ^{9} lo g _{9} ( x )}{2 ln ( 3 ) x ( 7 + x ^{9} ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{lo g _{9} ( x )}{7 + x ^{9}})

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= \frac{\frac{d}{d x} ( lo g _{9} ( x ) ) ( 7 + x ^{9} ) - \frac{d}{d x} ( 7 + x ^{9} ) lo g _{9} ( x )}{( 7 + x ^{9} ) ^{2}}

\frac{d}{d x} (lo g_{9} (x)) = \frac{1}{2 ln ( 3 ) x}

\frac{d}{d x} (7 + x^{9}) = 9 x^{8}

= \frac{\frac{1}{2 l n ( 3 ) x} ( 7 + x ^{9} ) - 9 x ^{8} lo g _{9} ( x )}{( 7 + x ^{9} ) ^{2}}

Simplify

\frac{\frac{1}{2 l n ( 3 ) x} ( 7 + x ^{9} ) - 9 x ^{8} lo g _{9} ( x )}{( 7 + x ^{9} ) ^{2}} : \frac{7 + x ^{9} - 18 ln ( 3 ) x ^{9} lo g _{9} ( x )}{2 ln ( 3 ) x ( 7 + x ^{9} ) ^{2}}

= \frac{7 + x ^{9} - 18 ln ( 3 ) x ^{9} lo g _{9} ( x )}{2 ln ( 3 ) x ( 7 + x ^{9} ) ^{2}}

g (x) = 9 e^{x} x

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

\frac{\partial}{\partial x} (\frac{x - y}{3 x y + 2})

\int \frac{2 + ln ( x )}{x ln ( x )} d x

y = 3 x^{2} + 3 x

What is the derivative of f(x)=(log_{9}(x))/(7+x^9) ?
The derivative of f(x)=(log_{9}(x))/(7+x^9) is (7+x^9-18ln(3)x^9log_{9}(x))/(2ln(3)x(7+x^9)^2)
What is the first derivative of f(x)=(log_{9}(x))/(7+x^9) ?
The first derivative of f(x)=(log_{9}(x))/(7+x^9) is (7+x^9-18ln(3)x^9log_{9}(x))/(2ln(3)x(7+x^9)^2)