What is the derivative of (3^x+8/(3^x-1)) ?

The derivative of (3^x+8/(3^x-1)) is -(ln(3)*3^{2+x})/((3^x-1)^2)

What is the first derivative of (3^x+8/(3^x-1)) ?

The first derivative of (3^x+8/(3^x-1)) is -(ln(3)*3^{2+x})/((3^x-1)^2)

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Popular >

\frac{d}{d x} (\frac{3 ^{x} + 8}{3 ^{x} - 1})

- \frac{ln ( 3 ) \cdot 3 ^{2 + x}}{( 3 ^{x} - 1 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{3 ^{x} + 8}{3 ^{x} - 1})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( 3 ^{x} + 8 ) ( 3 ^{x} - 1 ) - \frac{d}{d x} ( 3 ^{x} - 1 ) ( 3 ^{x} + 8 )}{( 3 ^{x} - 1 ) ^{2}}

\frac{d}{d x} (3^{x} + 8) = ln (3) \cdot 3^{x}

\frac{d}{d x} (3^{x} - 1) = ln (3) \cdot 3^{x}

= \frac{ln ( 3 ) \cdot 3 ^{x} ( 3 ^{x} - 1 ) - ln ( 3 ) \cdot 3 ^{x} ( 3 ^{x} + 8 )}{( 3 ^{x} - 1 ) ^{2}}

Simplify

\frac{ln ( 3 ) \cdot 3 ^{x} ( 3 ^{x} - 1 ) - ln ( 3 ) \cdot 3 ^{x} ( 3 ^{x} + 8 )}{( 3 ^{x} - 1 ) ^{2}} : - \frac{ln ( 3 ) \cdot 3 ^{2 + x}}{( 3 ^{x} - 1 ) ^{2}}

= - \frac{ln ( 3 ) \cdot 3 ^{2 + x}}{( 3 ^{x} - 1 ) ^{2}}

\int e^{- 0.25 x} d x

\int 2 x y^{2} d x

\frac{tan ( \frac{x}{2} ) + cot ( \frac{x}{2} )}{x}

3 cos^{2} (t)

\frac{d y}{d x} = - 7 x^{2}

What is the derivative of (3^x+8/(3^x-1)) ?
The derivative of (3^x+8/(3^x-1)) is -(ln(3)*3^{2+x})/((3^x-1)^2)
What is the first derivative of (3^x+8/(3^x-1)) ?
The first derivative of (3^x+8/(3^x-1)) is -(ln(3)*3^{2+x})/((3^x-1)^2)