What is the derivative of (2e^{3x}-3/(x+1)) ?

The derivative of (2e^{3x}-3/(x+1)) is (6e^{3x}x+4e^{3x}+3)/((x+1)^2)

What is the first derivative of (2e^{3x}-3/(x+1)) ?

The first derivative of (2e^{3x}-3/(x+1)) is (6e^{3x}x+4e^{3x}+3)/((x+1)^2)

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

\frac{6 e ^{3 x} x + 4 e ^{3 x} + 3}{( x + 1 ) ^{2}}

Solution steps

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

Apply the Quotient Rule:

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

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

\frac{d}{d x} (2 e^{3 x} - 3) = 6 e^{3 x}

\frac{d}{d x} (x + 1) = 1

= \frac{6 e ^{3 x} ( x + 1 ) - 1 \cdot ( 2 e ^{3 x} - 3 )}{( x + 1 ) ^{2}}

Simplify

\frac{6 e ^{3 x} ( x + 1 ) - 1 \cdot ( 2 e ^{3 x} - 3 )}{( x + 1 ) ^{2}} : \frac{6 e ^{3 x} x + 4 e ^{3 x} + 3}{( x + 1 ) ^{2}}

= \frac{6 e ^{3 x} x + 4 e ^{3 x} + 3}{( x + 1 ) ^{2}}

\int a^{2} - x^{\frac{2}{3}} d x

x \to 0 lim (- 0. 5^{2} + 1)

x \to 3 lim (- 3 x^{2})

y^{^{'}} = 5 y - t y^{2}

\int (3 x^{2} - 5 x^{\frac{3}{2}} + 6 x) d x

What is the derivative of (2e^{3x}-3/(x+1)) ?
The derivative of (2e^{3x}-3/(x+1)) is (6e^{3x}x+4e^{3x}+3)/((x+1)^2)
What is the first derivative of (2e^{3x}-3/(x+1)) ?
The first derivative of (2e^{3x}-3/(x+1)) is (6e^{3x}x+4e^{3x}+3)/((x+1)^2)