What is the derivative of y=(7ln(x))/(3x+4) ?

The derivative of y=(7ln(x))/(3x+4) is (7(3x+4-3xln(x)))/(x(3x+4)^2)

What is the first derivative of y=(7ln(x))/(3x+4) ?

The first derivative of y=(7ln(x))/(3x+4) is (7(3x+4-3xln(x)))/(x(3x+4)^2)

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derivative of

y = \frac{7 ln ( x )}{3 x + 4}

\frac{7 ( 3 x + 4 - 3 x ln ( x ) )}{x ( 3 x + 4 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{7 ln ( x )}{3 x + 4})

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 7 \frac{d}{d x} (\frac{ln ( x )}{3 x + 4})

Apply the Quotient Rule:

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

= 7 \cdot \frac{\frac{d}{d x} ( ln ( x ) ) ( 3 x + 4 ) - \frac{d}{d x} ( 3 x + 4 ) ln ( x )}{( 3 x + 4 ) ^{2}}

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

\frac{d}{d x} (3 x + 4) = 3

= 7 \cdot \frac{\frac{1}{x} ( 3 x + 4 ) - 3 ln ( x )}{( 3 x + 4 ) ^{2}}

Simplify

7 \cdot \frac{\frac{1}{x} ( 3 x + 4 ) - 3 ln ( x )}{( 3 x + 4 ) ^{2}} : \frac{7 ( 3 x + 4 - 3 x ln ( x ) )}{x ( 3 x + 4 ) ^{2}}

= \frac{7 ( 3 x + 4 - 3 x ln ( x ) )}{x ( 3 x + 4 ) ^{2}}

x \to \frac{1}{3} lim (3 x^{3} + 2 x^{2})

y = \frac{x}{x ^{2} + 1}, at x = - 1

\frac{\partial}{\partial x} (2 x \cdot y - \frac{1}{2} x^{2})

f (t) = 6 sin (t)

\int_{- 1}^{3} \frac{x}{x ^{2} + 4} d x

What is the derivative of y=(7ln(x))/(3x+4) ?
The derivative of y=(7ln(x))/(3x+4) is (7(3x+4-3xln(x)))/(x(3x+4)^2)
What is the first derivative of y=(7ln(x))/(3x+4) ?
The first derivative of y=(7ln(x))/(3x+4) is (7(3x+4-3xln(x)))/(x(3x+4)^2)