What is the derivative of y=(4ln(3x))/(4-5x) ?

The derivative of y=(4ln(3x))/(4-5x) is (4(4-5x+5xln(3x)))/(x(4-5x)^2)

What is the first derivative of y=(4ln(3x))/(4-5x) ?

The first derivative of y=(4ln(3x))/(4-5x) is (4(4-5x+5xln(3x)))/(x(4-5x)^2)

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derivative of

y = \frac{4 ln ( 3 x )}{4 - 5 x}

\frac{4 ( 4 - 5 x + 5 x ln ( 3 x ) )}{x ( 4 - 5 x ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{4 ln ( 3 x )}{4 - 5 x})

Take the constant out:

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

= 4 \frac{d}{d x} (\frac{ln ( 3 x )}{4 - 5 x})

Apply the Quotient Rule:

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

= 4 \cdot \frac{\frac{d}{d x} ( ln ( 3 x ) ) ( 4 - 5 x ) - \frac{d}{d x} ( 4 - 5 x ) ln ( 3 x )}{( 4 - 5 x ) ^{2}}

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

\frac{d}{d x} (4 - 5 x) = - 5

= 4 \cdot \frac{\frac{1}{x} ( 4 - 5 x ) - ( - 5 ) ln ( 3 x )}{( 4 - 5 x ) ^{2}}

Simplify

4 \cdot \frac{\frac{1}{x} ( 4 - 5 x ) - ( - 5 ) ln ( 3 x )}{( 4 - 5 x ) ^{2}} : \frac{4 ( 4 - 5 x + 5 x ln ( 3 x ) )}{x ( 4 - 5 x ) ^{2}}

= \frac{4 ( 4 - 5 x + 5 x ln ( 3 x ) )}{x ( 4 - 5 x ) ^{2}}

y^{^{''}} + y^{^{'}} - 2 y = 0, y (0) = 1, y^{^{'}} (0) = 1

(3 cos (t))^{^{'}}

5 x + \frac{7}{x}

\frac{d}{d x} (x^{3} - 9 x + 3)

f (x) = \frac{1}{x + 2}, [3, 6]

What is the derivative of y=(4ln(3x))/(4-5x) ?
The derivative of y=(4ln(3x))/(4-5x) is (4(4-5x+5xln(3x)))/(x(4-5x)^2)
What is the first derivative of y=(4ln(3x))/(4-5x) ?
The first derivative of y=(4ln(3x))/(4-5x) is (4(4-5x+5xln(3x)))/(x(4-5x)^2)