What is the derivative of y=ln(8-4x) ?

The derivative of y=ln(8-4x) is -1/(2-x)

What is the first derivative of y=ln(8-4x) ?

The first derivative of y=ln(8-4x) is -1/(2-x)

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derivative of

y = ln (8 - 4 x)

- \frac{1}{2 - x}

Solution steps

\frac{d}{d x} (ln (8 - 4 x))

Apply the chain rule:

\frac{1}{8 - 4 x} \frac{d}{d x} (8 - 4 x)

= \frac{1}{8 - 4 x} \frac{d}{d x} (8 - 4 x)

\frac{d}{d x} (8 - 4 x) = - 4

= \frac{1}{8 - 4 x} (- 4)

Simplify

\frac{1}{8 - 4 x} (- 4) : - \frac{1}{2 - x}

= - \frac{1}{2 - x}

f (x) = 21 x

y = 8 x sin (x), (\frac{π}{2}, 4 π)

y = \frac{1}{x ^{5}}, (1, 1)

\frac{\partial}{\partial x} ((x_{1} + y_{2})^{2})

3 x + 2

What is the derivative of y=ln(8-4x) ?
The derivative of y=ln(8-4x) is -1/(2-x)
What is the first derivative of y=ln(8-4x) ?
The first derivative of y=ln(8-4x) is -1/(2-x)