What is the derivative of y=ln(4t) ?

The derivative of y=ln(4t) is 1/t

What is the first derivative of y=ln(4t) ?

The first derivative of y=ln(4t) is 1/t

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derivative of

y = ln (4 t)

\frac{1}{t}

Solution steps

\frac{d}{d t} (ln (4 t))

Apply the chain rule:

\frac{1}{4 t} \frac{d}{d t} (4 t)

= \frac{1}{4 t} \frac{d}{d t} (4 t)

\frac{d}{d t} (4 t) = 4

= \frac{1}{4 t} \cdot 4

Simplify

\frac{1}{4 t} \cdot 4 : \frac{1}{t}

= \frac{1}{t}

y = 2 x^{2} - x^{4}, (1, 1)

\frac{d y}{d x} = - 2 x + 3 y - 5

\frac{\partial}{\partial z} (x + y + z)

\int e^{a t} d t

\frac{\partial}{\partial x} (\frac{1}{4} 2^{x} y^{3})

What is the derivative of y=ln(4t) ?
The derivative of y=ln(4t) is 1/t
What is the first derivative of y=ln(4t) ?
The first derivative of y=ln(4t) is 1/t