What is (\partial)/(\partial x)(ln(t+2cx)) ?

The answer to (\partial)/(\partial x)(ln(t+2cx)) is (2c)/(t+2cx)

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(\partial)/(\partial x)(ln(t+2cx))

\frac{\partial}{\partial x} (ln (t + 2 c x))

\frac{2 c}{t + 2 c x}

Solution steps

\frac{\partial}{\partial x} (ln (t + 2 c x))

Treat

c, t

as a constant

Apply the chain rule:

\frac{1}{t + 2 c x} \frac{\partial}{\partial x} (t + 2 c x)

= \frac{1}{t + 2 c x} \frac{\partial}{\partial x} (t + 2 c x)

\frac{\partial}{\partial x} (t + 2 c x) = 2 c

= \frac{1}{t + 2 c x} \cdot 2 c

Simplify

\frac{1}{t + 2 c x} \cdot 2 c : \frac{2 c}{t + 2 c x}

= \frac{2 c}{t + 2 c x}

f (x) = a x^{b} + d e^{- c x}

\int (4^{x} - sec^{2} (x)) d x

\int \frac{d}{1} (e^{- x} y) d x

x \to 0 lim (\frac{e ^{2 x} - 1}{x})

\frac{d y}{d x} = \frac{x + y ^{2}}{2 y}, y (0) = 1

What is (\partial)/(\partial x)(ln(t+2cx)) ?
The answer to (\partial)/(\partial x)(ln(t+2cx)) is (2c)/(t+2cx)