What is the d/(dt)(log_{3}(2t-x)) ?

The d/(dt)(log_{3}(2t-x)) is (2-\frac{dx)/(dt)}{ln(3)(2t-x)}

What is the first d/(dt)(log_{3}(2t-x)) ?

The first d/(dt)(log_{3}(2t-x)) is (2-\frac{dx)/(dt)}{ln(3)(2t-x)}

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\frac{d}{d t} (lo g_{3} (2 t - x))

\frac{2 - \frac{d x}{d t}}{ln ( 3 ) ( 2 t - x )}

Solution steps

\frac{d}{d t} (lo g_{3} (2 t - x))

Apply log rule:

lo g_{a} (b) = \frac{ln ( b )}{ln ( a )}

= \frac{d}{d t} (\frac{ln ( 2 t - x )}{ln ( 3 )})

Take the constant out:

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

= \frac{1}{ln ( 3 )} \frac{d}{d t} (ln (2 t - x))

Apply the chain rule:

\frac{1}{2 t - x} \frac{d}{d t} (2 t - x)

= \frac{1}{2 t - x} \frac{d}{d t} (2 t - x)

\frac{d}{d t} (2 t - x) = 2 - \frac{d x}{d t}

= \frac{1}{ln ( 3 )} \cdot \frac{1}{2 t - x} (2 - \frac{d x}{d t})

Simplify

\frac{1}{ln ( 3 )} \cdot \frac{1}{2 t - x} (2 - \frac{d x}{d t}) : \frac{2 - \frac{d x}{d t}}{ln ( 3 ) ( 2 t - x )}

= \frac{2 - \frac{d x}{d t}}{ln ( 3 ) ( 2 t - x )}

y = 2 x^{3} + 3 x^{2} - 12 x + 1

\frac{d}{d r} (r + 3 r)

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

cos (x sin (x))

\frac{d}{d x} (\frac{- 6 x}{1 - x})

What is the d/(dt)(log_{3}(2t-x)) ?
The d/(dt)(log_{3}(2t-x)) is (2-\frac{dx)/(dt)}{ln(3)(2t-x)}
What is the first d/(dt)(log_{3}(2t-x)) ?
The first d/(dt)(log_{3}(2t-x)) is (2-\frac{dx)/(dt)}{ln(3)(2t-x)}