What is the first d/(dt)(ln(t+1)) ?

The first d/(dt)(ln(t+1)) is 1/(t+1)

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d/(dt)(ln(t+1))

\frac{d}{d t} (ln (t + 1))

\frac{1}{t + 1}

Solution steps

\frac{d}{d t} (ln (t + 1))

Apply the chain rule:

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

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

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

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

Simplify

= \frac{1}{t + 1}

f (x) = \frac{2 x}{x + 2}, at x = 2

x y y^{^{'}} = 1 - x^{2}

\int \frac{1}{( x ^{2} - 64 ) ^{\frac{3}{2}}} d x

x \to 5 lim (c x^{2} + 6 x)

y = \frac{1}{2 x}