What is the derivative of g(t)=sqrt(1+ln(t)) ?

The derivative of g(t)=sqrt(1+ln(t)) is 1/(2tsqrt(1+ln(t)))

What is the first derivative of g(t)=sqrt(1+ln(t)) ?

The first derivative of g(t)=sqrt(1+ln(t)) is 1/(2tsqrt(1+ln(t)))

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derivative of

g (t) = 1 + ln (t)

\frac{1}{2 t 1 + ln ( t )}

Solution steps

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

Apply the chain rule:

\frac{1}{2 1 + ln ( t )} \frac{d}{d t} (1 + ln (t))

= \frac{1}{2 1 + ln ( t )} \frac{d}{d t} (1 + ln (t))

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

= \frac{1}{2 1 + ln ( t )} \cdot \frac{1}{t}

Simplify

\frac{1}{2 1 + ln ( t )} \cdot \frac{1}{t} : \frac{1}{2 t 1 + ln ( t )}

= \frac{1}{2 t 1 + ln ( t )}

What is the derivative of g(t)=sqrt(1+ln(t)) ?
The derivative of g(t)=sqrt(1+ln(t)) is 1/(2tsqrt(1+ln(t)))
What is the first derivative of g(t)=sqrt(1+ln(t)) ?
The first derivative of g(t)=sqrt(1+ln(t)) is 1/(2tsqrt(1+ln(t)))