What is the d/(dy)((ln(y))/(2sqrt(y))+1/(sqrt(y))) ?

The d/(dy)((ln(y))/(2sqrt(y))+1/(sqrt(y))) is -(ln(y))/(4y^{3/2)}

What is the first d/(dy)((ln(y))/(2sqrt(y))+1/(sqrt(y))) ?

The first d/(dy)((ln(y))/(2sqrt(y))+1/(sqrt(y))) is -(ln(y))/(4y^{3/2)}

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العربية

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

- \frac{ln ( y )}{4 y ^{\frac{3}{2}}}

Solution steps

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

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

\frac{d}{d y} (\frac{ln ( y )}{2 y}) = \frac{2 - ln ( y )}{4 y y}

\frac{d}{d y} (\frac{1}{y}) = - \frac{1}{2 y ^{\frac{3}{2}}}

= \frac{2 - ln ( y )}{4 y y} - \frac{1}{2 y ^{\frac{3}{2}}}

Simplify

\frac{2 - ln ( y )}{4 y y} - \frac{1}{2 y ^{\frac{3}{2}}} : - \frac{ln ( y )}{4 y ^{\frac{3}{2}}}

= - \frac{ln ( y )}{4 y ^{\frac{3}{2}}}

What is the d/(dy)((ln(y))/(2sqrt(y))+1/(sqrt(y))) ?
The d/(dy)((ln(y))/(2sqrt(y))+1/(sqrt(y))) is -(ln(y))/(4y^{3/2)}
What is the first d/(dy)((ln(y))/(2sqrt(y))+1/(sqrt(y))) ?
The first d/(dy)((ln(y))/(2sqrt(y))+1/(sqrt(y))) is -(ln(y))/(4y^{3/2)}