What is the derivative of ${ln [\frac{2 x - 3}{7 x + 8}]}^{\frac{1}{2}}$ $ln[(2x-3)/(7x+8)]^(1/2)$ ?

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Answer 1 · 2018-05-02T01:05:35

The answer $(\frac{1}{2 x - 3}) - (\frac{7}{14 x + 16})$ $(1/(2x-3))-(7/(14x+16))$

show below we will use the properties of ln

${ln [\frac{2 x - 3}{7 x + 8}]}^{\frac{1}{2}} = \frac{1}{2} \cdot ln [\frac{2 x - 3}{7 x + 8}]$ $ln[(2x-3)/(7x+8)]^(1/2)=1/2*ln[(2x-3)/(7x+8)]$

$\frac{1}{2} \cdot ln [\frac{2 x - 3}{7 x + 8}] = \frac{1}{2} [ln (2 x - 3) - ln (7 x + 8)]$ $1/2*ln[(2x-3)/(7x+8)]=1/2[ln(2x-3)-ln(7x+8)]$

$\frac{1}{2} ln (2 x - 3) - \frac{1}{2} ln (7 x + 8)$ $1/2ln(2x-3)-1/2ln(7x+8)$

now the derivative

$\frac{1}{2} \cdot (\frac{2}{2 x - 3}) - \frac{1}{2} \cdot (\frac{7}{7 x + 8})$ $1/2*(2/(2x-3))-1/2*(7/(7x+8))$

$(\frac{1}{2 x - 3}) - (\frac{7}{14 x + 16})$ $(1/(2x-3))-(7/(14x+16))$

What is the derivative of ln[(2x-3)/(7x+8)]^(1/2)ln[2x−37x+8]12ln[(2x-3)/(7x+8)]^(1/2)?