Answer
$\frac{dt}{dc}=\frac{3\sqrt {9c^2-8c}+ (9c-4)}{\sqrt {9c^2-8c}.(3c+\sqrt {9c^2-8c})}$
Work Step by Step
Given :
$t=ln(\frac{3c+\sqrt {9c^2-8c}}{2})$
$t=ln(3c+\sqrt {9c^2-8c})-ln2$
Taking differentiate w.r.t $c$.
$\frac{dt}{dc}=\frac{d}{dc}(ln(3c+\sqrt {9c^2-8c})-ln2)$
$=\frac{d}{dc}ln(3c+\sqrt {9c^2-8c})-\frac{d}{dc}(ln2)$
$=\frac{1}{3c+\sqrt {9c^2-8c}}\frac{d}{dc}(3c+\sqrt {9c^2-8c})-\frac{d}{dc}(ln2)$
$=\frac{1}{3c+\sqrt {9c^2-8c}}(3+\frac{1}{2}(9c^2-8c)^{\frac{-1}{2}}\frac{d}{dc}(9c^2-8c)-0)$
$=\frac{1}{3c+\sqrt {9c^2-8c}}(3+\frac{1}{2}(9c^2-8c)^{\frac{-1}{2}}(18c-8))$
$=\frac{3+\frac{(18c-8)}{2\sqrt {9c^2-8c}}}{3c+\sqrt {9c^2-8c}}$
$=\frac{3+\frac{2(9c-4)}{2\sqrt {9c^2-8c}}}{3c+\sqrt {9c^2-8c}}$
$=\frac{3+\frac{(9c-4)}{\sqrt {9c^2-8c}}}{3c+\sqrt {9c^2-8c}}$
$=\frac{\frac{3\sqrt {9c^2-8c}+ (9c-4)}{\sqrt {9c^2-8c}}}{3c+\sqrt {9c^2-8c}}$
$\frac{dt}{dc}=\frac{3\sqrt {9c^2-8c}+ (9c-4)}{\sqrt {9c^2-8c}.(3c+\sqrt {9c^2-8c})}$
