Calculate the value

In the following problem $a,b,$ b,and c represent REAL NUMBERS. 
The derivative of $g\left(x\right)=log _{a}\left(bx\right)+cx^{2}is$ 
$○$ $g^{1}\left(x\right)=\right)dfrac\left(b\right)log _{-}a\left(bx\right)\right)\left(N1n$ $a\right)+2cx1\right)$ 
$○$ $\left(g\left(x\right)=\right)dtracb$ $loga\left(bx\right)+2c\times \right)Nln$ a}\) 
$○$ $g^{'}\left(x\right)=\left(dfraC\left(a\right)log _{-}a\left(bx\right)\right)\left(ln$ $\right)+2c\times 1\right)\right)$ 
$○$ $g^{1}\left(x\right)=\left(dfrac\left(b$ $log _{-}a\left(bx\right)\right)\left(ln$ $a1\right)$ 
$○$ $\left(\left(g^{1}\left(x\right)=b$ $log _{-}a\left(bx\right)+2cx1\right)$ 
$○$ $g\left(x\right)=\left(dfrac\left(b$ $log _{-}a\left(bx\right)\right)\left(1ln$ $a|+2c1\right)$ 

$7\right)\dfrac {x} {4}=\dfrac {15} {20}$ $8,\right)$ $\dfrac {2} {5}x=-30$ 
$9.\right)$ $\dfrac {x} {-9}$ $=81$ $10.\right)$ $\dfrac {4} {9}=\dfrac {9} {x}$ 

Use a number line to compare the fractions. Use $sc<>,0r=$ 
$\dfrac {4} {9}$ $\dfrac {5} {9}$ 
$a$ $\dfrac {4} {9}<\dfrac {5} {9}$ b. $\dfrac {4} {9}=\dfrac {5} {9}$ $c$ $\dfrac {4} {9}>\dfrac {5} {9}$ 

I solved this question but I'm not sure if I'm right. Please provide a solution so that I can check my answer.

Express $5\dfrac {4} {9}$ as an improper fraction. 
$5\dfrac {4} {9}=\dfrac {5\times 9+4} {9}=\dfrac {45+4} {9}=\dfrac {49} {9}$ 

Which of the following rational numbers are 
equivalent? 
$0Ptionsy$ 
A \frac{5}{6}, \frac{30}{36} 
B $s\sqrt{rac\left(} -2\right)\left(3\right)\sqrt{1rac} \sqrt{4\right)16\right)4} $ 
C $s\sqrt{11aC\left(} -4\right)1-7b,\sqrt{1rac\left(16\sqrt{35\right)9} } $ 
D \frac{1}{2},\frac{3}{8} 

Calculator search results