Question
Differentiate the following functions with respect to x:$\frac{1+3^\text{x}}{1-3^\text{x}}$
✓
Answer
We have,$\frac{\text{d}}{\text{dx}}\Big(\frac{1+3^\text{x}}{1-3^\text{x}}\Big)$
Using quotient rule, we get
$\frac{(1-3^\text{x})\frac{\text{d}}{\text{dx}}(1+3^\text{x})-(1+3^\text{x})\frac{\text{d}}{\text{dx}}(1-3^\text{x})}{(1-3^\text{x})^2}$
$=\frac{(1-3^\text{x})3^\text{x}\log3+(1+3^\text{x})3^\text{x}\log3}{(1-3^\text{x})^2}$
$=\frac{3^\text{x}\log3-3^\text{x}\times3^\text{x}\log3+3^\text{x}\log3+3^\text{x}\times3^\text{x}\log3}{(1-3^\text{x})^2}$
$=\frac{2\times3^\text{x}\log3}{(1-3^\text{x})^2}$
Using quotient rule, we get
$\frac{(1-3^\text{x})\frac{\text{d}}{\text{dx}}(1+3^\text{x})-(1+3^\text{x})\frac{\text{d}}{\text{dx}}(1-3^\text{x})}{(1-3^\text{x})^2}$
$=\frac{(1-3^\text{x})3^\text{x}\log3+(1+3^\text{x})3^\text{x}\log3}{(1-3^\text{x})^2}$
$=\frac{3^\text{x}\log3-3^\text{x}\times3^\text{x}\log3+3^\text{x}\log3+3^\text{x}\times3^\text{x}\log3}{(1-3^\text{x})^2}$
$=\frac{2\times3^\text{x}\log3}{(1-3^\text{x})^2}$
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