Question

Find the derivative please

Answer 1

`\displaystyle{ y' = \frac{2}{1 - {x}^(2) }}`

Explanation:

Apply natural logarithm chain rule:

`\displaystyle{y = \ln(u) \to y' = (u')/(u)}`

Where u is a function. The ' symbol indicates that the function to be derived. From this formula, we will have:

`\displaystyle{y' = ( \left ((1 + x)/(1 - x) \right)')/( (1 + x)/(1 - x))}`

Now we have to derive quotient functions, we can use quotient rule for this:

`\displaystyle{y = (u)/(v) \to y' = \frac{u'v - uv'}{ {v}^(2) }}`

Where u and v are functions. Therefore, from numerator:

`\displaystyle{ \left( (1 + x)/(1 - x) \right)' = \frac{(1 + x)'(1 - x) - (1 + x)(1 - x)'}{ {(1 - x)}^(2) }}`

Derive using power rule and constant rule where:

`\displaystyle{y = {x}^(n) \to y' = n {x}^(n - 1) } \\ \\ \displaystyle{y = c \to y' = 0 \: \: \: \: ( c \in \mathbb{R})}`

Thus:

`\displaystyle{ \left( (1 + x)/(1 - x) \right)' = \frac{1(1 - x) - (1 + x)( - 1)}{ {(1 - x)}^(2) }} \\ \\ \displaystyle{ \left( (1 + x)/(1 - x) \right)' = \frac{1 - x- ( - 1 - x)}{ {(1 - x)}^(2) }} \\ \\ \displaystyle{ \left( (1 + x)/(1 - x) \right)' = \frac{1 - x + 1 + x}{ {(1 - x)}^(2) }} \\ \\ \displaystyle{ \left( (1 + x)/(1 - x) \right)' = \frac{2}{ {(1 - x)}^(2) }}`

Therefore, we will now have:

`\displaystyle{y' = \frac{ \frac{2}{ {(1 - x)}^(2) }}{ (1 + x)/(1 - x)}}`

Simplify expressions:

`\displaystyle{ \frac{ \frac{2}{ {(1 - x)}^(2) }}{ (1 + x)/(1 - x)} = \frac{2}{ {(1 - x)}^(2)} * (1 - x)/(1 + x) } \\ \\ \displaystyle{ \frac{ \frac{2}{ {(1 - x)}^(2) }}{ (1 + x)/(1 - x)} = (2)/( 1 - x) * (1 )/(1 + x) } \\ \\ \displaystyle{ \frac{ \frac{2}{ {(1 - x)}^(2) }}{ (1 + x)/(1 - x)} = (2)/( (1 - x)(1 + x))}`

Therefore, the derived function is:

`\displaystyle{ y' = \frac{2}{1 - {x}^(2) }}`