How do you find the derivative using quotient rule and chain rule for 1√1−x2?

1 Answer
Tiago Hands
Apr 22, 2015

Differentiating ln(1+x)

Say: p=ln(1+x)=lnu

dpdu=1u=11+x

u=1+x, dudx=1

Therefore:

dpdx=11+x

Differentiating ln(1−x)

Say:

q=ln(1−x)=lnu

dqdu=1u=11−x

u=1−x, dudx=−1

Therefore:

dqdx=−11−x

Differentiating y=1√1−x2

y=1√1−x2

lny=ln(1√1−x2)

lny=ln1−ln((1−x2)12)

lny=−12ln(1−x2)

lny=−12ln((1+x)(1−x))

lny=−12{ln(1+x)+ln(1−x)}

lny=−12ln(1+x)−12ln(1−x)

1y⋅dydx=−12{11+x}−12{−11−x}

1y⋅dydx=12(1−x)−12(1+x)

y⋅1y⋅dydx=y{12(1−x)−12(1+x)}

dydx=1√1−x2{12(1−x)−12(1+x)}

dydx=1√(1+x)(1−x){12(1−x)−12(1+x)}

An explanation provided by Harivogind S. as to why the result above is the same as the result he has provided...

{12(1−x)−12(1+x)}=(1+x)−(1−x)2⋅(1−x)⋅(1+x) - common denominator.

=2x2⋅(1−x2)
{12(1−x)−12(1+x)}=x1−x2
and,
1√(1+x)(1−x)=1√(1−x2)
Therefore -
dydx=1√(1+x)(1−x){12(1−x)−12(1+x)}=1√1−x2⋅[x1−x2]
dydx=x(1−x2)32

*Quotient rule was not required in these workings as your fraction didn't contain both a numerator and denominator that were both functions of x.

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