Think different to be brilliant

Calculus Level 3

Given that $f \left( \frac{z-1}{z+1} \right) = \ln{z}$ , what is $f' \left( \frac{1}{2} \right)$ ?

\frac{3}{8}

\frac{3}{2}

None of these

\frac{8}{3}

\frac{2}{3}

1 solution

Rishabh Jain
Feb 24, 2016

I don't know why you love alphabet z so much.. Anyways here's the solution... Differentiate both sides with respect to z: $f'(\dfrac{z-1}{z+1})(\dfrac{z-1}{z+1})'=\dfrac{1}{z}$ $\implies (\dfrac{2}{(z+1)^2} )f'(\dfrac{z-1}{z+1})=\dfrac 1z$ $\implies f'(\dfrac{z-1}{z+1})=\dfrac{(z+1)^2}{2z}$ We note $\dfrac{z-1}{z+1}=\dfrac{1}{2}$ when $z=3$ . Hence putting $z=3$ : $\huge f'(\dfrac 12)=\boxed{\dfrac 83}$

I did in the following way:

i took $x = \dfrac{z-1}{z+1}$ implies $z = \dfrac{1+x}{1-x}$

Therefore According to question $f(x) = \ln \dfrac{1+x}{1-x}$

$f'(x) = \dfrac{1}{1+x} - \dfrac{1}{1-x}$

$f'(\dfrac{1}{2})$ = $\dfrac{2}{3} + 2 = \dfrac{8}{3}$

Anik Mandal - 5 years, 3 months ago

That's a great way to approach!

Pulkit Gupta - 5 years, 3 months ago

I think just like most of us are accustomed to letter x, Zeeshan is more into z :-)

Nice solution,upvoted.

Pulkit Gupta - 5 years, 3 months ago

Probably because his name starts from z... BTW thanks :-)

Rishabh Jain - 5 years, 3 months ago

Yeah! Since my school days, you know!

Zeeshan Ali - 5 years, 3 months ago

Think different to be brilliant

1 solution

0 pending reports