# Implicit & Logarithmic differentiation problems.

• Oct 23rd 2012, 11:21 PM
jamesrb
Implicit & Logarithmic differentiation problems.
#1. Find the derivative of: $e^x^y=sin(y^2)$

$lne^x^y=lnsin(y^2)$
$xylne=lnsin(y^2)$
$xy*1=lnsin(y^2)$
$y+xy'=lnsin(y^2)$
I need to use 2 chain rules on $lnsin(y^2)$
$y+xy'=[{\frac{1}{sin(y^2)}]*[cos(y^2)*(2yy')]$
$y+xy'=\frac{[cos(y^2)*(2yy')]}{\frac{1}{sin(y^2)}}}$
$y=\frac{[cos(y^2)*(2yy')]}{\frac{1}{sin(y^2)}}}-xy'$
$y=y'[\frac{[cos(y^2)*(2y)]}{\frac{1}{sin(y^2)}}}-x]$
$y'=\frac{y}{\frac{[cos(y^2)*(2y)]}{\frac{1}{sin(y^2)}}-x}}$

I'm sure there is something wrong in there :)

#2. Find the derivative using logartihmic differentiation:
$y=\frac{x(x+1)^3}{(3x+1)^2}$
$lny=ln\frac{x(x+1)^3}{(3x+1)^2}$
Now I am unsure of what to do. Can I, and if so should I, rewrite the right side as:
$lny=[ln(x(x+1)^3]-[ln(3x+1)^2]$
Or should I use a combination of the quotient, product and multiple chain rules on what I had before:
$lny=ln\frac{x(x+1)^3}{(3x+1)^2}$
You don't have to show me any work past this point, I would like to figure this our on my own, I just need a push in the right direction.
• Oct 23rd 2012, 11:45 PM
earboth
Re: Implicit & Logarithmic differentiation problems.
Quote:

Originally Posted by jamesrb
[B]...

#2. Find the derivative using logartihmic differentiation:
$y=\frac{x(x+1)^3}{(3x+1)^2}$
$lny=ln\frac{x(x+1)^3}{(3x+1)^2}$
Now I am unsure of what to do. Can I, and if so should I, rewrite the right side as:
$lny=[ln(x(x+1)^3]-[ln(3x+1)^2]$
Or should I use a combination of the quotient, product and multiple chain rules on what I had before:
$lny=ln\frac{x(x+1)^3}{(3x+1)^2}$
You don't have to show me any work past this point, I would like to figure this our on my own, I just need a push in the right direction.

You were on the right track:

$lny=ln\frac{x(x+1)^3}{(3x+1)^2} = \underbrace{\ln(x)+3\ln(x+1)}_{numerator}-\overbrace{2\ln(3x+1)}^{denominator}$

Now the differentiation of the RHS is quite simple.
• Oct 24th 2012, 12:01 AM
jamesrb
Re: Implicit & Logarithmic differentiation problems.
Quote:

Originally Posted by earboth
You were on the right track:

$lny=ln\frac{x(x+1)^3}{(3x+1)^2} = \underbrace{\ln(x)+3\ln(x+1)}_{numerator}-\overbrace{2\ln(3x+1)}^{denominator}$

Now the differentiation of the RHS is quite simple.

Thanks, I think I got it:
$y'=y(\frac{1}{x}+\frac{3}{x+1}-\frac{6}{3x+1})$
• Oct 24th 2012, 06:12 AM
Prove It
Re: Implicit & Logarithmic differentiation problems.
Quote:

Originally Posted by jamesrb
#1. Find the derivative of: $e^x^y=sin(y^2)$

$lne^x^y=lnsin(y^2)$
$xylne=lnsin(y^2)$
$xy*1=lnsin(y^2)$
$y+xy'=lnsin(y^2)$
I need to use 2 chain rules on $lnsin(y^2)$
$y+xy'=[{\frac{1}{sin(y^2)}]*[cos(y^2)*(2yy')]$
$y+xy'=\frac{[cos(y^2)*(2yy')]}{\frac{1}{sin(y^2)}}}$
$y=\frac{[cos(y^2)*(2yy')]}{\frac{1}{sin(y^2)}}}-xy'$
$y=y'[\frac{[cos(y^2)*(2y)]}{\frac{1}{sin(y^2)}}}-x]$
$y'=\frac{y}{\frac{[cos(y^2)*(2y)]}{\frac{1}{sin(y^2)}}-x}}$

I'm sure there is something wrong in there :)

#2. Find the derivative using logartihmic differentiation:
$y=\frac{x(x+1)^3}{(3x+1)^2}$
$lny=ln\frac{x(x+1)^3}{(3x+1)^2}$
Now I am unsure of what to do. Can I, and if so should I, rewrite the right side as:
$lny=[ln(x(x+1)^3]-[ln(3x+1)^2]$
Or should I use a combination of the quotient, product and multiple chain rules on what I had before:
$lny=ln\frac{x(x+1)^3}{(3x+1)^2}$
You don't have to show me any work past this point, I would like to figure this our on my own, I just need a push in the right direction.

I think in Question 1, attempting logarithmic differentiation is a waste of time.

\displaystyle \begin{align*} e^{x\,y} &= \sin{\left( y^2 \right)} \\ \frac{d}{dx} \left( e^{x\,y} \right) &= \frac{d}{dx}\left[ \sin{\left( y^2 \right)} \right] \\ e^{x\,y} \, \frac{d}{dx} \left( x\, y\right) &= \frac{d}{dy}\left[ \sin{\left( y^2 \right)} \right] \frac{dy}{dx} \\ e^{x\,y} \left( x\,\frac{dy}{dx} + y \right) &= 2y \cos{\left( y^2 \right)} \, \frac{dy}{dx} \\ x\, e^{x\,y} \, \frac{dy}{dx} + y \, e^{x\,y} &= 2y\cos{\left( y^2 \right)} \, \frac{dy}{dx} \\ y\, e^{x\,y} &= 2y\cos{\left( y^2 \right)} \, \frac{dy}{dx} - x\, e^{x\,y} \, \frac{dy}{dx} \\ y\, e^{x\,y} &= \left[ 2y\cos{\left( y^2 \right)} - x\, e^{x\,y} \right] \frac{dy}{dx} \\ \frac{y\, e^{x\,y}}{2y\cos{\left( y^2 \right)} - x \, e^{x\,y} } &= \frac{dy}{dx} \end{align*}