Solve for 
![y=\displaystyle \sqrt[5]{\frac{4^{x}\coth\left(e^{-x}\right)}{\sin\left(x^{2}\ln x\right)}}](http://l.wordpress.com/latex.php?latex=y%3D%5Cdisplaystyle+%5Csqrt%5B5%5D%7B%5Cfrac%7B4%5E%7Bx%7D%5Ccoth%5Cleft%28e%5E%7B-x%7D%5Cright%29%7D%7B%5Csin%5Cleft%28x%5E%7B2%7D%5Cln+x%5Cright%29%7D%7D&bg=ffffff&fg=333333&s=0 "y=\displaystyle \sqrt[5]{\frac{4^{x}\coth\left(e^{-x}\right)}{\sin\left(x^{2}\ln x\right)}}")
Solution
If we proceed directly in number 1, we get
![\displaystyle \frac{dy}{dx}=\frac{ y^{-4}}{5}\cdot\frac{\sin\left(x^{2}\ln x\right)\cdot\left[4^{x}e^{-x}\mathrm{csch}^{2}\left(e^{-x}\right)+4^{x}\ln4\coth\left(e^{-x}\right)\right]}{\sin^{2}\left(x^{2}\ln x\right)}](http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle+%5Cfrac%7Bdy%7D%7Bdx%7D%3D%5Cfrac%7B+y%5E%7B-4%7D%7D%7B5%7D%5Ccdot%5Cfrac%7B%5Csin%5Cleft%28x%5E%7B2%7D%5Cln+x%5Cright%29%5Ccdot%5Cleft%5B4%5E%7Bx%7De%5E%7B-x%7D%5Cmathrm%7Bcsch%7D%5E%7B2%7D%5Cleft%28e%5E%7B-x%7D%5Cright%29%2B4%5E%7Bx%7D%5Cln4%5Ccoth%5Cleft%28e%5E%7B-x%7D%5Cright%29%5Cright%5D%7D%7B%5Csin%5E%7B2%7D%5Cleft%28x%5E%7B2%7D%5Cln+x%5Cright%29%7D&bg=ffffff&fg=333333&s=0 "\displaystyle \frac{dy}{dx}=\frac{ y^{-4}}{5}\cdot\frac{\sin\left(x^{2}\ln x\right)\cdot\left[4^{x}e^{-x}\mathrm{csch}^{2}\left(e^{-x}\right)+4^{x}\ln4\coth\left(e^{-x}\right)\right]}{\sin^{2}\left(x^{2}\ln x\right)}")
![\displaystyle \rule{0.0001em}{0.0001ex} \phantom{\frac{dy}{dx}=} -\frac{y^{-4}}{5}\cdot\frac{4^{x}\coth\left( e^{-x} \right)\cos\left(x^{2}\ln x\right)\left[ x+2x\ln x\right]}{\sin^{2}\left(x^{2}\ln x\right)}](http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle+%5Crule%7B0.0001em%7D%7B0.0001ex%7D+%5Cphantom%7B%5Cfrac%7Bdy%7D%7Bdx%7D%3D%7D+-%5Cfrac%7By%5E%7B-4%7D%7D%7B5%7D%5Ccdot%5Cfrac%7B4%5E%7Bx%7D%5Ccoth%5Cleft%28+e%5E%7B-x%7D+%5Cright%29%5Ccos%5Cleft%28x%5E%7B2%7D%5Cln+x%5Cright%29%5Cleft%5B+x%2B2x%5Cln+x%5Cright%5D%7D%7B%5Csin%5E%7B2%7D%5Cleft%28x%5E%7B2%7D%5Cln+x%5Cright%29%7D&bg=ffffff&fg=333333&s=0 "\displaystyle \rule{0.0001em}{0.0001ex} \phantom{\frac{dy}{dx}=} -\frac{y^{-4}}{5}\cdot\frac{4^{x}\coth\left( e^{-x} \right)\cos\left(x^{2}\ln x\right)\left[ x+2x\ln x\right]}{\sin^{2}\left(x^{2}\ln x\right)}")
Meanwhile, if we first simplify the expression by exploiting the properties of logarithms, we obtain
![\displaystyle \ln y = \frac{1}{5} \left[ x\ln4 + \ln\coth \left( e^{-x}\right) - \ln \sin \left( x^{2}\ln x \right) \right]](http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle+%5Cln+y+%3D+%5Cfrac%7B1%7D%7B5%7D+%5Cleft%5B+x%5Cln4+%2B+%5Cln%5Ccoth+%5Cleft%28+e%5E%7B-x%7D%5Cright%29+-+%5Cln+%5Csin+%5Cleft%28+x%5E%7B2%7D%5Cln+x+%5Cright%29+%5Cright%5D+&bg=ffffff&fg=333333&s=0 "\displaystyle \ln y = \frac{1}{5} \left[ x\ln4 + \ln\coth \left( e^{-x}\right) - \ln \sin \left( x^{2}\ln x \right) \right]")
from which the derivative is easier to compute compared to the previous:
![\displaystyle \frac{dy}{dx}=\frac{1}{5}y\left[\ln4+\frac{e^{-x}\mathrm{csch}^{2}\left(e^{-x}\right)}{\coth\left(e^{-x}\right)}-\frac{\cos\left(x^{2}\ln x\right)\left[x+2x\ln x\right]}{\sin\left(x^{2}\ln x\right)}\right].](http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle+%5Cfrac%7Bdy%7D%7Bdx%7D%3D%5Cfrac%7B1%7D%7B5%7Dy%5Cleft%5B%5Cln4%2B%5Cfrac%7Be%5E%7B-x%7D%5Cmathrm%7Bcsch%7D%5E%7B2%7D%5Cleft%28e%5E%7B-x%7D%5Cright%29%7D%7B%5Ccoth%5Cleft%28e%5E%7B-x%7D%5Cright%29%7D-%5Cfrac%7B%5Ccos%5Cleft%28x%5E%7B2%7D%5Cln+x%5Cright%29%5Cleft%5Bx%2B2x%5Cln+x%5Cright%5D%7D%7B%5Csin%5Cleft%28x%5E%7B2%7D%5Cln+x%5Cright%29%7D%5Cright%5D.&bg=ffffff&fg=333333&s=0 "\displaystyle \frac{dy}{dx}=\frac{1}{5}y\left[\ln4+\frac{e^{-x}\mathrm{csch}^{2}\left(e^{-x}\right)}{\coth\left(e^{-x}\right)}-\frac{\cos\left(x^{2}\ln x\right)\left[x+2x\ln x\right]}{\sin\left(x^{2}\ln x\right)}\right].")
In number 2, we just differentiate the equation implicitly with respect to 
Solving for
![\displaystyle \rule{0.0001em}{0.0001ex} \phantom{\frac{dy}{dx}} =\frac{y\ln3\left[e^{x}\sinh\left(e^{x}+\log_{3}y\right)-2x \ln 2 \cdot2^{x^{2}}\right]}{ye^{y}\ln3-\sinh\left(e^{x}+\log_{3}y\right)}](http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle+%5Crule%7B0.0001em%7D%7B0.0001ex%7D+%5Cphantom%7B%5Cfrac%7Bdy%7D%7Bdx%7D%7D+%3D%5Cfrac%7By%5Cln3%5Cleft%5Be%5E%7Bx%7D%5Csinh%5Cleft%28e%5E%7Bx%7D%2B%5Clog_%7B3%7Dy%5Cright%29-2x+%5Cln+2+%5Ccdot2%5E%7Bx%5E%7B2%7D%7D%5Cright%5D%7D%7Bye%5E%7By%7D%5Cln3-%5Csinh%5Cleft%28e%5E%7Bx%7D%2B%5Clog_%7B3%7Dy%5Cright%29%7D&bg=ffffff&fg=333333&s=0 "\displaystyle \rule{0.0001em}{0.0001ex} \phantom{\frac{dy}{dx}} =\frac{y\ln3\left[e^{x}\sinh\left(e^{x}+\log_{3}y\right)-2x \ln 2 \cdot2^{x^{2}}\right]}{ye^{y}\ln3-\sinh\left(e^{x}+\log_{3}y\right)}")
March 23, 2008 at 10:36 pm |
sir kulang po ata ng ln2 yung number 2.