Implicit diff.: y = [(2x)^(2/3) (x + 5)^(3/4)] / (3 - x)^4

[Sophie]

Could someone see if I have made any errors on the following question. Thanks

Use implicit differentiation to find dy/dx in each case:

\(\displaystyle \L\\\begin{array}{l}
y = \frac{{\left( {2x} \right)^{2/3} \left( {x + 5} \right)^{3/4} }}{{\left( {3 - x} \right)^4 }} \\
\\
y\left( {3 - x} \right)^4 = \left( {2x} \right)^{2/3} \left( {x + 5} \right)^{3/4} \\
\\
\frac{{dy}}{{dx}} \\
\\
y'\left( {3 - x} \right)^4 + y4(3 - x)^3 \left( 1 \right) = \frac{2}{3}\left( {2x} \right)^{ - 1/3} 2\left( {x + 5} \right)^{3/4} + \left( {2x} \right)^{2/3} \frac{3}{4}\left( {x + 5} \right)^{ - 1/4} \left( 1 \right) \\
\\
y'\left( {3 - x} \right)^4 = \frac{2}{3}\left( {2x} \right)^{ - 1/3} 2\left( {x + 5} \right)^{3/4} + \left( {2x} \right)^{2/3} \frac{3}{4}\left( {x + 5} \right)^{ - 1/4} - y4(3 - x)^3 \\
\\
y'\left( {3 - x} \right)^4 = \left( {2x} \right)^{ - 1/3} \left( {x + 5} \right)^{ - 1/4} \left( {\frac{4}{3}\left( {x + 5} \right) + \frac{3}{4}\left( {2x} \right)} \right) - y4(3 - x)^3 \\
\\
y' = \frac{{\left( {2x} \right)^{ - 1/3} \left( {x + 5} \right)^{ - 1/4} \left( {\frac{4}{3}\left( {x + 5} \right) + \frac{3}{4}\left( {2x} \right)} \right) - y4(3 - x)^3 }}{{\left( {3 - x} \right)^4 }} \\
\\
y' = \frac{{\left( {2x} \right)^{ - 1/3} \left( {x + 5} \right)^{ - 1/4} \left( {\frac{4}{3}\left( {x + 5} \right) + \frac{3}{4}\left( {2x} \right)} \right)}}{{\left( {3 - x} \right)^4 }} - \frac{{y4}}{{\left( {3 - x} \right)}} \\
\\
y' = \frac{{\left( {2x} \right)^{ - 1/3} \left( {x + 5} \right)^{ - 1/4} \left( {\frac{{17x}}{6} + \frac{{20}}{3}} \right)}}{{\left( {3 - x} \right)^4 }} - \frac{{y4}}{{\left( {3 - x} \right)}} \\
\\
y' = \frac{{\left( {\frac{{17x}}{6} + \frac{{20}}{3}} \right)}}{{\left( {\sqrt[3]{{2x}}} \right)\left( {\sqrt[4]{{x + 5}}} \right)\left( {3 - x} \right)^4 }} - \frac{{y4}}{{\left( {3 - x} \right)}} \\
\end{array}\)

Thanks Sophie

 

[mark07]

Only mistake I see is at the first step of differentiation: derivative of 3-x is -1, not 1. Other than that, at the end you are also supposed to substitute y back in there maybe?

As a side note, if you learned logarithms, you can simplify it first, and then differentiation becomes a lot easier.

\(\displaystyle \L \ln y = \ln (2x)^{2/3} + \ln (x+5)^{3/4} - \ln (3-x)^4\)

\(\displaystyle \L = \frac{2}{3} (\ln 2 + \ln x ) + \frac{3}{4} \ln (x+5) -4 \ln (3-x)\)

Take derivative now

\(\displaystyle \L \frac{y'}{y} = \frac{2}{3x} + \frac{3}{4(x+5)} + \frac{4}{3-x}\)

and from that y'=?

 

[Sophie]

Thanks

I am not surprised to have made a mistake with so many steps. Logarithims are in my next chapter.

Thanks Sophie