Logarithmic Differientation

Question

Using logarithmic differentiation I need to find the implicit derivative of: $$y=\sqrt[18]{(x^{10}+1)^3(x^7-3)^8}$$

The result I came up with was this not regarding $y$

$$ y'=(y) \frac {24(17x^{16}-30x^9+7x^6)}{18(x^{10}+1)^3(x^7-3)^8(x^{10}+1)(x^7-3)}$$

It seems the answer was still wrong and I cannot find the problem

Answer 1

For expressions containing only products, quotients and powers, logarithmic differentiation is effecitively very useful and makes life much easier

Considering $$y=\sqrt[18]{(x^{10}+1)^3(x^7-3)^8}$$ $$\log(y)=\frac 1{6} \log(x^{10}+1)+\frac 4{9} \log(x^7-3)$$ $$\frac{y'}y=\frac 1{6} \frac {10x^9}{x^{10}+1}+\frac 4{9}\frac {7x^6} {x^7-3}=\frac{x^6 \left(43 x^{10}-45 x^3+28\right) }{9(x^{10}+1)(x^7-3)}$$ $$y'=y\frac{y'}y=\frac{x^6 \left(43 x^{10}-45 x^3+28\right) }{9(x^{10}+1)(x^7-3)}(x^{10}+1)^{1/6} (x^7-3)^{4/9}$$ $$y'=\frac{x^6 \left(43 x^{10}-45 x^3+28\right) }{9(x^{10}+1)^{5/6}(x^7-3)^{5/9} }$$

If we make the same for the more general case $$y=\frac{P(x)^a \,Q(x)^b }{ R(x)^{c}}$$ we should arrive to $$y'=P(x)^{a-1} Q(x)^{b-1} R(x)^{-c-1} \left(a Q(x) R(x) P'(x)+b P(x) R(x) Q'(x)-c P(x) Q(x) R'(x)\right)$$ and this is valid for any set of functions $P(x), Q(x), R(x)$

Answer 2

Hint: $$y=\sqrt[18]{(x^{10}+1)^3(x^7-3)^8}=[(x^{10}+1)^3(x^7-3)^8]^{\frac{1}{18}}=[(x^{10}+1)^{\frac{1}{6}}][(x^7-3)^{\frac{4}{9}}]$$ this might be easier to see, from here you can use the product rule and the chain rule to compute the derivative.