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Question Number 141274 by mnjuly1970 last updated on 17/May/21

$$.......elementary.....calculus........$$ $$iff\left(x\right):=\frac{\sqrt[3]{(x^3+x-2)(2x^2+x-3)(1-x^3)}}{{(2+cos\left(\pi x\right))}^3}$$ $$thenf{}^{\prime }\left(0\right)=?\mathrm{\&}f{}^{\prime }\left(1\right)=?$$ $$$$

Answered by mindispower last updated on 17/May/21

$$ln\left(f\left(x\right)\right)=\frac{1}{3}(ln(2-(x^3+x))+ln(3-2x^2-x)+ln(1-x^3))-3ln(2+cos\left(\pi x\right))$$ $$\frac{f^{\prime }\left(x\right)}{f\left(x\right)}=\frac{1}{3}(\frac{-3x^2-1}{2-x^3-x}-\frac{4x+1}{3-2x^2-x}-\frac{3x^2}{1-x^3}+\frac{3\pi sin\left(\pi x\right)}{2+cos\left(\pi x\right)}).\frac{1}{f\left(x\right)}$$ $$f\left(0\right)=\frac{\sqrt[3]{6}}{27}$$ $$\frac{f^{\prime }\left(0\right)}{f\left(0\right)}=(\frac{-1}{2}-\frac{1}{3}).\frac{1}{3}=\frac{-5}{18}$$ $$f^{\prime }\left(0\right)=-\frac{5}{18}.f\left(0\right)=-\frac{5\sqrt[3]{6}}{18.27}$$

Commented bymnjuly1970 last updated on 17/May/21

$$grateful....$$

Commented bymindispower last updated on 17/May/21

$$pleasur$$

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