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Question Number 147612 by ZiYangLee last updated on 22/Jul/21
Given that f(x)=(((x^3 +1)^2 (√(1+x^2 )))/(1+(√x))). By using logarithmatic differentiation, find the value of f ′(1).
$$\mathrm{Given}\:\mathrm{that}\:{f}\left({x}\right)=\frac{\left({x}^{\mathrm{3}} +\mathrm{1}\right)^{\mathrm{2}} \sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{\mathrm{1}+\sqrt{{x}}}. \\ $$$$\mathrm{By}\:\mathrm{using}\:\mathrm{logarithmatic}\:\mathrm{differentiation}, \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{f}\:'\left(\mathrm{1}\right). \\ $$
Answered by mathmax by abdo last updated on 22/Jul/21
another way we have ln(f(x))=2ln(x^3 +1)+(1/2)ln(1+x^2 )−ln(1+(√x)) ⇒ ⇒((f^′ (x))/(f(x)))=((6x^2 )/(x^3 +1)) +(x/(1+x^2 ))−(1/(2(√x)(1+(√x)))) ⇒ ((f^′ (1))/(f(1)))=(6/2)+(1/2)−(1/4)=3+(1/4)=((13)/4) ⇒f^′ (1)=((13)/4)f(1) ⇒f^′ (1)=((13)/4)×((4(√2))/2)=((13(√2))/2)
$$\mathrm{another}\:\mathrm{way}\:\:\mathrm{we}\:\mathrm{have} \\ $$$$\mathrm{ln}\left(\mathrm{f}\left(\mathrm{x}\right)\right)=\mathrm{2ln}\left(\mathrm{x}^{\mathrm{3}} +\mathrm{1}\right)+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)−\mathrm{ln}\left(\mathrm{1}+\sqrt{\mathrm{x}}\right)\:\Rightarrow \\ $$$$\Rightarrow\frac{\mathrm{f}^{'} \left(\mathrm{x}\right)}{\mathrm{f}\left(\mathrm{x}\right)}=\frac{\mathrm{6x}^{\mathrm{2}} }{\mathrm{x}^{\mathrm{3}} \:+\mathrm{1}}\:+\frac{\mathrm{x}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{x}}\left(\mathrm{1}+\sqrt{\mathrm{x}}\right)}\:\Rightarrow \\ $$$$\frac{\mathrm{f}^{'} \left(\mathrm{1}\right)}{\mathrm{f}\left(\mathrm{1}\right)}=\frac{\mathrm{6}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{4}}=\mathrm{3}+\frac{\mathrm{1}}{\mathrm{4}}=\frac{\mathrm{13}}{\mathrm{4}}\:\Rightarrow\mathrm{f}^{'} \left(\mathrm{1}\right)=\frac{\mathrm{13}}{\mathrm{4}}\mathrm{f}\left(\mathrm{1}\right) \\ $$$$\Rightarrow\mathrm{f}^{'} \left(\mathrm{1}\right)=\frac{\mathrm{13}}{\mathrm{4}}×\frac{\mathrm{4}\sqrt{\mathrm{2}}}{\mathrm{2}}=\frac{\mathrm{13}\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$
Answered by mathmax by abdo last updated on 22/Jul/21
f(x)=(((x^3 +1)^2 (√(1+x^2 )))/(1+(√x))) ⇒ f^′ (x)=(({6x^2 (x^3 +1)(√(1+x^2 )) +(x/( (√(1+x^2 ))))(x^3 +1)^2 }(1+(√x))−((1/(2(√x))))(x^3 +1)^2 (√(1+x^2 )))/((1+(√x))^2 )) ⇒f^′ (1)=(((12(√2)+(1/( (√2)))×4)(2)−(1/2)(4)(√2))/4) =((2(14(√2))−2(√2))/4)=((26(√2))/4)=((13(√2))/2)
$$\mathrm{f}\left(\mathrm{x}\right)=\frac{\left(\mathrm{x}^{\mathrm{3}} +\mathrm{1}\right)^{\mathrm{2}} \sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }}{\mathrm{1}+\sqrt{\mathrm{x}}}\:\Rightarrow \\ $$$$\mathrm{f}^{'} \left(\mathrm{x}\right)=\frac{\left\{\mathrm{6x}^{\mathrm{2}} \left(\mathrm{x}^{\mathrm{3}} +\mathrm{1}\right)\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\:+\frac{\mathrm{x}}{\:\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }}\left(\mathrm{x}^{\mathrm{3}} +\mathrm{1}\right)^{\mathrm{2}} \right\}\left(\mathrm{1}+\sqrt{\mathrm{x}}\right)−\left(\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{x}}}\right)\left(\mathrm{x}^{\mathrm{3}} +\mathrm{1}\right)^{\mathrm{2}} \sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }}{\left(\mathrm{1}+\sqrt{\mathrm{x}}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow\mathrm{f}^{'} \left(\mathrm{1}\right)=\frac{\left(\mathrm{12}\sqrt{\mathrm{2}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}×\mathrm{4}\right)\left(\mathrm{2}\right)−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{4}\right)\sqrt{\mathrm{2}}}{\mathrm{4}} \\ $$$$=\frac{\mathrm{2}\left(\mathrm{14}\sqrt{\mathrm{2}}\right)−\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{4}}=\frac{\mathrm{26}\sqrt{\mathrm{2}}}{\mathrm{4}}=\frac{\mathrm{13}\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$$ \\ $$

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