Question Number 204055 by hardmath last updated on 05/Feb/24
$$\mathrm{y}\:=\:\frac{\mathrm{2}}{\mathrm{3}}\:\mathrm{arctg}\:\left(\mathrm{x}^{\mathrm{4}} \right) \\ $$$$\mathrm{find}:\:\:\mathrm{y}^{'} \:=\:? \\ $$
Answered by AST last updated on 05/Feb/24
![y^′ =(2/3)[(d/dx)tan^(−1) (x^4 )] z=tan^(−1) (x^4 )⇒p=tan(z)=x^4 (dp/dz)=(d/dz)[sin(z){cos(z)}^(−1) ]=1+((sin^2 (z))/(cos^2 (z)))=1+tan^2 z (dp/dx)=4x^3 ⇒(dz/dx)=(dz/dp)×(dp/dx) ⇒y^′ =[1/(1+tan^2 (tan^(−1) (x^4 )))]×4x^3 =((8x^3 )/(3+3x^8 ))](https://www.tinkutara.com/question/Q204059.png)
$${y}^{'} =\frac{\mathrm{2}}{\mathrm{3}}\left[\frac{{d}}{{dx}}{tan}^{−\mathrm{1}} \left({x}^{\mathrm{4}} \right)\right] \\ $$$${z}={tan}^{−\mathrm{1}} \left({x}^{\mathrm{4}} \right)\Rightarrow{p}={tan}\left({z}\right)={x}^{\mathrm{4}} \\ $$$$\frac{{dp}}{{dz}}=\frac{{d}}{{dz}}\left[{sin}\left({z}\right)\left\{{cos}\left({z}\right)\right\}^{−\mathrm{1}} \right]=\mathrm{1}+\frac{{sin}^{\mathrm{2}} \left({z}\right)}{{cos}^{\mathrm{2}} \left({z}\right)}=\mathrm{1}+{tan}^{\mathrm{2}} {z} \\ $$$$\frac{{dp}}{{dx}}=\mathrm{4}{x}^{\mathrm{3}} \Rightarrow\frac{{dz}}{{dx}}=\frac{{dz}}{{dp}}×\frac{{dp}}{{dx}} \\ $$$$\Rightarrow{y}^{'} =\left[\mathrm{1}/\left(\mathrm{1}+{tan}^{\mathrm{2}} \left({tan}^{−\mathrm{1}} \left({x}^{\mathrm{4}} \right)\right)\right)\right]×\mathrm{4}{x}^{\mathrm{3}} =\frac{\mathrm{8}{x}^{\mathrm{3}} }{\mathrm{3}+\mathrm{3}{x}^{\mathrm{8}} } \\ $$
Answered by Mathspace last updated on 06/Feb/24
$${y}^{'} =\frac{\mathrm{2}}{\mathrm{3}}×\frac{\left({x}^{\mathrm{4}} \right)^{'} }{\mathrm{1}+\left({x}^{\mathrm{4}} \right)^{\mathrm{2}} }=\frac{\mathrm{2}}{\mathrm{3}}×\frac{\mathrm{4}{x}^{\mathrm{3}} }{\mathrm{1}+{x}^{\mathrm{8}} } \\ $$$$=\frac{\mathrm{8}}{\mathrm{3}}×\frac{{x}^{\mathrm{3}} }{\mathrm{1}+{x}^{\mathrm{8}} } \\ $$