y = (2/3) arctg (x^4 ) find: y^′ = ?


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Question Number 204055 by hardmath last updated on 05/Feb/24

$y = \frac{2}{3} arctg (x^{4})$ $find : y^{^{'}} = ?$
	Answered by deleteduser1 last updated on 05/Feb/24
	$y^′ =(2/3)[(d/dx)tan^(−1) (x^4 )] z=tan^(−1) (x^4 )⇒p=tan(z)=x^4 (dp/dz)=(d/dz)[sin(z){cos(z)}^(−1) ]=1+((sin^2 (z))/(cos^2 (z)))=1+tan^2 z (dp/dx)=4x^3 ⇒(dz/dx)=(dz/dp)×(dp/dx) ⇒y^′ =[1/(1+tan^2 (tan^(−1) (x^4 )))]×4x^3 =((8x^3 )/(3+3x^8 ))$
	$y^{^{'}} = \frac{2}{3} [\frac{d}{d x} {t a n}^{- 1} (x^{4})]$ $z = {t a n}^{- 1} (x^{4}) \Rightarrow p = t a n (z) = x^{4}$ $\frac{d p}{d z} = \frac{d}{d z} [s i n (z) {c o s (z)}^{- 1}] = 1 + \frac{{s i n}^{2} (z)}{{c o s}^{2} (z)} = 1 + {t a n}^{2} z$ $\frac{d p}{d x} = 4 x^{3} \Rightarrow \frac{d z}{d x} = \frac{d z}{d p} \times \frac{d p}{d x}$ $\Rightarrow y^{^{'}} = [1 / (1 + {t a n}^{2} ({t a n}^{- 1} (x^{4})))] \times 4 x^{3} = \frac{8 x^{3}}{3 + 3 x^{8}}$
	Answered by Mathspace last updated on 06/Feb/24

	$y^{^{'}} = \frac{2}{3} \times \frac{{(x^{4})}^{^{'}}}{1 + {(x^{4})}^{2}} = \frac{2}{3} \times \frac{4 x^{3}}{1 + x^{8}}$ $= \frac{8}{3} \times \frac{x^{3}}{1 + x^{8}}$
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