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Question Number 121866 by danielasebhofoh last updated on 12/Nov/20

	Answered by TANMAY PANACEA last updated on 12/Nov/20
	$y=cos(x^x^x lnx) y=cosu→(dy/du)=−sinu=−sin(x^x^x lnx) u=x^x^x lnx u=pq (du/dx)=p(dq/dx)+q(dp/dx)=x^x^x ((1/x))+lnx×(dp/dx) red coloured to be found... p=x^x^x lnp=x^x lnx (1/p)×(dp/dx)=x^x ×(1/x)+lnx×(d/dx)(x^x ) (1/p)×(dp/dx)=x^(x−1) +lnx×x^x ×(1+lnx) (dp/dx)=x^x^x [x^(x−1) +lnx×x^x ×(1+lnx)] (du/dx)=[x^x^x ×(1/x)+lnx×x^x^x {x^(x−1) +lnx×x^x ×(1+lnx)}] so answer (dy/dx)=(dy/du)×(du/dx) =−sin(x^x^x lnx)×[x^x^x ×(1/x)+lnx×x^x^x {x^(x−1) +lnx×x^x (1+lnx)}] ★★ k=x^x →lnk=xlnx (1/k)×(dk/dx)=x×(1/x)+lnx=(dk/dx)=x^x (1+lnx)★★$
	$y = c o s (x^{x^{x}} l n x)$ $y = c o s u \to \frac{d y}{d u} = - s i n u = - s i n (x^{x^{x}} l n x)$ $u = x^{x^{x}} l n x$ $u = p q$ $\frac{d u}{d x} = p \frac{d q}{d x} + q \frac{d p}{d x} = x^{x^{x}} (\frac{1}{x}) + l n x \times \frac{d p}{d x}$ $r e d c o l o u r e d t o b e f o u n d . . .$ $p = x^{x^{x}}$ $l n p = x^{x} l n x$ $\frac{1}{p} \times \frac{d p}{d x} = x^{x} \times \frac{1}{x} + l n x \times \frac{d}{d x} (x^{x})$ $\frac{1}{p} \times \frac{d p}{d x} = x^{x - 1} + l n x \times x^{x} \times (1 + l n x)$ $\frac{d p}{d x} = x^{x^{x}} [x^{x - 1} + l n x \times x^{x} \times (1 + l n x)]$ $\frac{d u}{d x} = [x^{x^{x}} \times \frac{1}{x} + l n x \times x^{x^{x}} {x^{x - 1} + l n x \times x^{x} \times (1 + l n x)}]$ $s o a n s w e r$ $\frac{d y}{d x} = \frac{d y}{d u} \times \frac{d u}{d x}$ $= - s i n (x^{x^{x}} l n x) \times [x^{x^{x}} \times \frac{1}{x} + l n x \times x^{x^{x}} {x^{x - 1} + l n x \times x^{x} (1 + l n x)}]$ $★ ★ k = x^{x} \to l n k = x l n x$ $\frac{1}{k} \times \frac{d k}{d x} = x \times \frac{1}{x} + l n x = \frac{d k}{d x} = x^{x} (1 + l n x) ★ ★$
	Commented by danielasebhofoh last updated on 12/Nov/20


	Answered by Dwaipayan Shikari last updated on 12/Nov/20

	$\frac{d}{d x} (c o s (x^{x^{x}} l o g x))$ $= - s i n (x^{x^{x}} l o g x) (\frac{1}{x} . x^{x^{x}} + \frac{d (x^{x^{x}})}{d x})$ $= - x^{x^{x}} s i n (x^{x^{x}} l o g x) (\frac{1}{x} + x^{x} ({l o g}^{2} x + l o g x) + x^{x - 1})$ $f (x) = x^{x^{x}}$ $l o g (f (x)) = x^{x} l o g x$ $\frac{f^{'} (x)}{f (x)} = x^{x} (l o g x + 1) l o g x + x^{x - 1}$ $f^{'} (x) = x^{x^{x}} . x^{x} (l o g x + 1) l o g x + x^{x - 1} . x^{x^{x}}$
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