how is solution y=(cosx)^((3x^2 −1)^e^x ) (dy/dx)=?


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Question Number 187916 by mustafazaheen last updated on 23/Feb/23

$h o w i s s o l u t i o n$ $y = {(c o s x)}^{{(3 x^{2} - 1)}^{e^{x}}}$ $\frac{d y}{d x} = ?$
	Answered by mr W last updated on 23/Feb/23
	$generally y=f(x)^(g(x)) =e^(g(x)ln f(x)) y′=f(x)^(g(x)) [g′(x)ln f(x)+((g(x)f′(x))/(f(x)))] y=(cos x)^((3x^2 −1)^e^x ) y′=(cos x)^((3x^2 −1)^e^x ) {[(3x^2 −1)^e^x ]^, ln (cos x)−(((3x^2 −1)^e^x sin x)/(cos x))} =(cos x)^((3x^2 −1)^e^x ) {[(3x^2 −1)^e^x (e^x ln (3x^2 −1)+((6xe^x )/(3x^2 −1)))]ln (cos x)−(((3x^2 −1)^e^x sin x)/(cos x))} =(cos x)^((3x^2 −1)^e^x ) {[(3x^2 −1)^e^x e^x (ln (3x^2 −1)+((6x)/(3x^2 −1)))]ln (cos x)−(((3x^2 −1)^e^x sin x)/(cos x))}$
	$g e n e r a l l y$ $y = f {(x)}^{g (x)} = e^{g (x) \ln f (x)}$ $y^{'} = f {(x)}^{g (x)} [g^{'} (x) \ln f (x) + \frac{g (x) f^{'} (x)}{f (x)}]$ $y = {(\cos x)}^{{(3 x^{2} - 1)}^{e^{x}}}$ $y^{'} = {(\cos x)}^{{(3 x^{2} - 1)}^{e^{x}}} {{[{(3 x^{2} - 1)}^{e^{x}}]}^{,} \ln (\cos x) - \frac{{(3 x^{2} - 1)}^{e^{x}} \sin x}{\cos x}}$ $= {(\cos x)}^{{(3 x^{2} - 1)}^{e^{x}}} {[{(3 x^{2} - 1)}^{e^{x}} (e^{x} \ln (3 x^{2} - 1) + \frac{6 {x e}^{x}}{3 x^{2} - 1})] \ln (\cos x) - \frac{{(3 x^{2} - 1)}^{e^{x}} \sin x}{\cos x}}$ $= {(\cos x)}^{{(3 x^{2} - 1)}^{e^{x}}} {[{(3 x^{2} - 1)}^{e^{x}} e^{x} (\ln (3 x^{2} - 1) + \frac{6 x}{3 x^{2} - 1})] \ln (\cos x) - \frac{{(3 x^{2} - 1)}^{e^{x}} \sin x}{\cos x}}$
	Commented by mustafazaheen last updated on 23/Feb/23

	$T h a n k s M r$
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