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Question Number 112310 by mohammad17 last updated on 07/Sep/20
Answered by bemath last updated on 07/Sep/20
Q2/A) 5x+6y = 5×5+6×2 5x+6y = 37
$$\left.\mathrm{Q2}/\mathrm{A}\right)\:\mathrm{5x}+\mathrm{6y}\:=\:\mathrm{5}×\mathrm{5}+\mathrm{6}×\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{5x}+\mathrm{6y}\:=\:\mathrm{37} \\ $$
Answered by mathmax by abdo last updated on 07/Sep/20
Q_1 ) i suppose the question is (dy/dx) here y =cosx tan^3 x ⇒(dy/dx) =−sinx tan^3 x +cosx(3tan^2 x)(1+tan^2 x) =−sinx tan^3 x +3cosxtan^2 x(1+tan^2 x) y =cos^2 ((1/x)) ⇒(dy/dx) =2cos((1/x))(−(1/x^2 ) sin((1/x))) =((−2)/x^2 ) cos((1/x))sin((1/x))
$$\left.\mathrm{Q}_{\mathrm{1}} \right)\:\mathrm{i}\:\mathrm{suppose}\:\mathrm{the}\:\mathrm{question}\:\mathrm{is}\:\frac{\mathrm{dy}}{\mathrm{dx}}\:\mathrm{here} \\ $$$$\mathrm{y}\:=\mathrm{cosx}\:\mathrm{tan}^{\mathrm{3}} \mathrm{x}\:\Rightarrow\frac{\mathrm{dy}}{\mathrm{dx}}\:=−\mathrm{sinx}\:\mathrm{tan}^{\mathrm{3}} \mathrm{x}\:+\mathrm{cosx}\left(\mathrm{3tan}^{\mathrm{2}} \mathrm{x}\right)\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \mathrm{x}\right) \\ $$$$=−\mathrm{sinx}\:\mathrm{tan}^{\mathrm{3}} \mathrm{x}\:+\mathrm{3cosxtan}^{\mathrm{2}} \mathrm{x}\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \mathrm{x}\right) \\ $$$$\mathrm{y}\:=\mathrm{cos}^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{x}}\right)\:\Rightarrow\frac{\mathrm{dy}}{\mathrm{dx}}\:=\mathrm{2cos}\left(\frac{\mathrm{1}}{\mathrm{x}}\right)\left(−\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\:\mathrm{sin}\left(\frac{\mathrm{1}}{\mathrm{x}}\right)\right)\:=\frac{−\mathrm{2}}{\mathrm{x}^{\mathrm{2}} }\:\mathrm{cos}\left(\frac{\mathrm{1}}{\mathrm{x}}\right)\mathrm{sin}\left(\frac{\mathrm{1}}{\mathrm{x}}\right) \\ $$
Answered by mathmax by abdo last updated on 07/Sep/20
Q_2 ) det(AM^→ ,u^→ ) =0 with A(5,2) and u^→ (−6,5) ⇒ determinant (((x−5 −6)),((y−2 5)))=0 ⇒5(x−5)+6(y−2) =0 ⇒ 5x−25 +6y−12 =0 ⇒5x +6y −27 =0
$$\left.\mathrm{Q}_{\mathrm{2}} \right)\:\:\mathrm{det}\left(\mathrm{A}\overset{\rightarrow} {\mathrm{M}}\:,\overset{\rightarrow} {\mathrm{u}}\right)\:=\mathrm{0}\:\:\mathrm{with}\:\:\:\:\mathrm{A}\left(\mathrm{5},\mathrm{2}\right)\:\:\:\:\mathrm{and}\:\overset{\rightarrow} {\mathrm{u}}\left(−\mathrm{6},\mathrm{5}\right)\:\Rightarrow \\ $$$$\begin{vmatrix}{\mathrm{x}−\mathrm{5}\:\:\:\:\:\:\:\:\:\:\:−\mathrm{6}}\\{\mathrm{y}−\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{5}}\end{vmatrix}=\mathrm{0}\:\Rightarrow\mathrm{5}\left(\mathrm{x}−\mathrm{5}\right)+\mathrm{6}\left(\mathrm{y}−\mathrm{2}\right)\:=\mathrm{0}\:\Rightarrow \\ $$$$\mathrm{5x}−\mathrm{25}\:+\mathrm{6y}−\mathrm{12}\:=\mathrm{0}\:\Rightarrow\mathrm{5x}\:+\mathrm{6y}\:−\mathrm{27}\:=\mathrm{0} \\ $$
Answered by abdomsup last updated on 08/Sep/20
y^2 =x^3 +sin(xy) by derivation we get 2yy^′ =3x^2 +(y+xy^′ )cos(xy) ⇒ 2yy^′ −xy^′ cos(xy) =3x^2 +y cos(xy) ⇒(2y−xcos(xy))y^′ =3x^2 +ycos(xy) ⇒(dy/dx) =((3x^2 +ycos(xy))/(2y−xcos(xy)))
$${y}^{\mathrm{2}} \:={x}^{\mathrm{3}} \:+{sin}\left({xy}\right)\:{by}\:{derivation} \\ $$$${we}\:{get}\:\mathrm{2}{yy}^{'} \:\:=\mathrm{3}{x}^{\mathrm{2}} \:+\left({y}+{xy}^{'} \right){cos}\left({xy}\right)\:\Rightarrow \\ $$$$\mathrm{2}{yy}^{'} \:−{xy}^{'} \:{cos}\left({xy}\right)\:=\mathrm{3}{x}^{\mathrm{2}} \:+{y}\:{cos}\left({xy}\right) \\ $$$$\Rightarrow\left(\mathrm{2}{y}−{xcos}\left({xy}\right)\right){y}^{'} =\mathrm{3}{x}^{\mathrm{2}} \:+{ycos}\left({xy}\right) \\ $$$$\Rightarrow\frac{{dy}}{{dx}}\:\:=\frac{\mathrm{3}{x}^{\mathrm{2}} \:+{ycos}\left({xy}\right)}{\mathrm{2}{y}−{xcos}\left({xy}\right)} \\ $$$$ \\ $$

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