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Question Number 105132 by mohammad17 last updated on 26/Jul/20

	Answered by Dwaipayan Shikari last updated on 26/Jul/20
	$1)∫5^((x^2 +6)) (x+3)dx =(1/2)∫5^u du {u=x^2 +6x (du/dx)=2x+6 =(1/2) (5^u /(log5))+C=(1/(2log5))5^(x^2 +6x) +C 2)∫e^x cosxdx=I =e^x sinx−∫e^x sinx =e^x sinx+e^x cosx−∫e^x cosx=I 2I=e^x (sinx+cosx) I=(e^x /2)(sinx+cosx)+C$
	$1) \int 5^{(x^{2} + 6)} (x + 3) d x$ $= \frac{1}{2} \int 5^{u} d u {u = x^{2} + 6 x \frac{d u}{d x} = 2 x + 6$ $= \frac{1}{2} \frac{5^{u}}{l o g 5} + C = \frac{1}{2 l o g 5} 5^{x^{2} + 6 x} + C$ $2) \int e^{x} c o s x d x = I$ $= e^{x} s i n x - \int e^{x} s i n x$ $= e^{x} s i n x + e^{x} c o s x - \int e^{x} c o s x = I$ $2 I = e^{x} (s i n x + c o s x)$ $I = \frac{e^{x}}{2} (s i n x + c o s x) + C$
	Answered by Dwaipayan Shikari last updated on 26/Jul/20

	$b) \lim_{x \to 0} \frac{t a n x}{x} = \frac{s i n x}{x} . \frac{1}{c o s x} = \frac{x}{x} . 1 = 1 a s s i n x \to x$
	Answered by Dwaipayan Shikari last updated on 26/Jul/20

	$x y + y + 1 = x$ $\frac{d y}{d x} x + y + \frac{d y}{d x} = 1$ $\frac{d y}{d x} = \frac{1 - y}{1 + x}$ $2) x y = 1$ $l o g x + l o g y = 0$ $\frac{1}{x} + \frac{1}{y} \frac{d y}{d x} = 0$ $\frac{d y}{d x} = - \frac{y}{x} = - \frac{1}{x^{2}}$
	Answered by Dwaipayan Shikari last updated on 26/Jul/20

	$a) \lim_{x \to 0} \frac{e^{x} + e^{- x} - 2}{1 - c o s 2 x} = \lim_{x \to 0} \frac{e^{x} - e^{- x}}{2 s i n 2 x} = \frac{e^{2 x} - 1}{2 e^{x} s i n 2 x}$ $= \frac{1}{4 x} . \frac{e^{2 x} - 1}{e^{x}} = \frac{1}{2}$
	Answered by Dwaipayan Shikari last updated on 26/Jul/20

	$\frac{d}{d x} (s i n (\sqrt{1 + c o s 10 x})) = \frac{d}{d x} (s i n (\sqrt{2} c o s 5 x) = 5 \sqrt{2} c o s (\sqrt{2} c o s 5 x) (- s i n 5 x)$
	Answered by mathmax by abdo last updated on 26/Jul/20

	$1) let f (x) = \sin (\sqrt{1 + \cos (10 x)}) \Rightarrow f (x) = \sin (\sqrt{2} \cos (5 x)) \Rightarrow$ $f^{^{'}} (x) = - 5 \sqrt{2} \sin (5 x) \cos (\sqrt{2} \cos (5 x))$
	Answered by mathmax by abdo last updated on 26/Jul/20

	$2) let g (x) = 3 \arcsin (\frac{x}{3}) + \sqrt{9 - x^{2}} \Rightarrow$ $g^{^{'}} (x) = 3 \times \frac{1}{3 \sqrt{1 - \frac{x^{2}}{9}}} + \frac{- 2 x}{2 \sqrt{9 - x^{2}}} = \frac{3}{\sqrt{9 - x^{2}}} - \frac{x}{\sqrt{9 - x^{2}}} = \frac{3 - x}{\sqrt{9 - x^{2}}}$
	Answered by mathmax by abdo last updated on 26/Jul/20

	$I = \int 5^{x^{2} + 6 x} (x + 3) dx changement x + 3 = t give I = \int 5^{{(t - 3)}^{2} + 6 (t - 3)} t dt$ $= \int t 5^{t^{2} - 6 t + 9 + 6 t - 18} dt = \int t 5^{t^{2} - 9} dt = \int t e^{\ln 5 (t^{2} - 9)} dt$ $= \frac{1}{2 \ln 5} e^{\ln 5 (t^{2} - 9)} + C = \frac{1}{2 \ln 5} \times 5^{t^{2} - 9} + c = \frac{1}{2 \ln 5} \times 5^{{(x + 3)}^{2} - 9} + C$
	Answered by 1549442205PVT last updated on 26/Jul/20

	$3 a) \underset{x \to 0}{Lim} \frac{e^{x} + e^{- x} - 2}{1 - \cos (2 x)} = \underset{x \to 0}{Lim} \frac{e^{x} - e^{- x}}{2 \sin (2 x)} =$ $= \underset{x \to 0}{Lim} \frac{e^{x} + e^{x}}{4 \cos 2 x} = \frac{2}{4} = \frac{1}{2}$ $b) \underset{x \to 0}{Lim} \frac{tanx}{x} = \underset{x \to 0}{Lim} \frac{sinx}{x} \times \underset{x \to 0}{Lim} \frac{1}{cosx} = 1 \times 1 = 1$
	Answered by 1549442205PVT last updated on 26/Jul/20
	$1a)Set y=sin(√(1+cos(10x)))⇒y^2 =sin^2 (√(1+cos(10x))) Derivative both two sides of above equation by x we get 2y.y′=2sin(√(1+cos(10x))).((√(1+cos(10x))))′ =2sin(√(1+cos(10x))).(((1+cos(10x))′)/(2(√(1+cos(10x))))) =2sin(√(1+cos(10x))).((−10sin(10x))/(2(√(1+cos(10x))))) =−10sin(√(1+cos(10x))).((sin(10x))/(√(1+cos(10x)))) ⇒(dy/dx)=y′=((−10sin(√(1+cos(10x))).((sin(10x))/(√(1+cos(10x)))))/(2y)) =−((5sin(10x))/(2(√(1+cos(10x)))))=((−5sin(10x))/(2(√(2cos^2 (5x))))) =((−5sin5xcos5x)/((√(2 )) ∣cos5x∣))= { ((((−5(√2))/2)sin5x if x∈(((−π+4kπ)/(10));((π+4kπ)/(10))))),((((5(√2))/2)sin5x if x∈([((π+4kπ)/(10));((3π+4kπ)/(10))))) :} 1b)Set y=3sin^(−1) ((x/3))+(√(9−x^2 )) y′=(3/(√(1−(x^2 /9))))×(1/3)+((−x)/(√(9−x^2 )))=((3−x)/(√(9−x^2 )))$
	$1 a) Set y = \sin \sqrt{1 + \cos (10 x)} \Rightarrow y^{2} = \sin^{2} \sqrt{1 + \cos (10 x)}$ $Derivative both two sides of above$ $equation by x we get$ $2 y . y^{'} = 2 \sin \sqrt{1 + \cos (10 x)} . {(\sqrt{1 + \cos (10 x)})}^{'}$ $= 2 \sin \sqrt{1 + \cos (10 x)} . \frac{{(1 + \cos (10 x))}^{'}}{2 \sqrt{1 + \cos (10 x)}}$ $= 2 \sin \sqrt{1 + \cos (10 x)} . \frac{- 10 \sin (10 x)}{2 \sqrt{1 + \cos (10 x)}}$ $= - 10 \sin \sqrt{1 + \cos (10 x)} . \frac{\sin (10 x)}{\sqrt{1 + \cos (10 x)}}$ $\Rightarrow \frac{dy}{dx} = y^{'} = \frac{- 10 \sin \sqrt{1 + \cos (10 x)} . \frac{\sin (10 x)}{\sqrt{1 + \cos (10 x)}}}{2 y}$ $= - \frac{5 \sin (10 x)}{2 \sqrt{1 + \cos (10 x)}} = \frac{- 5 \sin (10 x)}{2 \sqrt{{2 \cos}^{2} (5 x)}}$ $= \frac{- 5 \sin 5 xcos 5 x}{\sqrt{2} ∣ \cos 5 x ∣} = {\begin{cases} \frac{- 5 \sqrt{2}}{2} \sin 5 x if x \in (\frac{- π + 4 k π}{10}; \frac{π + 4 k π}{10}) \\ \frac{5 \sqrt{2}}{2} \sin 5 x if x \in ([\frac{π + 4 k π}{10}; \frac{3 π + 4 k π}{10}) \end{cases}$ $1 b) Set y = {3 \sin}^{- 1} (\frac{x}{3}) + \sqrt{9 - x^{2}}$ $y^{'} = \frac{3}{\sqrt{1 - \frac{x^{2}}{9}}} \times \frac{1}{3} + \frac{- x}{\sqrt{9 - x^{2}}} = \frac{3 - x}{\sqrt{9 - x^{2}}}$
	Answered by 1549442205PVT last updated on 26/Jul/20

	$II - 2) F = \int e^{x} cosxdx = \int \cos {xde}^{x} = e^{x} \cos x + \int e^{x} sinxdx$ $= e^{x} cosx + \int {sinxde}^{x} = e^{x} cosx + e^{x} sinx - \int e^{x} cosxdx$ $F = e^{x} (cosx + sinx) - F \Rightarrow 2 F = e^{x} (cosx + sinx)$ $\Rightarrow F = \frac{e^{x} (cosx + sinx)}{2} + C$ $II - 1) \int 5^{x^{2} + 6 x} (x + 3) = \frac{1}{2} \int 5^{x^{2} + 6 x} d (x^{2} + 6 x)$ $= \frac{5^{x^{2} + 6 x}}{2 \ln 5} + C$
	Answered by 1549442205PVT last updated on 26/Jul/20

	$4 a) xy + y + 1 = x \Leftrightarrow y (x + 1) + 1 = x \Rightarrow y = \frac{x - 1}{x + 1} = 1 - \frac{2}{x + 1}$ $y^{'} = \frac{2}{{(x + 1)}^{2}}$ $b) xy = 1 \Rightarrow y = \frac{1}{x} \Rightarrow y^{'} = - \frac{1}{x^{2}}$
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