Question Number 170488 by leicianocosta last updated on 25/May/22
Answered by FelipeLz last updated on 25/May/22
![1) a(t) = (d/dt)[v(t)] = (d/dt)[2+3t+5t^2 ] = 3+10t a(5) = 53 m∙s^(−2) 2) y = f′(x_0 )(x−x_0 )+y_0 y_0 = f(x_0 ) → y_0 = f(π) = [3sin(π)+4cos(π)]^5 = [−4]^5 = −1024 u = 3sin(x)+4cos(x) → f′(x) = (d/du)[u^5 ](d/dx)[3sin(x)+4cos(x)] = 5[3sin(x)+4cos(x)]^4 [3cos(x)−4sin(x)] f′(π) = 5[3sin(π)+4cos(π)]^4 [3cos(π)−4sin(π)] = 5[−4]^4 [−3] = −3840 y = −3840x+3840π−1024 3) a) (df/dx) = (d/dx)[x^2 +1]∙tan(x)+(x^2 +1)(d/dx)[tan(x)] = 2xtan(x)+(x^2 +1)sec^2 (x) b) (df/dx) = (((d/dx)[x^2 +3x+1]∙(x−2)−(x^2 +3x+1)(d/dx)[x−2])/((x−2)^2 )) = (((2x+3)(x−2)−(x^2 +3x+1))/(x^2 −4x+4)) = ((x^2 −4x−7)/(x^2 −4x+4)) 4) u = x^2 +5x f′(x) = (d/du)[e^u ](d/dx)[x^2 +5x] = (2x+5)e^(x^2 +5x) f′(−1) = 3e^(−4) 5) (x_0 , y_0 ) ∣ f′(x_0 ) = 0 f′(x) = (((√(x−1))−(x+1)∙(1/(2(√(x−1)))))/(x−1)) = (((2(x−1)−(x+1))/(2(√(x−1))))/(x−1)) = ((x−3)/(2(((x−1)^2 ))^(1/3) )) ((x_0 −3)/(2(((x_0 −1)^2 ))^(1/3) )) = 0 ⇒ x_0 = 3 y_0 = f(3) = ((3+1)/( (√(3−1)))) = (4/( (√2))) = 2(√2) (3, 2(√2)) 6) F(x) = ∫_a ^x g(t)dt = G(x)−G(a) ∴ (dF/dx) = (dG/dx) = g(x) a) (dF/dx) = 5x+2 b) (dF/dx) = (√x)](Q170490.png)
$$\left.\mathrm{1}\right)\: \\ $$$$\:\:\:\:{a}\left({t}\right)\:=\:\frac{{d}}{{dt}}\left[{v}\left({t}\right)\right]\:=\:\frac{{d}}{{dt}}\left[\mathrm{2}+\mathrm{3}{t}+\mathrm{5}{t}^{\mathrm{2}} \right]\:=\:\mathrm{3}+\mathrm{10}{t} \\ $$$$\:\:\:\:{a}\left(\mathrm{5}\right)\:=\:\mathrm{53}\:{m}\centerdot{s}^{−\mathrm{2}} \\ $$$$\left.\mathrm{2}\right) \\ $$$$\:\:\:\:{y}\:=\:{f}'\left({x}_{\mathrm{0}} \right)\left({x}−{x}_{\mathrm{0}} \right)+{y}_{\mathrm{0}} \\ $$$$\:\:\:\:\:\:\:\:\:\:{y}_{\mathrm{0}} \:=\:{f}\left({x}_{\mathrm{0}} \right)\:\rightarrow\:{y}_{\mathrm{0}} \:=\:{f}\left(\pi\right)\:=\:\left[\mathrm{3sin}\left(\pi\right)+\mathrm{4cos}\left(\pi\right)\right]^{\mathrm{5}} \:=\:\left[−\mathrm{4}\right]^{\mathrm{5}} \:=\:−\mathrm{1024} \\ $$$$\:\:\:\:\:\:\:\:\:\:{u}\:=\:\mathrm{3sin}\left({x}\right)+\mathrm{4cos}\left({x}\right)\:\rightarrow\:{f}'\left({x}\right)\:=\:\frac{{d}}{{du}}\left[{u}^{\mathrm{5}} \right]\frac{{d}}{{dx}}\left[\mathrm{3sin}\left({x}\right)+\mathrm{4cos}\left({x}\right)\right]\:=\:\mathrm{5}\left[\mathrm{3sin}\left({x}\right)+\mathrm{4cos}\left({x}\right)\right]^{\mathrm{4}} \left[\mathrm{3cos}\left({x}\right)−\mathrm{4sin}\left({x}\right)\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{f}'\left(\pi\right)\:=\:\mathrm{5}\left[\mathrm{3sin}\left(\pi\right)+\mathrm{4cos}\left(\pi\right)\right]^{\mathrm{4}} \left[\mathrm{3cos}\left(\pi\right)−\mathrm{4sin}\left(\pi\right)\right]\:=\:\mathrm{5}\left[−\mathrm{4}\right]^{\mathrm{4}} \left[−\mathrm{3}\right]\:=\:−\mathrm{3840} \\ $$$$\:\:\:\:{y}\:=\:−\mathrm{3840}{x}+\mathrm{3840}\pi−\mathrm{1024} \\ $$$$\left.\mathrm{3}\right) \\ $$$$\left.\:\:\:\:\:{a}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\frac{{df}}{{dx}}\:=\:\frac{{d}}{{dx}}\left[{x}^{\mathrm{2}} +\mathrm{1}\right]\centerdot\mathrm{tan}\left({x}\right)+\left({x}^{\mathrm{2}} +\mathrm{1}\right)\frac{{d}}{{dx}}\left[\mathrm{tan}\left({x}\right)\right]\:=\:\mathrm{2}{x}\mathrm{tan}\left({x}\right)+\left({x}^{\mathrm{2}} +\mathrm{1}\right)\mathrm{sec}^{\mathrm{2}} \left({x}\right) \\ $$$$\left.\:\:\:\:\:{b}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\frac{{df}}{{dx}}\:=\:\frac{\frac{{d}}{{dx}}\left[{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{1}\right]\centerdot\left({x}−\mathrm{2}\right)−\left({x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{1}\right)\frac{{d}}{{dx}}\left[{x}−\mathrm{2}\right]}{\left({x}−\mathrm{2}\right)^{\mathrm{2}} }\:=\:\frac{\left(\mathrm{2}{x}+\mathrm{3}\right)\left({x}−\mathrm{2}\right)−\left({x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{1}\right)}{{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{4}}\:=\:\frac{{x}^{\mathrm{2}} −\mathrm{4}{x}−\mathrm{7}}{{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{4}} \\ $$$$\left.\mathrm{4}\right) \\ $$$$\:\:\:\:\:{u}\:=\:{x}^{\mathrm{2}} +\mathrm{5}{x}\:\:\: \\ $$$$\:\:\:\:{f}'\left({x}\right)\:=\:\frac{{d}}{{du}}\left[{e}^{{u}} \right]\frac{{d}}{{dx}}\left[{x}^{\mathrm{2}} +\mathrm{5}{x}\right]\:=\:\left(\mathrm{2}{x}+\mathrm{5}\right){e}^{{x}^{\mathrm{2}} +\mathrm{5}{x}} \\ $$$$\:\:\:\:\:{f}'\left(−\mathrm{1}\right)\:=\:\mathrm{3}{e}^{−\mathrm{4}} \\ $$$$\left.\mathrm{5}\right) \\ $$$$\:\:\:\:\left({x}_{\mathrm{0}} ,\:{y}_{\mathrm{0}} \right)\:\mid\:{f}'\left({x}_{\mathrm{0}} \right)\:=\:\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:{f}'\left({x}\right)\:=\:\frac{\sqrt{{x}−\mathrm{1}}−\left({x}+\mathrm{1}\right)\centerdot\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}−\mathrm{1}}}}{{x}−\mathrm{1}}\:=\:\frac{\frac{\mathrm{2}\left({x}−\mathrm{1}\right)−\left({x}+\mathrm{1}\right)}{\mathrm{2}\sqrt{{x}−\mathrm{1}}}}{{x}−\mathrm{1}}\:=\:\frac{{x}−\mathrm{3}}{\mathrm{2}\sqrt[{\mathrm{3}}]{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }} \\ $$$$\:\:\:\:\:\:\:\:\:\frac{{x}_{\mathrm{0}} −\mathrm{3}}{\mathrm{2}\sqrt[{\mathrm{3}}]{\left({x}_{\mathrm{0}} −\mathrm{1}\right)^{\mathrm{2}} }}\:=\:\mathrm{0}\:\Rightarrow\:{x}_{\mathrm{0}} \:=\:\mathrm{3} \\ $$$$\:\:\:\:\:\:\:\:\:{y}_{\mathrm{0}} \:=\:{f}\left(\mathrm{3}\right)\:=\:\frac{\mathrm{3}+\mathrm{1}}{\:\sqrt{\mathrm{3}−\mathrm{1}}}\:=\:\frac{\mathrm{4}}{\:\sqrt{\mathrm{2}}}\:=\:\mathrm{2}\sqrt{\mathrm{2}} \\ $$$$\:\:\:\:\:\left(\mathrm{3},\:\mathrm{2}\sqrt{\mathrm{2}}\right) \\ $$$$\left.\mathrm{6}\right) \\ $$$$\:\:\:\:{F}\left({x}\right)\:=\:\underset{{a}} {\overset{{x}} {\int}}{g}\left({t}\right){dt}\:=\:{G}\left({x}\right)−{G}\left({a}\right)\:\therefore\:\frac{{dF}}{{dx}}\:=\:\frac{{dG}}{{dx}}\:=\:{g}\left({x}\right) \\ $$$$\left.\:\:\:\:{a}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\frac{{dF}}{{dx}}\:=\:\mathrm{5}{x}+\mathrm{2} \\ $$$$\left.\:\:\:\:{b}\right) \\ $$$$\:\:\:\:\:\:\:\:\frac{{dF}}{{dx}}\:=\:\sqrt{{x}} \\ $$