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Question Number 170488 by leicianocosta last updated on 25/May/22

Answered by FelipeLz last updated on 25/May/22

$$1)$$$$a\left(t\right)=\frac{d}{dt}\left[v\left(t\right)\right]=\frac{d}{dt}[2+3t+5t^2]=3+10t$$$$a\left(5\right)=53m\cdot s^{-2}$$$$2)$$$$y=f^{\prime }\left(x_0\right)(x-x_0)+y_0$$$$y_0=f\left(x_0\right)\to y_0=f\left(\pi \right)={[3\sin \left(\pi \right)+4\cos \left(\pi \right)]}^5={[-4]}^5=-1024$$$$u=3\sin \left(x\right)+4\cos \left(x\right)\to f^{\prime }\left(x\right)=\frac{d}{du}\left[u^5\right]\frac{d}{dx}[3\sin \left(x\right)+4\cos \left(x\right)]=5{[3\sin \left(x\right)+4\cos \left(x\right)]}^4[3\cos \left(x\right)-4\sin \left(x\right)]$$$$f^{\prime }\left(\pi \right)=5{[3\sin \left(\pi \right)+4\cos \left(\pi \right)]}^4[3\cos \left(\pi \right)-4\sin \left(\pi \right)]=5{[-4]}^4[-3]=-3840$$$$y=-3840x+3840\pi -1024$$$$3)$$$$a)$$$$\frac{df}{dx}=\frac{d}{dx}[x^2+1]\cdot \tan \left(x\right)+(x^2+1)\frac{d}{dx}\left[\tan \left(x\right)\right]=2x\tan \left(x\right)+(x^2+1){\sec }^2\left(x\right)$$$$b)$$$$\frac{df}{dx}=\frac{\frac{d}{dx}[x^2+3x+1]\cdot (x-2)-(x^2+3x+1)\frac{d}{dx}[x-2]}{{(x-2)}^2}=\frac{(2x+3)(x-2)-(x^2+3x+1)}{x^2-4x+4}=\frac{x^2-4x-7}{x^2-4x+4}$$$$4)$$$$u=x^2+5x$$$$f^{\prime }\left(x\right)=\frac{d}{du}\left[e^u\right]\frac{d}{dx}[x^2+5x]=(2x+5)e^{x^2+5x}$$$$f^{\prime }(-1)=3e^{-4}$$$$5)$$$$(x_0,y_0)\mid f^{\prime }\left(x_0\right)=0$$$$f^{\prime }\left(x\right)=\frac{\sqrt{x-1}-(x+1)\cdot \frac{1}{2\sqrt{x-1}}}{x-1}=\frac{\frac{2(x-1)-(x+1)}{2\sqrt{x-1}}}{x-1}=\frac{x-3}{2\sqrt[3]{{(x-1)}^2}}$$$$\frac{x_0-3}{2\sqrt[3]{{(x_0-1)}^2}}=0\Rightarrow x_0=3$$$$y_0=f\left(3\right)=\frac{3+1}{\sqrt{3-1}}=\frac{4}{\sqrt{2}}=2\sqrt{2}$$$$(3,2\sqrt{2})$$$$6)$$$$F\left(x\right)=\underset{a}{\stackrel{x}{\int }}g\left(t\right)dt=G\left(x\right)-G\left(a\right)\therefore \frac{dF}{dx}=\frac{dG}{dx}=g\left(x\right)$$$$a)$$$$\frac{dF}{dx}=5x+2$$$$b)$$$$\frac{dF}{dx}=\sqrt{x}$$

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