To find the first derivative and second derivative of the following functions.
Now,
a)
[tex]y=4x^7+5x^3[/tex]The first derivative is given by,
[tex]\begin{gathered} y^{\prime}=\frac{d}{dx}(4x^7+5x^3) \\ =4\times7x^{7-1}+5\times3x^{3-1} \\ =28x^6+15x^2 \end{gathered}[/tex]The second derivative is,
[tex]\begin{gathered} y^{\prime\prime}=\frac{d^{}}{dx^{}}(28x^6+15x^2) \\ =28\times6x^{6-1}+15\times2x^{2-1} \\ =168x^5+30x \end{gathered}[/tex]b)
[tex]y=\frac{2}{x^4}[/tex]The first derivative is,
[tex]\begin{gathered} y^{\prime}=\frac{d}{dx}(\frac{2}{x^4}) \\ =\frac{d}{dx}(2x^{-4}) \\ =2\times-4x^{-4-1} \\ =-8x^{-5} \\ =-\frac{8}{x^5} \end{gathered}[/tex]The second derivative is,
[tex]\begin{gathered} y^{\prime\prime}=\frac{d}{dx}(-\frac{8}{x^5}) \\ =\frac{d}{dx}(-8x^{-5}) \\ =-8\times-5x^{-5-1} \\ =40x^{-6} \\ =\frac{40}{x^6} \end{gathered}[/tex]c)
[tex]x=36\sqrt[3]{t}[/tex]The first derivative is,
[tex]\begin{gathered} x^{\prime}=\frac{d}{dt}(36\sqrt[3]{t}) \\ =\frac{d}{dt}(36\times t^{\frac{1}{3}}) \\ =36\times\frac{1}{3}\times t^{\frac{1}{3}-1} \\ =12\times t^{\frac{1-3}{3}} \\ =12\times t^{-\frac{2}{3}} \\ =\frac{12}{\sqrt[3]{t^2}} \end{gathered}[/tex]The second derivative is,
[tex]\begin{gathered} x^{\prime\prime}=\frac{d}{dt}(\frac{12}{\sqrt[3]{t^2}}) \\ =\frac{d}{dt}(12\times t^{-\frac{2}{3}}) \\ =12\times-\frac{2}{3}\times t^{-\frac{2}{3}-1} \\ =-8t^{-\frac{5}{3}} \\ =-\frac{8}{t\sqrt[3]{t^2}} \end{gathered}[/tex]d)
[tex]x=4e^{5t+3}[/tex]The first derivative is,
[tex]\begin{gathered} x^{\prime}=\frac{d}{dt}(4e^{5t+3}) \\ =4\times e^{5t+3}\times(5+0) \\ =20e^{5t+3} \end{gathered}[/tex]The second derivative is,
[tex]\begin{gathered} x^{\prime\prime}=\frac{d}{dt}(20e^{5t+3}) \\ =20\times e^{5t+3}\times(5+0)_{} \\ =100e^{5t+3} \end{gathered}[/tex]e)
[tex]y=16\ln (2x)[/tex]The first derivative is,
[tex]\begin{gathered} y^{\prime}=\frac{d}{dt}(16\ln (2x)) \\ =16\times\frac{1}{2x}\times2 \\ =\frac{16}{x} \end{gathered}[/tex]The second derivative is,
[tex]\begin{gathered} y^{\prime\prime}=\frac{d}{dt}(\frac{16}{x}) \\ =\frac{d}{dt}(16x^{-1}) \\ =16\times-1x^{-1-1} \\ =-16x^{-2} \\ =-\frac{16}{x^2} \end{gathered}[/tex]f)
[tex]x=4\sin (3\theta-1)+5\cos (2\theta+7)[/tex]The first derivate is,
[tex]\begin{gathered} x^{\prime}=\frac{d}{d\theta}(4\sin (3\theta-1)+5\cos (2\theta+7)) \\ =4\times\cos (3\theta-1)\times(3-0)+5\times-\sin (2\theta+7)\times(2+0) \\ =12\cos (3\theta-1)-10\sin (2\theta+7) \end{gathered}[/tex]The second derivative is,
[tex]\begin{gathered} x^{\prime\prime}=\frac{d}{d\theta}(12\cos (3\theta-1)-10\sin (2\theta+7)) \\ =12\times-\sin (3\theta-1)\times(3-0)-10\times\cos (2\theta+7)\times(2+0) \\ =-36\sin (3\theta-1)-20\cos (2\theta+7) \end{gathered}[/tex]