Question

f(x) = 2x ^ 2 - 3x ^ 2 + 2x - 1

Answer 1

1a)

$\begin{gathered} f(x)=2x^3-3x^2+2x-1 \\ \Rightarrow f^{\prime}(x)=2\cdot3x^{3-1}-3\cdot2x^{2-1}+2x^{1-1}=6x^2-6x+2^{} \\ \Rightarrow f^{\prime}(x)=6x^2-6x+2 \end{gathered}$

1b)

$G(t)=2\sqrt[]{t}+\frac{3}{\sqrt[]{t}}=2t^{\frac{1}{2}}+3t^{-\frac{1}{2}}$$\begin{gathered} \Rightarrow G^{\prime}(t)=\frac{1}{2}\cdot2t^{\frac{1}{2}-1}+3(-\frac{1}{2})t^{-\frac{1}{2}-1}=t^{-\frac{1}{2}}-\frac{3}{2}t^{-\frac{3}{2}} \\ \Rightarrow G^{\prime}(t)=\frac{1}{\sqrt[]{t}}-\frac{3}{2\sqrt[2]{t^3}} \end{gathered}$

1c)

$g(t)=\frac{4}{t^4}-\frac{3}{t^3}+\frac{2}{t}=4t^{-4}-3t^{-3}+2t^{-1}$$\begin{gathered} \Rightarrow g^{\prime}(t)=4(-4)t^{-4-1}-3(-3)t^{-3-1}+2(-1)t^{-1-1}=-16t^{-5}+9t^{-4}-2t^{-2} \\ \Rightarrow g^{\prime}(t)=-\frac{16}{t^5}+\frac{9}{t^4}-\frac{2}{t^2}^{} \end{gathered}$

1d)

$f(x)=\frac{3}{x^3}+\frac{4}{\sqrt[]{x}}+1=3x^{-3}+4x^{-\frac{1}{2}}+1$$\begin{gathered} \Rightarrow f^{\prime}(x)=3(-3)x^{-3-1}+4(-\frac{1}{2})x^{-\frac{1}{2}-1}=-9x^{-4}-2x^{-\frac{3}{2}}=-\frac{9}{x^4}-\frac{2}{\sqrt[2]{x^3}^{}} \\ \Rightarrow f^{\prime}(x)=-\frac{9}{x^4}-\frac{2}{\sqrt[2]{x^3}^{}} \end{gathered}$

1e)

$f(x)=\frac{x^3+2x^2+x-1}{x}=x^2+2x+1-\frac{1}{x}=x^2+2x+1-x^{-1}$$\begin{gathered} \Rightarrow f^{\prime}(x)=2x+2-(-1)x^{-1-1}=2x+2+x^{-2}=2x+2+\frac{1}{x^2} \\ \Rightarrow f^{\prime}(x)=2x+2+\frac{1}{x^2} \end{gathered}$