MA005: Calculus I

Practice 1: $\quad \mathrm{f}(\mathrm{x})=5 \mathrm{x}-4$ and $\mathrm{g}(\mathrm{x})=\mathrm{x}^{2}$ so $\mathrm{f}^{\prime}(\mathrm{x})=5$ and $\mathrm{g}^{\prime}(\mathrm{x})=2 \mathrm{x} . \operatorname{fog}(\mathrm{x})=\mathrm{f}(\mathrm{g}(\mathrm{x}))=\mathrm{f}\left(\mathrm{x}^{2}\right)=5 \mathrm{x}^{2}-4$

$D(\operatorname{fog}(x))=f^{\prime}(g(x)) \cdot g^{\prime}(x)=5 \cdot 2 x=10 x$ or $D(\operatorname{fog}(x))=D\left(5 x^{2}-4\right)=10 x$

$\begin{align*}\begin{aligned}\operatorname{g} \circ \mathrm{f}(\mathrm{x})=& \mathrm{g}(\mathrm{f}(\mathrm{x}))=\mathrm{g}(5 \mathrm{x}-4)=(5 \mathrm{x}-4)^{2}=25 \mathrm{x}^{2}-40 \mathrm{x}+16 \\& \mathrm{D}(\mathrm{g} \circ \mathrm{f}(\mathrm{x}))=\mathrm{g}^{\prime}(\mathrm{f}(\mathrm{x})) \cdot \mathrm{f}^{\prime}(\mathrm{x})=2(5 \mathrm{x}-4) \cdot 5=50 \mathrm{x}-40 \text { or } \\& \mathrm{D}(\mathrm{g} \circ \mathrm{f}(\mathrm{x}))=\mathrm{D}\left(25 \mathrm{x}^{2}-40 \mathrm{x}+16\right)=50 \mathrm{x}-40\end{aligned}\end{align*}$

Practice 2: $\quad \frac{d}{d x}\left(\sin \left(4 x+e^{x}\right)\right)=\cos \left(4 x+e^{x}\right) \cdot D\left(4 x+e^{x}\right)=\cos \left(4 x+e^{x}\right) \cdot\left(4+e^{x}\right)$

Practice 3: Fill in the table in Example 6 for $( f ∘ g )(x)$ and $( f ∘ g ) '(x)$ at $x=1, 2$ and $3$.

$\begin{array}{c|c|c|c|c|c|c}x & \mathrm{f}(\mathrm{x}) & \mathrm{g}(\mathrm{x}) & \mathrm{f}^{\prime}(\mathrm{x}) & \mathrm{g} '(\mathrm{x}) & (\mathrm{fog})(\mathrm{x}) & (\mathrm{fog})^{\prime}(\mathrm{x})=\mathrm{f}^{\prime}(\mathrm{g}(\mathrm{x})) \cdot \mathrm{g}^{\prime}(\mathrm{x}) \\\hline-1 & 2 & 3 & 1 & 0 & 0 & 0 \\0 & -1 & 1 & 3 & 2 & 1 & -2 \\1 & 1 & 0 & -1 & 3 & -1 & \mathrm{f}^{\prime}(\mathrm{g}(1)) \cdot \mathrm{g}^{\prime}(1)=\mathrm{f}^{\prime}(0) \cdot(3)=(3)(3)=\mathbf{9} \\2 & 3 & -1 & 0 & 1 & \mathbf{2} & \mathrm{f}^{\prime}(\mathrm{g}(2)) \cdot \mathrm{g}^{\prime}(2)=\mathrm{f}^{\prime}(-1) \cdot(1)=(1)(1)=\mathbf{1} \\3 & 0 & 2 & 2 & -1 & \mathbf{3} & \mathrm{f}^{\prime}\left(\mathrm{g}(3) \cdot \mathrm{g}^{\prime}(3)=\mathrm{f}^{\prime}(2) \cdot(-1)=(0)(-1)=\mathbf{0}\right.\end{array}$

Practice 4: $\mathbf{D}(\sin (7 x-1))=\cos (7 x-1) \mathbf{D}(7 x-1)=7 \cdot \cos (7 x-1)$.

$\frac{d}{d x} \sin (a x+b)=\cos (a x+b) D(a x+b)=a \cdot \cos (a x+b) \quad \frac{d}{d t}\left(e^{3 t}\right)=e^{3 t} \frac{d}{d t}(3 t)=3 \cdot e^{3 t}$

Practice 5: $\quad D(\sin (g(x)))=\cos (g(x)) \cdot g^{\prime}(x) .$ At $x=\pi, \cos (g(\pi)) \cdot g^{\prime}(\pi) \approx \cos (0.86) \cdot(-1) \approx-0.65$. 

$D(g(\sin (x)))=g^{\prime}(\sin (x)) \cdot \cos (x) .$ At $x=\pi, g^{\prime}(\sin (\pi)) \cdot \cos (\pi)=g^{\prime}(0) \cdot(-1) \approx-2$

Practice 6:

$\begin{align*}\begin{aligned}&\mathbf{D}(\sin (\cos (5 \mathrm{x})))=\cos (\cos (5 \mathrm{x})) \cdot \mathbf{D}(\cos (5 \mathrm{x}))=\cos (\cos (5 \mathrm{x})) \cdot(-\sin (5 \mathrm{x})) \cdot \mathbf{D}(5 \mathrm{x})=-5 \cdot \sin (5 \mathrm{x}) \cdot \cos (\cos (5 \mathrm{x})) \\&\frac{\mathrm{d}}{\mathrm{dx}} \mathrm{e}^{\cos (3 \mathrm{x})}=\mathrm{e}^{\cos (3 \mathrm{x})} \mathrm{D}(\cos (3 \mathrm{x}))=\mathrm{e}^{\cos (3 \mathrm{x})}(-\sin (3 \mathrm{x})) \mathbf{D}(3 \mathrm{x})=-3 \cdot \sin (3 \mathrm{x}) \cdot \mathrm{e}^{\cos (3 \mathrm{x})}\end{aligned}\end{align*}$

Practice 7:

$\begin{align*}\begin{aligned}&D\left(\arctan \left(x^{3}\right)\right)=\frac{1}{1+\left(x^{3}\right)^{2}} D\left(x^{3}\right)=\frac{3 x^{2}}{1+x^{6}} \\&\frac{d}{d x}\left(\arctan \left(e^{x}\right)\right)=\frac{1}{1+\left(e^{x}\right)^{2}} \mathbf{D}\left(e^{x}\right)=\frac{e^{x}}{1+e^{2 x}}\end{aligned}\end{align*}$

Practice 8: $\quad D(\sinh (5 x-7))=\cosh (5 x-7) D(5 x-7)=5 \cdot \cosh (5 x-7)$

$\begin{align*}\begin{aligned}&\frac{\mathrm{d}}{\mathbf{d x}} \ln \left(3+\mathrm{e}^{2 \mathrm{x}}\right)=\frac{1}{3+\mathrm{e}^{2 \mathrm{x}}} \mathbf{D}\left(3+\mathrm{e}^{2 \mathrm{x}}\right)=\frac{2 \mathrm{e}^{2 \mathrm{x}}}{3+\mathrm{e}^{2 \mathrm{x}}} \\&\mathbf{D}(\arcsin (1+3 \mathrm{x}))=\frac{1}{\sqrt{1-(1+3 \mathrm{x})^{2}}} \mathbf{D}(1+3 \mathrm{x})=\frac{\mathbf{3}}{\sqrt{1-(1+3 \mathrm{x})^{2}}}\end{aligned}\end{align*}$