MA005: Calculus I

Practice 1: $\mathrm{D}\left(\mathrm{x}^{2}+\mathrm{y}^{2}\right)=2 \mathrm{x}+2 \mathrm{y} \cdot \mathrm{y}^{\prime}$

                   $\frac{\mathrm{d}}{\mathrm{dx}}(\sin (2+3 \mathrm{y}))=\cos (2+3 \mathrm{y}) \cdot \mathrm{D}(2+3 \mathrm{y})=\cos (2+3 \mathrm{y}) \cdot 3 \mathbf{y}^{\prime}$

Practice 2: Explicitly: $y=\left(3 x^{2}+15\right)^{1 / 3}$ so $y^{\prime}=\frac{1}{3}\left(3 x^{2}+15\right)^{-2 / 3} \cdot D\left(3 x^{2}+15\right)=\frac{1}{3}\left(3 x^{2}+15\right)^{-2 / 3} \cdot(6 x)$

                                   When $(x, y)=(2,3), y^{\prime}=\frac{1}{3}\left(3(2)^{2}+15\right)^{-2 / 3} \cdot(6 \cdot 2)=4(27)^{-2 / 3}=\frac{4}{9}$

                    Implicitly: $D\left(y^{3}-3 x^{2}\right)=D(15)$ so $3 y^{2} \cdot y^{\prime}-6 x=0$ and $y^{\prime}=\frac{2 x}{y^{2}}$.

                                    When $(x, y)=(2,3), y^{\prime}=\frac{2 \cdot(2)}{(3)^{2}}=\frac{4}{9}$.

Practice 3: $y+\sin (y)=x^{3}-x$

Practice 4: To find where the parabola crosses the y-axis, we can set $\mathrm{x}=0$ and solve for the values of $\mathrm{y}$.

Replacing $x$ with 0 in $x^{2}+2 x y+y^{2}+3 x-7 y+2=0$, we have $y^{2}-7 y+2=0$ so $y=\frac{7 \pm \sqrt{(-7)^{2}-4(1)(2)}}{2(1)}=\frac{7 \pm \sqrt{41}}{2} \approx 0.3$ and $6.7.$ The parabola crosses the $\mathrm{y}$-axis approximately at the points $(0,0.3)$ and $(0,6.7)$.

From Example 4, we know that $y^{\prime}=\frac{-2 x-2 y-3}{2 x+2 y-7}$, so

at the point $(0,0.3)$, the slope is approximately $\frac{0-0.6-3}{0+0.6-7} \approx 0.56$, and

at the point $(0,6.7)$, the slope is approximately $\frac{0-13.4-3}{0+13.4-7} \approx-2.56$.

Practice 5: $f^{\prime}(x)=f(x) \cdot D(\ln (f(x)))$ and $f(x)=(2 x+1)^{3}\left(3 x^{2}-4\right)^{7}(x+7)^{4}$

Practice 6: $\mathrm{f}^{\prime}(\mathrm{x})=\mathrm{f}(\mathrm{x}) \cdot \mathrm{D}(\ln (\mathrm{f}(\mathrm{x})))$ and $\mathrm{f}(\mathrm{x})=\mathrm{x}^{\sin (\mathrm{x})} \mathrm{so}$