Calculus Exercise (6) : Higher Order Derivatives

1. Let $y=5 x^{3}+7 x^{2}+6$. Find $\dfrac{d y}{d x}, \dfrac{d^{2} y}{d x^{2}}, \dfrac{d^{3} y}{d x^{3}}$.





$\begin{aligned} y &=5 x^{3}+7 x^{2}+6 . \\\\ \dfrac{d y}{d x} &=\dfrac{d}{d x}\left(5 x^{3}+7 x^{2}+6\right) \\\\ &=15 x^{2}+14 x \\\\ \dfrac{d^{2} y}{d x^{2}} &=\dfrac{d}{d x}\left(15 x^{2}+14 x\right) \\\\ &=30 x+14 \\\\ \dfrac{d^{3} y}{d x^{3}} &=\dfrac{d}{d x}(30 x+14) \\\\ &=30 \end{aligned}$



2. Find $\dfrac{d y}{d x}$ and $\dfrac{d^{2} y}{d x^{2}}$ for each of the following functions.
   (a) $y=\dfrac{x}{x-1}\\\\ $
   (b) $y=x \sqrt{x+2}\\\\ $
   (c) $y=\dfrac{x+1}{x^{2}}\\\\ $
   (d) $y=\left(3 x^{2}-2 x+1\right)^{2}\\\\ $
   (e) $y=(3 x+2)^{20}\\\\ $
   (f) $y=x(2 x-1)^{6}\\\\ $
   (g) $y=\dfrac{x+1}{x-1}\\\\ $
   (h) $y=\dfrac{3 x^{2}}{x+3}\\\\ $
   (i) $y=\dfrac{3}{\sqrt{x+2}}\\\\ $
   





$ \begin{aligned} \text{ (a) } \quad\quad y &=\dfrac{x}{x-1} \\\\ \dfrac{d y}{d x} &=\dfrac{(x-1) \dfrac{d}{d x}(x)-x \dfrac{d}{d x}(x-1)}{(x-1)^{2}} \\\\ &=\dfrac{x-1-x}{(x-1)^{2}} \\\\ &=-\dfrac{1}{(x-1)^{2}} \\\\ \dfrac{d^{2} y}{d x^{2}} &=\dfrac{d}{d x}\left(\dfrac{-1}{(x-1)^{2}}\right) \\\\ &=\dfrac{d}{d x}\left[-(x-1)^{-2}\right] \\\\ &=\dfrac{2}{(x-1)^{3}}\\\\ \text{ (b) } \quad\quad y &=x \sqrt{x+2} \\\\ \dfrac{d y}{d x} &=x \dfrac{d}{d x} \sqrt{x+2}+\sqrt{x+2} \dfrac{d}{d x}(x) \\\\ &=\dfrac{x}{2 \sqrt{x+2}}+\sqrt{x+2} \\\\ &=\dfrac{3 x+4}{2 \sqrt{x+2}} \\\\ \dfrac{d^{2} y}{d x^{2}} &=\dfrac{1}{2}\left(\dfrac{\sqrt{x+2} \dfrac{d}{d x}(3 x+4)-(3 x+4) \dfrac{d}{d x} \sqrt{x+2}}{}\right) \\\\ &=\dfrac{3 \sqrt{x+2}-\dfrac{3 x+4}{2 \sqrt{x+2}}}{2(x+2)} \\\\ &=\dfrac{6 x+12-3 x-4}{4(x+2)^{\frac{3}{2}}}\\\\ \text{ (c) } \quad\quad y &=\dfrac{x+1}{x^{2}} \\\\ \dfrac{d y}{d x} &=\dfrac{x^{2} \dfrac{d}{d x}(x+1)-(x+1) \dfrac{d}{d x}\left(x^{2}\right)}{x^{4}} \\\\ &=\dfrac{x^{2}-2 x(x+1)}{x^{4}} \\\\ &=-\dfrac{x+2}{x^{3}} \\\\ \dfrac{d^{2} y}{d x^{2}} &=-\dfrac{x^{3} \dfrac{d}{d x}(x+2)-(x+2) \dfrac{d}{d x}\left(x^{3}\right)}{x^{6}} \\\\ &=\dfrac{-x^{3}+3 x^{2}(x+2)}{x^{6}} \\\\ &=\dfrac{2(x+3)}{x^{4}}\\\\ \text{ (d) } \quad\quad y &=\left(3 x^{2}-2 x+1\right)^{2} \\\\ \dfrac{d y}{d x} &=2\left(3 x^{2}-2 x+1\right) \dfrac{d}{d x}\left(3 x^{2}-2 x+1\right) \\\\ &=4(3 x-1)\left(3 x^{2}-2 x+1\right) \\\\ \dfrac{d^{2} y}{d x^{2}} &=4\left[(3 x-1) \dfrac{d}{d x}\left(3 x^{2}-2 x+1\right)+\left(3 x^{2}-2 x+1\right) \dfrac{d}{d x}(3 x-1)\right] \\\\ &=4(3 x-1)(6 x-2)+3\left(3 x^{2}-2 x+1\right) \\\\ &=4\left(27 x^{2}-18 x+5\right)\\\\ \text{ (e) } \quad\quad y &=(3 x+2)^{20} \\\\ \dfrac{d y}{d x} &=20(3 x+2)^{19} \dfrac{d}{d x}(3 x+2) \\\\ &=60(3 x+2)^{19} \\\\ \dfrac{d^{2} y}{d x^{2}} &=1140(3 x+2)^{18} \dfrac{d}{d x}(3 x+2) \\\\ &=3420(3 x+2)^{18}\\\\ \text{ (f) } \quad\quad y &=x(2 x-1)^{6} \\\\ \dfrac{d y}{d x} &=x \dfrac{d}{d x}(2 x-1)^{6}+(2 x-1)^{6} \dfrac{d}{d x}(x) \\\\ &=6 x(2 x-1)^{5} \dfrac{d}{d x}(2 x-1)+(2 x-1)^{6} \\\\ &=12 x(2 x-1)^{5}+(2 x-1)^{6} \\\\ &=(2 x-1)^{5}(14 x-1)\\\\ \dfrac{d^{2} y}{d x^{2}} &=(2 x-1)^{5} \dfrac{d}{d x}(14 x-1)+(14 x-1) \dfrac{d}{d x}(2 x-1)^{5} \\\\ &=14(2 x-1)^{5}+5(14 x-1)(2 x-1)^{4} \dfrac{d}{d x}(2 x-1) \\\\ &=14(2 x-1)^{5}+10(14 x-1)(2 x-1)^{4} \\\\ &=24(2 x-1)^{4}(7 x-1)\\\\ \text{ (g) } \quad\quad y &=\dfrac{x+1}{x-1} \\\\ \dfrac{d y}{d x} &=\dfrac{(x-1) \dfrac{d}{d x}(x+1)-x \dfrac{d}{d x}(x-1)}{(x-1)^{2}} \\\\ &=\dfrac{(x-1)-(x+1)}{(x-1)^{2}} \\\\ &=-\dfrac{2}{(x-1)^{2}} \\\\ \dfrac{d^{2} y}{d x^{2}} &=\dfrac{d}{d x}\left(\dfrac{-2}{(x-1)^{2}}\right) \\\\ &=\dfrac{d}{d x}\left[-2(x-1)^{-2}\right] \\\\ &=\dfrac{4}{(x-1)^{3}}\\\\ \text{ (h) } \quad\quad y &=\dfrac{3 x^{2}}{x+3} \\\\ \dfrac{d y}{d x} &=\dfrac{(x+3) \dfrac{d}{d x}\left(3 x^{2}\right)-3 x^{2} \dfrac{d}{d x}(x+3)}{(x+3)^{2}} \\\\ &=\dfrac{6 x(x+3)-3 x^{2}}{(x+3)^{2}} \\\\ &=\dfrac{3\left(x^{2}+6 x\right)}{(x+3)^{2}} \\\\ \dfrac{d^{2} y}{d x^{2}} &=3\left(\dfrac{(x+3)^{2} \dfrac{d}{d x}\left(x^{2}+6 x\right)-\left(x^{2}+6 x\right) \dfrac{d}{d x}(x+3)^{2}}{(x+3)^{4}}\right) \\\\ &=3\left(\dfrac{2(x+3)^{3}-2(x+3)\left(x^{2}+6 x\right)}{(x+3)^{4}}\right) \\\\ &=\dfrac{6(x+3)\left(x^{2}+6 x+9-x^{2}-6 x\right)}{(x+3)^{3}}\\\\ \text{ (i) } \quad\quad y &=\dfrac{3}{\sqrt{x+2}} \\\\ \dfrac{d y}{d x} &=\dfrac{d}{d x}\left(3(x+2)^{-\frac{1}{2}}\right) \\\\ &=-\dfrac{3}{2(x+2)^{\frac{3}{2}}} \\\\ \dfrac{d^{2} y}{d x^{2}} &=-\dfrac{3}{2} \dfrac{d}{d x}(x+2)^{-\frac{3}{2}} \\\\ &=\dfrac{9}{4(x+2)^{\frac{5}{2}}} \end{aligned}$



3. If $f(x)=x^{3}-2 x^{2}+3 x+1$, find $f^{\prime}(x)$ and $f^{\prime \prime}(x)$.





$\begin{aligned} f(x)&=x^{3}-2 x^{2}+3 x+1 \\\\ f^{\prime}(x)&=3 x^{2}-4 x+3 \\\\ f^{\prime \prime}(x)&=6 x-4 \end{aligned}$



4. If $y=3 x^{2}+4 x$, prove that $x^{2} \dfrac{d^{2} y}{d x^{2}}-2 x \dfrac{d y}{d x}+2 y=0$.





$\begin{aligned} y =&\ 3 x^{2}+4 x \\\\ \dfrac{d y}{d x} =&\ 6 x+4 \\\\ \dfrac{d^{2} y}{d x^{2}} =&\ 6 \\\\ \therefore \quad & x^{2} \dfrac{d^{2} y}{d x^{2}}-2 x \dfrac{d y}{d x}+2 y \\\\ =&\ x^{2}(6)-2 x(6 x+4)+2\left(3 x^{2}+4 x\right) \\\\ =&\ 6 x^{2}-12 x^{2}-8 x+6 x^{2}+8 x \\\\ =&\ 0 \end{aligned}$



5. If $y=\dfrac{2 x^{2}+3}{x}$, prove that $x^{2} \dfrac{d^{2} y}{d x^{2}}+x \dfrac{d y}{d x}=y$.





$\begin{aligned} y &=\dfrac{2 x^{2}+3}{x} \\\\ \dfrac{d y}{d x} &=\dfrac{x(4 x)-\left(2 x^{2}+3\right)}{x^{2}} \\\\ &=\dfrac{2 x^{2}-3}{. .2} \\\\ \dfrac{d^{2} y}{d x^{2}} &=\dfrac{x^{2}(4 x)-2 x\left(2 x^{2}-3\right)}{x^{4}} \\\\ &=\dfrac{6}{x^{3}} \\\\ \therefore x^{2} \dfrac{d^{2} y}{d x^{2}}+x \dfrac{d y}{d x} &=x^{2}\left(\dfrac{6}{x^{3}}\right)+x\left(\dfrac{2 x^{2}-3}{x^{2}}\right) \\\\ &=\dfrac{6+2 x^{2}-3}{x} \\\\ &=\dfrac{2 x^{2}+3}{x} \\\\ &=y \end{aligned}$



6. If $y=x^{2}+2 x+3$, show that $\left(\dfrac{d y}{d x}\right)^{2}+\left(\dfrac{d^{2} y}{d x^{2}}\right)^{3}=4 y$.





$\begin{aligned} y =&\ x^{2}+2 x+3 \\\\ \dfrac{d y}{d x} =&\ 2 x+2 \\\\ \dfrac{d^{2} y}{d x^{2}} =&\ 2 \\\\ \therefore\quad &\left(\dfrac{d y}{d x}\right)^{2}+\left(\dfrac{d^{2} y}{d x^{2}}\right)^{3} \\\\ =&\ (2 x+2)^{2}+2^{3} \\\\ =&\ 4 x^{2}+8 x+12 \\\\ =&\ 4\left(x^{2}+2 x+3\right) \\\\ =&\ 4 y \end{aligned}$



7. If $y=\dfrac{x^{4}-3}{x^{2}}$, show that $x^{2} y^{\prime \prime}+x y^{\prime}=4 y$.





$\begin{aligned} y &=\dfrac{x^{4}-3}{x^{2}} \\\\ y^{\prime} &=\dfrac{x^{2}\left(4 x^{3}\right)-2 x\left(x^{4}-3\right)}{x^{4}} \\\\ &=\dfrac{2\left(x^{4}+3\right)}{x^{3}} \\\\ y^{\prime \prime} &=2\left[\dfrac{x^{3}\left(4 x^{3}\right)-\left(3 x^{2}\right)\left(x^{4}+3\right)}{x^{3}}\right] \\\\ x^{2} y^{\prime \prime}+x y^{\prime} &=x^{2} \cdot \dfrac{2\left(x^{4}-9\right)}{r^{4}}+x \cdot \dfrac{2\left(x^{4}+3\right)}{x^{3}} \\\\ &=\dfrac{2\left(x^{4}-9\right)}{\left.x^{4}-3\right)} \\\\ &=4 y \end{aligned}$



8. If $y=2 x^{3}-\dfrac{3}{x}$, show that $x^{2} y^{\prime \prime}-x y^{\prime}-3 y=0$.





$\begin{aligned} y &=2 x^{3}-\dfrac{3}{x} \\\\ y^{\prime} &=6 x^{2}+\dfrac{3}{x^{2}} \\\\ y^{\prime \prime} &=12 x-\dfrac{6}{x^{3}} \\\\ x^{2} y^{\prime \prime}-x y^{\prime}-3 y &=x^{2}\left(12 x-\dfrac{6}{x^{3}}\right)-x\left(6 x^{2}+\dfrac{3}{x^{2}}\right)-3\left(2 x^{3}-\dfrac{3}{x}\right) \\\\ &=12 x^{3}-\dfrac{6}{x}-6 x^{3}-\dfrac{3}{x}-6 x^{3}+\dfrac{3}{x} \\\\ &=0 \end{aligned}$



9. If $y=x^{2}+x+1$, show that $\left(\dfrac{d y}{d x}\right)^{2}+\dfrac{d^{2} y}{d x^{2}}=4 y-1$.





$\begin{aligned} y &=x^{2}+x+1 \\\\ \dfrac{d y}{d x} &=2 x+1 \\\\ \dfrac{d^{2} y}{d x^{2}} &=2 \\\\ \left(\dfrac{d y}{d x}\right)^{2}+\dfrac{d^{2} y}{d x^{2}} &=(2 x+1)^{2}+2 \\\\ &=4 x^{2}+4 x+3 \\\\ &=4 x^{2}+4 x+4-1 \\\\ \therefore\ \left(\dfrac{d y}{d x}\right)^{2}+\dfrac{d^{2} y}{d x^{2}} &=4\left(x^{2}+x+1\right)-1 \\\\ &=4 y-1 \end{aligned}$



10. Given that $y=(2 x-3)^{3}$, find the value of $x$ when $\dfrac{d^2y}{dx^{2}}=0$.





$\begin{aligned} y &=(2 x-3)^{3} \\\\ \dfrac{d y}{d x} &=3(2 x-3)^{2} \dfrac{d}{d x}(2 x-3) \\\\ &=3(2 x-3)^{2}(2) \\\\ &=6(2 x-3)^{2} \\\\ \dfrac{d^{2} y}{d x^{2}} &=12(2 x-3) \dfrac{d}{d x}(2 x-3) \\\\ &=12(2 x-3)(2) \\\\ &=2 4(2 x-3)\\\\ \dfrac{d^{2} y}{d x^{2}} &=0 \\\\ 2 4(2 x-3) &=0 \\\\ x &=\dfrac{3}{2} \end{aligned}$



11. Given that $f(x)=p x^{3}+(1-3 p) x^{2}-4$. When $x=2,{f}^{\prime \prime}(x)=-1$. Find the value of $p$.





$\begin{aligned} f(x) &=p x^{3}+(1-3 p) x^{2}-4 \\\\ f^{\prime}(x) &=3 p x^{2}+2(1-3 p) x \\\\ f^{\prime \prime}(x) &=6 p x+2(1-3 p) \\\\ \text { When } x &=2 \\\\ f^{\prime \prime}(x) &=-1 \\\\ \therefore f^{\prime \prime}(-2) &=-1 \\\\ 6 p(-2)+2(1-3 p) &=-1 \\\\ -12 p+2-6 p &=-1 \\\\ -18 p &=-3 \\\\ p &=\dfrac{1}{6} \end{aligned}$



12. The displacement of a particle in metres at time $t$ seconds is modelled by the function $$ f(t)=\dfrac{t^{2}+2}{\sqrt{t}} $$ The acceleration of the particle in $\mathrm{m} \mathrm{s}^{-2}$ is the second derivative of this function. Find an expression for the acceleration of the particle at time $t$ seconds.





$\begin{aligned} f(t)&= \dfrac{t^{2}+2}{\sqrt{t}} \\\\ f^{\prime}(t)&= \dfrac{\sqrt{t} \dfrac{d}{d t}\left(t^{2}+2\right)-\left(t^{2}+2\right) \dfrac{d}{d t}(\sqrt{t})}{t} \\\\ &= \dfrac{\sqrt{t}(2 t)-\dfrac{\left(t^{2}+2\right)}{2 \sqrt{t}}}{t} \\\\ &= \dfrac{4 t^{2}-t^{2}-2}{2 t^{3 / 2}} \\\\ &= \dfrac{3 t^{2}-2}{2 t^{3 / 2}}\\\\ \therefore\ f^{\prime}(t) &=\dfrac{3}{2} t^{1 / 2}-t^{-3 / 2} \\\\ f^{\prime \prime}(t) &=\dfrac{3}{4} t^{-1 / 2}+\dfrac{3}{2} t^{-5 / 2} \\\\ &=\dfrac{3}{4 t^{1 / 2}}+\dfrac{3}{2 t^{5 / 2}} \\\\ &=\dfrac{3}{4}\left(\dfrac{1}{t^{1 / 2}}+\dfrac{2}{t^{5 / 2}}\right) \\\\ &=\dfrac{3\left(t^{2}+2\right)}{4 t^{5 / 2}}\\\\ \end{aligned}$ $\therefore$ The acceleration of the particle at time $t$ seconds is $\dfrac{3\left(t^{2}+2\right)}{4 t^{5 / 2}}.$

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