Calculus Exercise (6) : Higher Order Derivatives

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


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



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}}\\\\ $
Answer $
\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}$


If $f(x)=x^{3}-2 x^{2}+3 x+1$, find $f^{\prime}(x)$ and $f^{\prime \prime}(x)$.
Answer
$\begin{aligned}
f(x)&=x^{3}-2 x^{2}+3 x+1 \\\\
f^{\prime}(x)&=3 x^{2}-4 x+3 \\\\
f^{\prime \prime}(x)&=6 x-4
\end{aligned}$




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


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



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

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



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

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

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


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

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




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


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

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


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

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



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 $p$.

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




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.

ANSWER
$\begin{aligned}
f(t)&= \dfrac{t^{2}+2}{\sqrt{t}} \\\\
f^{\prime}(t)&= \dfrac{\sqrt{t} \dfrac{d}{d t}\left(t^{2}+2\right)-\left(t^{2}+2\right) \dfrac{d}{d t}(\sqrt{t})}{t} \\\\
&= \dfrac{\sqrt{t}(2 t)-\dfrac{\left(t^{2}+2\right)}{2 \sqrt{t}}}{t} \\\\
&= \dfrac{4 t^{2}-t^{2}-2}{2 t^{3 / 2}} \\\\
&= \dfrac{3 t^{2}-2}{2 t^{3 / 2}}\\\\
\therefore\ f^{\prime}(t) &=\dfrac{3}{2} t^{1 / 2}-t^{-3 / 2} \\\\
f^{\prime \prime}(t) &=\dfrac{3}{4} t^{-1 / 2}+\dfrac{3}{2} t^{-5 / 2} \\\\
&=\dfrac{3}{4 t^{1 / 2}}+\dfrac{3}{2 t^{5 / 2}} \\\\
&=\dfrac{3}{4}\left(\dfrac{1}{t^{1 / 2}}+\dfrac{2}{t^{5 / 2}}\right) \\\\
&=\dfrac{3\left(t^{2}+2\right)}{4 t^{5 / 2}}\\\\
\end{aligned}$


$\therefore$ 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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