Calculus Exercise (5) : Change Rule, Product Rule and Quotient Rule

1. Differentiate the following with respect to $x$.
   (a) $\left(2 x^{2}+3 x\right)^{10}\\\\ $
   (b) $\dfrac{1}{3-2 x}\\\\ $
   (c) $\sqrt{9-x^{2}}\\\\ $
   (d) $\left(x^{2}+\dfrac{3}{x}\right)^{5}\\\\ $





$\begin{aligned} \text{(a) }\quad & \frac{d}{d x}\left(2 x^{2}+3 x\right)^{10}\\\\ =& \ 10\left(2 x^{2}+3 x\right)^{9} \frac{d}{d x}\left(2 x^{2}+3 x\right)\\\\ =& \ 10(4 x+3)\left(2 x^{2}+3 x\right)^{9}\\\\ \text{(b) } \quad & \frac{d}{d x}\left(\frac{1}{3-2 x}\right)\\\\ =& \frac{d}{d x}(3-2 x)^{-1}\\\\ =& -1(3-2 x)^{-2} \frac{d}{d x}(3-2 x)\\\\ =& \frac{2}{(3-2 x)^{2}}\\\\ \text{(c) } \quad & \frac{d}{d x} \sqrt{9-x^{2}}\\\\ =& \frac{1}{2 \sqrt{9-x^{2}}} \frac{d}{d x}\left(9-x^{2}\right)\\\\ =& -\frac{x}{\sqrt{9-x^{2}}}\\\\ \text{(d) } \quad & \frac{d}{d x}\left(x^{2}+\frac{3}{x}\right)^{5}\\\\ =& 5\left(x^{2}+\frac{3}{x}\right)^{4} \frac{d}{d x}\left(x^{2}+\frac{3}{x}\right)\\\\ =& 5\left(x^{2}+\frac{3}{x}\right)^{4}\left(2 x-\frac{3}{x^{2}}\right) \end{aligned}$



2. Differentiate $\sqrt{x+7}\left(x^{2}+2\right)^{7}$ with respect to $x$.





$\begin{aligned} &\frac{d}{d x}\left[\sqrt{x+7}\left(x^{2}+2\right)^{7}\right] \\\\ =&\ \sqrt{x+7} \frac{d}{d x}\left(x^{2}+2\right)^{7}+\left(x^{2}+2\right)^{7} \frac{d}{d x} \sqrt{x+7} \\\\ =&\ 7\left(x^{2}+2\right)^{6} \sqrt{x+7} \frac{d}{d x}\left(x^{2}+2\right)+\frac{\left(x^{2}+2\right)^{7}}{2 \sqrt{x+7}} \frac{d}{d x}(x+7) \\\\ =&\ 14 x\left(x^{2}+2\right)^{6} \sqrt{x+7}+\frac{\left(x^{2}+2\right)^{7}}{2 \sqrt{x+7}} \end{aligned}$



3. Differentiate $\dfrac{x^{2}}{\sqrt{x+1}}$ with respect to $x$.





$\begin{aligned} & \frac{d}{d x}\left[\frac{x^{2}}{\sqrt{x+1}}\right] \\\\ =&\ \frac{\sqrt{x+1} \frac{d}{d x}\left(x^{2}\right)-x^{2} \frac{d}{d x} \sqrt{x+1}}{x+1} \\\\ =&\ \frac{2 x \sqrt{x+1}-\frac{x^{2}}{\sqrt{x+1}}}{x+1} \\\\ =&\ \frac{2 x^{2}+2 x-x^{2}}{(x+1) \sqrt{x+1}} \\\\ =&\ \frac{x^{2}+2 x}{(x+1)^{\frac{3}{2}}} \end{aligned}$



4. Differentiate the following with respect to $x.\\\\ $
   (a) $\left(2 x^{2}+3\right)^{4}\left(x^{2}-3 x\right)^{5}\\\\ $
   (b) $\left(3+x^{2}\right) \sqrt{3-x^{2}}\\\\ $
   (c) $\dfrac{\sqrt{x+3}}{x+1}\\\\ $
   (d) $\dfrac{3 x-5}{2 x^{2}+7}\\\\ $
   (e) $\dfrac{2 x-7}{\sqrt{x+7}}\\\\ $
   (f) $\sqrt{\dfrac{x^{2}+1}{x^{2}-1}}\\\\ $
   





$\begin{aligned} \text { (a) }\quad &\dfrac{d}{d x}\left[\left(2 x^{2}+3\right)^{4}\left(x^{2}-3 x\right)^{5}\right] \\\\ =&\ \left(2 x^{2}+3\right)^{4} \dfrac{d}{d x}\left(x^{2}-3 x\right)^{5}+\left(x^{2}-3 x\right)^{5} \dfrac{d}{d x}\left(2 x^{2}+3\right)^{4} \\\\ =&\ 5\left(x^{2}-3 x\right)^{4}\left(2 x^{2}+3\right)^{4} \dfrac{d}{d x}\left(x^{2}-3 x\right)+4\left(2 x^{2}+3\right)^{3}\left(x^{2}-3 x\right)^{5} \dfrac{d}{d x}\left(2 x^{2}+3\right) \\\\ =&\ 3\left(2 x^{2}+3\right)^{3}\left(x^{2}-3 x\right)^{4}\left(12 x^{3}-26 x^{2}+10 x-15\right)\\\\ \text { (b) }\quad &\dfrac{d}{d x}\left[\left(3+x^{2}\right) \sqrt{3-x^{2}}\right] \\\\ =&\ \left(3+x^{2}\right) \dfrac{d}{d x} \sqrt{3-x^{2}}+\sqrt{3-x^{2}} \dfrac{d}{d x}\left(3+x^{2}\right) \\\\ =&\ \dfrac{\left(3+x^{2}\right)}{2 \sqrt{3-x^{2}}} \dfrac{d}{d x}\left(3-x^{2}\right)+2 x \sqrt{3-x^{2}} \\\\ =&\ -\dfrac{x\left(3+x^{2}\right)}{\sqrt{3-x^{2}}}+2 x \sqrt{3-x^{2}} \\\\ =&\ \dfrac{3 x\left(1-x^{2}\right)}{\sqrt{3-x^{2}}}\\\\ \text { (c) }\quad &\dfrac{d}{d x}\left[\dfrac{\sqrt{x+3}}{x+1}\right]\\\\ =&\ \dfrac{(x+1) \dfrac{d}{d x} \sqrt{x+3}-\sqrt{x+3} \dfrac{d}{d x}(x+1)}{(x+1)^{2}}\\\\ =&\ \dfrac{\dfrac{x+1}{2 \sqrt{x+3}}-\sqrt{x+3}}{(x+1)^{2}} \\\\ =&\ -\dfrac{x+5}{2(x+1)^{2} \sqrt{x+3}}\\\\ \text { (d) } \quad &\dfrac{d}{d x}\left[\dfrac{3 x-5}{2 x^{2}+7}\right]\\\\ =&\ \dfrac{\left(2 x^{2}+7\right) \dfrac{d}{d x}(3 x-5)-(3 x-5) \dfrac{d}{d x}\left(2 x^{2}+7\right)}{\left(2 x^{2}+7\right)^{2}} \\\\ =&\ \dfrac{3\left(2 x^{2}+7\right)-4 x(3 x-5)}{\left(2 x^{2}+7\right)^{2}} \\\\ =&\ \dfrac{21+20 x-6 x^{2}}{\left(2 x^{2}+7\right)^{2}} \\\\ \text { (e) } \quad & \dfrac{d}{d x}\left[\dfrac{2 x-7}{\sqrt{x+7}}\right]\\\\ =&\ \dfrac{\sqrt{x+7} \dfrac{d}{d x}(2 x-7)-(2 x-7) \dfrac{d}{d x} \sqrt{x+7}}{x+7} \\\\ =&\ \dfrac{2 \sqrt{x+7}-\dfrac{2 x-7}{\sqrt{x+7}}}{x+7} \\\\ =&\ \dfrac{21}{(x+7)^{\dfrac{3}{2}}}\\\\ \text { (e) } \quad & \dfrac{d}{d x}\left[\sqrt{\dfrac{x^{2}+1}{x^{2}-1}}\right]\\\\ =&\ \dfrac{d}{d x}\left[\dfrac{\sqrt{x^{2}+1}}{\sqrt{x^{2}-1}}\right] \\\\ =&\ \dfrac{\sqrt{x^{2}-1} \cdot \dfrac{d}{d x} \sqrt{x^{2}+1}-\sqrt{x^{2}+1} \cdot \dfrac{d}{d x} \sqrt{x^{2}-1}}{x^{2}-1} \\\\ =&\ \dfrac{\dfrac{\sqrt{x^{2}-1}}{2 \sqrt{x^{2}+1}} \cdot \dfrac{d}{d x}\left(x^{2}+1\right)-\dfrac{\sqrt{x^{2}+1}}{2 \sqrt{x^{2}-1}} \cdot \dfrac{d}{d x}\left(x^{2}-1\right)}{x^{2}-1}=\dfrac{2 x}{\left(1-x^{2}\right) \sqrt{x^{4}-1}} \\\\ =&\ \dfrac{\dfrac{x \sqrt{x^{2}-1}}{\sqrt{x^{2}+1}}-\dfrac{x \sqrt{x^{2}+1}}{\sqrt{x^{2}-1}}}{x^{2}-1}\\\\ =&\ \dfrac{-2 x}{\left(x^{2}-1\right) \sqrt{x^{4}-1}}=\dfrac{2 x}{\left(1-x^{2}\right)} \end{aligned}$



5. Calculate the gradient of the curve $y=\dfrac{3 x^{2}-8}{5-2 x}$ at the point $(2,4)$.





$\begin{aligned} \text { Curve: } y &=\frac{3 x^{2}-8}{5-2 x} \\\\ \frac{d y}{d x} &=\frac{(5-2 x)(6 x)-\left(3 x^{2}-8\right)(-2)}{(5-2 x)^{2}} \\\\ \left.\frac{d y}{d x}\right|_{(2,4)} &=\frac{(5-4)(12)+2(12-8)}{(5-4)^{2}} \\\\ &=20\\\\ \end{aligned}$ $\therefore$ The gradient of the curve $y=\dfrac{3 x^{2}-8}{5-2 x}$ at the point $(2,4)$ is 20 .



6. Calculate the gradient of the curve $y=x \sqrt{x+3}$. Find the coordinate of the point at which the gradient is zero.





Curve: $y=x \sqrt{x+3}$ $\begin{aligned} &\\ &\frac{d y}{d x}=\frac{x}{2 \sqrt{x+3}}+\sqrt{x+3}\\\\ \end{aligned}$ When the gradient of the curve is zero, $\begin{aligned} &\\ &\frac{x}{2 \sqrt{x+3}}+\sqrt{x+3}=0 \\\\ &\therefore 2(x+3)=-x \Rightarrow x=-2 \\\\ &\therefore \text { When } x=-2, y=-2 \sqrt{-2+3}=-2\\\\ \end{aligned}$ The coordinate of the point at which the gradient of the curve is zero is $(-2,-2)$.



7. Find the equation of the normal to the curve $y=\dfrac{x^{2}+1}{x-1}$ at the point on the curve where $x=2$.





$\begin{aligned} \text { Curve: } y &=\frac{x^{2}+1}{x-1} \\\\ \text { When } x &=2, y=5 \\\\ \text { Let }\left(x_{1}, y_{1}\right) &=(2,5) \\\\ \frac{d y}{d x} &=\frac{2 x(x-1)-\left(x^{2}+1\right)}{(x-1)^{2}} \\\\ &=\frac{x^{2}-2 x-1}{(x-1)^{2}} \\\\ \therefore m &=\frac{d y}{d x} \\\\ &=-1 \end{aligned}$ The equation of normal at $\left(x_{1}, y_{1}\right)$ is $\begin{aligned} y-y_{1} &=-\frac{1}{m}\left(x-x_{1}\right) \\\\ y-5 &=1(x-2) \\\\ y &=x+3 \end{aligned}$



8. Find the equation of the normal to the curve $y=\dfrac{4 x+1}{x-1}$ at the point on the curve where $y=5$.





$\begin{aligned} \text { Curve: } y &=\frac{4 x+1}{x-1} \\\\ \text { When } y &=5, \\\\ \frac{4 x+1}{x-1} &=5 \\\\ \therefore x &=6 \\\\ \text { Let }\left(x_{1}, y_{1}\right) &=(6,5) \\\\ \frac{d y}{d x} &=\frac{4(x-1)-(4 x+1)}{(x-1)^{2}} \\\\ &=\frac{-5}{(x-1)^{2}} \\\\ \left.\frac{d y}{d x}\right|_{(6,5)} &=-\frac{1}{5}\\\\ \end{aligned}$ The equation of normal at $\left(x_{1}, y_{1}\right)$ is $\begin{aligned} &\\ y-y_{1} &=-\frac{1}{m}\left(x-x_{1}\right) \\\\ y-5 &=5(x-6) \\\\ 5 x-y &=25 \end{aligned}$



9. If $f(x)=\dfrac{k+x}{1-2 x}$ where $k$ is a constant, find $f^{\prime}(x)$ in terms of $x$ and $k$. Given that $f^{\prime}(3)=0.52$, show that $k=6$.





$\begin{aligned} f(x) &=\frac{k+x}{1-2 x} \\\\ f^{\prime}(x) &=\frac{(1-2 x) \frac{d}{d x}(k+x)-(k+x) \frac{d}{d x}(1-2 x)}{(1-2 x)^{2}} \\\\ &=\frac{(1-2 x)-(k+x)(-2)}{(1-2 x)^{2}} \\\\ \therefore\ f^{\prime}(x) &=\frac{1+2 k}{(1-2 x)^{2}} \\\\ \text { Since } f^{\prime}(3) &=0.52, \\\\ \frac{1+2 k}{(1-6)^{2}} &=0.52 \\\\ \frac{1+2 k}{25} &=0.52 \\\\ 1+2 k &=13 \\\\ \therefore\ k &=6 \end{aligned}$



10. Find the equation of the normal to the curve $y=6-(x-2)^{4}$ at the point on the curve where $x=1$.





$\begin{aligned} \text{ Curve }: y&=6-(x-2)^{4}\\\\ \text{ When } x&=1, y=5\\\\ \text{ Let } \left(x_{1}, y_{1}\right)&=(1,5)\\\\ \frac{d y}{d x} &=-4(x-2)^{3} \\\\ \therefore\ m &=\left.\frac{d y}{d x}\right|_{(1,5)} \\\\ &=4\\\\ \end{aligned}$ The equation of normal at $\left(x_{1}, y_{1}\right)$ is $\begin{aligned} &\\ y-y_{1} &=-\frac{1}{m}\left(x-x_{1}\right) \\\\ y-5 &=-\frac{1}{4}(x-1) \\\\ x+4 y &=21 \end{aligned}$



11. The tangent to the curve $y=\left(\dfrac{x}{2}-1\right)^{6}$, at the point where $x=4$, meets the $y$-axis at $A$. Find the coordinates of $A$.





$\begin{aligned} \text { Curve: } y &= \left(\frac{x}{2}-1\right)^{6} \\\\ \text { When } x &= 4 \\\\ y &= \left(\frac{4}{2}-1\right)^{6} \\\\ &= 1 \\\\ \text { Let }\left(x_{1}, y_{1}\right) &= (4,1) \\\\ \frac{d y}{d x} &= 6\left(\frac{x}{2}-1\right)^{5} \cdot \frac{1}{2} \\\\ &= 3\left(\frac{x}{2}-1\right)^{5} \\\\ m &= \left.\frac{d y}{d x}\right|_{(4,1)} \\\\ &= 3\left(\frac{4}{2}-1\right)^{5} \\\\ &= 3\\\\ \end{aligned}$ $\therefore$ The equation of tangent at $\left(x_{1}, y_{1}\right)$ is $\begin{aligned} &\\ y-y_{1} &=m\left(x-x_{1}\right) \\\\ y-1 &=3(x-4) \\\\ y &=3 x-11\\\\ \end{aligned}$ When this tangent meets $y$-axis, $x=0$ that implies $y=-11\\\\ $. $\therefore$ The point $A$ is $(0,-11)$.



12. Show that the tangent to the curve $y=(x+2 a)^{3}$ at the point where $y=a^{3}$ is $y=3 a^{2} x+4 a^{3} .$





$\begin{aligned} \text { Curve : } y &=(x+2 a)^{3} \\\\ \text { When } y &=a^{3}, \\\\ (x+2 a)^{3} &=a^{3} \\\\ x+2 a &=a \\\\ x &=-a\\\\ \end{aligned}$ $\therefore\left(-a, a^{3}\right)$ is the point on the curve where the tangent exists. $\begin{aligned} &\\ \frac{d y}{d x} &=\frac{d}{d x}(x+2 a)^{3} \\\\ &=3(x+2 a)^{2} \frac{d}{d x}(x+2 a) \\\\ &=3(x+2 a)^{2}(1) \\\\ &=3(x+2 a)^{2}\\\\ \end{aligned}$ The gradient of the tangent at $\left(-a, a^{3}\right)$ on the curve is $\begin{aligned} &\\ m &=\left.\frac{d y}{d x}\right|_{\left(-a, a^{3}\right)} \\\\ &=3(-a+2 a)^{2} \\\\ &=3 a^{2}\\\\ \end{aligned}$ $\therefore$ The equation of tangent at $\left(-a, a^{3}\right)$ is $\begin{aligned} &\\ y-a^{3} &=m(x-(-a)) \\\\ y-a^{3} &=3 a^{2}(x-(-a)) \\\\ y &=3 a^{2} x+4 a^{3} \end{aligned}$



13. Find equations of the tangent lines to the curve $y=\dfrac{x+1}{x-1}$ that are parallel to the line $2 y+x=6$.





$\begin{aligned} \text{Curve }: y&= \frac{x+1}{x-1}\\\\ \frac{d y}{d x}&= \frac{(x-1)-(x+1)}{(x-1)^{2}}\\\\ &= -\frac{2}{(x-1)^{2}}\\\\ \text{Line }: 2 y+x&= 6 \\\\ y&= -\frac{1}{2} x+3\\\\ \end{aligned}$ Since the tangents are parallel to the given line, $\begin{aligned} &\\ -\frac{2}{(x-1)^{2}}&= -\frac{1}{2}\\\\ (x-1)^{2}&= 4\\\\ x-1&= \pm 2 \\\\ x&= -1 or x= 3 \\\\ \text{ When } x&= -1, y= 0 \\\\ \text{ When } x&= 3, y= 2\\\\ \end{aligned}$ Let $\left(x_{1}, y_{1}\right)= (-1,0)$ and $\left(x_{2}, y_{2}\right)= (3,2)\\\\ $ $\therefore$ The equation of tangent at $\left(x_{1}, y_{1}\right)$ is $\begin{aligned} &\\ y-y_{1}&= m\left(x-x_{1}\right)\\\\ y-0&= -\frac{1}{2}(x+1) \\\\ x+2 y&= -1 \end{aligned}$ $\therefore$ The equation of tangent at $\left(x_{2}, y_{2}\right)$ is $\begin{aligned} &\\ y-y_{1}&= m\left(x-x_{1}\right) \\\\ y-2&= -\frac{1}{2}(x-3) \\\\ x+2 y&= 7 \end{aligned}$



14. The equation of a curve is $y=\dfrac{x^{2}+6}{x-3}$. Find the gradient of tangent to the curve at the point $x=4$. Hence, find the equation of the normal to the curve at $x=4$.





$\begin{aligned} y &=\frac{x^{2}+6}{x-3} \\\\ \frac{d y}{d x} &=\frac{(x-3)(2 x)-\left(x^{2}+6\right)(1)}{(x-3)^{2}} \\\\ &=\frac{2 x^{2}-6 x-x^{2}-6}{(x-3)^{2}} \\\\ &=\frac{x^{2}-6 x-6}{(x-3)^{2}}\\\\ \text{ When } x&=4,\\\\ \frac{d y}{d x} &=\frac{4^{2}-6(4)-6}{(4-3)^{2}} \\\\ &=-14\\\\ \therefore\ \text{ Gradient of tomegent } &=-14\\\\ \therefore\ \text{ Gradient of normal } &=\frac{1}{14}\\\\ \text{ When } x&=4,\\\\ y&=\frac{4^{2}+6}{4-3}\\\\ &=22 \end{aligned}$



15. A water trough of length $12 \mathrm{~m}$ has a vertical cross-section in the shape of an equilateral triangle of sides $x \mathrm{~m}$ and height $h \mathrm{~m}$. Express $V$, the volume of water that the trough can hold in terms of $h$. When the height of water in the trough is $1.8 \mathrm{~m}$, its depth is decreasing at a rate of $0.2 \mathrm{~m} / \mathrm{s}$. Find the rate of change in the volume of water in the trough at this instant.
    
    





$\begin{aligned} h &=\dfrac{\sqrt{3}}{2} x \\\\ x &=\dfrac{2}{\sqrt{3}} h \\\\ \text { area of cross-section } &=\dfrac{1}{2} x h \\\\ &=\dfrac{1}{2} \cdot \dfrac{2}{\sqrt{3}} h \cdot h \\\\ &=\dfrac{h^{2}}{\sqrt{3}} \\\\ \text { volume of trough } &=V \\\\ \therefore V &=\dfrac{12 h^{2}}{\sqrt{3}} \\\\ &=\dfrac{4 \sqrt{3}}{3} h^{2} \mathrm{~m}^{3} \\\\ \text { When } h &=1.8 \mathrm{~m}, \\\\ \dfrac{d h}{d t} &=0.2 \mathrm{~m} / \mathrm{s}\\\\ \text { Since } V &=\dfrac{4 \sqrt{3}}{3} h^{2}, \\\\ \dfrac{d V}{d h} &=\dfrac{2 \sqrt{3}}{3} h \\\\ \text { When } h &=1.8 \mathrm{~m} \\\\ \dfrac{d V}{d h} &=\dfrac{2 \sqrt{3}}{3}(1.8) \\\\ &=2.08 \\\\ \dfrac{d V}{d t} &=\dfrac{d V}{d h} \cdot \dfrac{d h}{d t} \\\\ &=2.08 \times 0.2 \\\\ &=0.416 \mathrm{~m}^{3} / \mathrm{s} \end{aligned}$



16. The figure shows part of the curve $y=2 x^{2}+3$. The point $B(x, y)$ is a variable point that moves along the curve for $0 < x <6$. $C$ is a point on the $x$-axis such that $B C$ is parallel to the $y$-axis and $A(6,0)$ lies on the $x$-axis. Express the area of the triangle $A B C, S \mathrm{~cm}^{2}$, in terms of $x$, and find an expression for $\dfrac{dS}{dx} .$ Given that when $x=2$, $S$ is increasing at the rate of $0.8 \mathrm{~unit}^{2} / \mathrm{s}$, find the corresponding rate of change of $x$ at this instant.
    
    





$\begin{aligned} \text { Curve: } y &=2 x^{2}+3 \\\\ \therefore \frac{d y}{d x} &=4 x \\\\ \text { Area of } \triangle A B C &=\frac{1}{2} A C \cdot B C \\\\ S &=\frac{1}{2}(6-x)(y) \\\\ S &=\frac{1}{2}(6-x)\left(2 x^{2}+3\right) \\\\ &=\frac{1}{2}\left(-2 x^{3}+6 x^{2}-3 x+18\right) \\\\ \frac{d S}{d x} &=\frac{1}{2}\left(-6 x^{2}+12 x-3\right) \\\\ &=-\frac{3}{2}\left(2 x^{2}-6 x+1\right)\\\\ \text { When } x&=2,\\\\ \frac{d S}{d x} &=-\frac{3}{2}(8-12+1) \\\\ &=\frac{9}{2} \\\\ \frac{d S}{d t} &=0.08 \text { unit / s}\quad \text { (given) } \\\\ \frac{d S}{d t} &=\frac{d S}{d x} \cdot \frac{d x}{d t} \\\\ 0.08 &=\frac{9}{2} \cdot \frac{d x}{d t} \\\\ \frac{d x}{d t} &=\frac{0.16}{9} \\\\ &=0.018 \text { unit/s } \end{aligned}$

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