Calculus Exercise (7) : Differentiation of Implicit Functions

Explicit Function

A function in which the dependent variable can be written explicitly in terms of the independent variable.

For examples, $y = 3x^2 - 1, y = \sin 2x$, etc. are explicit functions where $y$ is called the dependent variable and $x$ is called the independent variable.

Implicit Functions

An implicit function is a function in which one variable cannot be explicitly expressed in terms of the other.

For examples, $x^2y + xy^3= 3, x + y^2 = cos xy,$ etc. are implicit functions where $y = f (x)$ but sometimes $y$ cannot be explicitly expressed in terms of $x$.

Differentiation of Implicit Functions

Since $y$ is a function of $x$, it is often easier to differentiate an implicit function by differentiating each term in turn by applying the chain, product and quotient rules.

Problems

1. Find $\dfrac{d y}{d x}.\\\\ $
   (a) $x y=5\\\\ $
   (b) $x(x+y)=y^{2}\\\\ $
   (c) $x^{2}-x y^{2}-y^{3}=2\\\\ $
   (d) $x^{3}-4 x y+y^{2}=14\\\\ $
   (e) $\dfrac{1}{x^{2}}+\dfrac{1}{y^{2}}=\dfrac{1}{4}\\\\ $
   (f) $x y-y^{2}+3 x=2\\\\ $
   (g) $\dfrac{1}{x^{3}}+\dfrac{1}{y^{3}}=\dfrac{1}{8}\\\\ $
   (h) $x^{2} y-x y^{2}+3 x=2\\\\ $
   (i) $2 y+5-x^{2}-y^{3}=0$





$\begin{aligned} \text{ (a) }\quad &x y=5\\\\ &\text { Differentiate with respect to } x.\\\\ &x \dfrac{d y}{d x}+y=0 \\\\ &\dfrac{d y}{d x}=-\dfrac{y}{x}\\\\ \text{ (b) }\quad &x(x+y)=y^{2}\\\\ &x^{2}+x y=y^{2}\\\\ &\text { Differentiate with respect to } x.\\\\ &2 x+x \dfrac{d y}{d x}+y=2 y \dfrac{d y}{d x} \\\\ &(2 y-x) \dfrac{d y}{d x}=2 x+y \\\\ &\dfrac{d y}{d x}=\dfrac{2 x+y}{2 y-x}\\\\ \text{ (c) }\quad & x^{2}-x y^{2}-y^{3}=2\\\\ &\text { Differentiate with respect to } x.\\\\ &2 x-2 x y \dfrac{d y}{d x}-y^{2}-3 y^{2} \dfrac{d y}{d x}=0 \\\\ &2 x-y^{2}=2 x y \dfrac{d y}{d x}+3 y^{2} \dfrac{d y}{d x} \\\\ &\left(2 x y-3 y^{2}\right) \dfrac{d y}{d x}=2 x-y^{2} \\\\ &\dfrac{d y}{d x}=\dfrac{2 x-y^{2}}{2 x y-3 y^{2}}\\\\ \text{ (d) }\quad & x^{3}-4 x y+y^{2}=14\\\\ &\text { Differentiate with respect to } x.\\\\ &3 x^{2}-4 x \dfrac{d y}{d x}-4 y+2 y \dfrac{d y}{d x}=0 \\\\ &3 x^{2}-4 y=4 x \dfrac{d y}{d x}-2 y \dfrac{d y}{d x} \\\\ &(4 x-2 y) \dfrac{d y}{d x}=3 x^{2}-4 y \\\\ &\dfrac{d y}{d x}=\dfrac{3 x^{2}-4 y}{4 x-2 y}\\\\ \text{ (e) }\quad & \dfrac{1}{x^{2}}+\dfrac{1}{y^{2}}=\dfrac{1}{4} \\\\ &\text { Differentiate with respect to } x.\\\\ &-\dfrac{2}{x^{3}}-\dfrac{2}{y^{3}} \dfrac{d y}{d x}=0 \\\\ &\dfrac{d y}{d x}=-\dfrac{y^{3}}{x^{3}}\\\\ \text{ (f) }\quad & x y-y^{2}+3 x=2 \\\\ &\text { Differentiate with respect to } x.\\\\ &x \dfrac{d y}{d x}+y-2 y \dfrac{d y}{d x}+3=0 \\\\ &(2 y-x) \dfrac{d y}{d x}=y+3 \\\\ &\dfrac{d y}{d x}=\dfrac{y+3}{2 y-x}\\\\ \text{ (g) }\quad & \dfrac{1}{x^{3}}+\dfrac{1}{y^{3}}=\dfrac{1}{8} \\\\ &\text { Differentiate with respect to } x.\\\\ &-\dfrac{3}{x^{4}}-\dfrac{3}{y^{4}} \dfrac{d y}{d x}=0 \\\\ &\dfrac{d y}{d x}=-\dfrac{y^{4}}{x^{4}}\\\\ \text{ (h) }\quad & x^{2} y-x y^{2}+3 x=2 \\\\ &\text { Differentiate with respect to } x.\\\\ &x^{2} \dfrac{d y}{d x}+2 x y-2 x y \dfrac{d y}{d x}-y^{2}+3=0 \\\\ &\left(x^{2}-2 x y\right) \dfrac{d y}{d x}=y^{2}-2 x y-3 \\\\ &\dfrac{d y}{d x}=\dfrac{y^{2}-2 x y-3}{x^{2}-2 x y}\\\\ \text{ (i) }\quad & 2 y+5-x^{2}-y^{3}=0 \\\\ &\text { Differentiate with respect to } x.\\\\ &2 \dfrac{d y}{d x}-2 x-3 y \dfrac{d y}{d x}=0 \\\\ &(2-3 y) \dfrac{d y}{d x}=2 x \\\\ &\dfrac{d y}{d x}=\dfrac{2 x}{2-3 y} \end{aligned}$



2. Given that $x^{2}+y^{2}=1$, show that $y y^{\prime \prime}+\left(y^{\prime}\right)^{2}+1=0$.





$\begin{aligned} &x^{2}+y^{2}=1\\\\ &\text { Differentiate with respect to } x.\\\\ &2 x+2 y y^{\prime}=0 \\\\ &y y^{\prime}+x=0\\\\ &\text { Differentiate with respect to } x.\\\\ &y y^{\prime \prime}+\left(y^{\prime}\right)^{2}+1=0 \end{aligned}$



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





$\begin{aligned} &x^{2}-y^{2}=3\\\\ &2 x-2 y y^{\prime}=0 \\\\ &x-y y^{\prime}=0 \\\\ &y y^{\prime}=x\\\\ &\text { Differentiate with respect to } x.\\\\ &y y^{\prime \prime}+y^{\prime} \cdot y^{\prime}=1 \\\\ &\therefore\ y^{2} y^{\prime \prime}+y y^{\prime} \cdot y^{\prime}=y \\\\ &\therefore\ y^{2} y^{\prime \prime}+x y^{\prime}=y \quad\left[\because y y^{\prime}=x\right] \end{aligned}$



4. Find the equation of the tangent line to the curve $3 x^{2}+2 y^{2}=3 x y+12$ at the point $(2,3)$.





$\begin{aligned} \text{ Curve } : & 3 x^{2}+2 y^{2}=3 x y+12\\\\ \text{ Differentiate } & \text{ with respect to } x.\\\\ (6 x+4 y) \dfrac{d y}{d x}&=3 x \dfrac{d y}{d x}+3 y \\\\ (4 y-3 x) \dfrac{d y}{d x}&=3 y-6 x \\\\ \therefore \dfrac{d y}{d x}&=\dfrac{3 y-6 x}{4 y-3 x} \\\\ \text { Let }\left(x_{1}, y_{1}\right)&=(2,3)\\\\ m&=\left.\dfrac{d y}{d x}\right|_{(2,3)}\\\\ &=\dfrac{3(3)-6(2)}{4(3)-3(2)}\\\\ &=-\dfrac{1}{2}\\\\ \end{aligned}$ $\therefore$ The equation of tangent at $\left(x_{1}, y_{1}\right)$ is $\begin{aligned} &\\\\ y-y_{1}&=m\left(x-x_{1}\right) \\\\ y-3&=-\dfrac{1}{2}(x-2) \\\\ x+2 y&=8 \end{aligned}$



5. Show that the equation of the tangent to the curve $x^{2}+x y+y=0$ at the point $(a, b)$ is $x(2 a+b)+y(a+1)+b=0$.





$\begin{aligned} \quad &\text{ Curve } : x^{2}+x y+y=0\\\\ \quad &\text{Differentiate with respect to } x.\\\\ \quad &2 x+x \frac{d y}{d x}+y+\frac{d y}{d x}=0 \\\\ \quad &(x+1) \frac{d y}{d x}=-(2 x+y) \\\\ \quad &\frac{d y}{d x}=-\frac{2 x+y}{x+1}\\\\ \end{aligned}$ The gradient of tangent at $(a, b)$ is $\begin{aligned} &\\ \quad \left.\frac{d y}{d x}\right|_{(a, b)}&=-\frac{2 a+b}{a+1}\\\\ \end{aligned}$ Since $(a, b)$ lies on the curve, $\begin{aligned} &\\ \quad & a^{2}+a b+b =0\\\\ \end{aligned}$ Hence the equation of tangent at $(a, b)$ is $\begin{aligned} &\\ \quad & y-b=-\frac{2 a+b}{a+1}(x-a) \\\\ \quad & y(a+1)-a b-b=-x(2 a+b)+2 a^{2}+a b \\\\ \quad & x(2 a+b)+y(a+1)-2 a^{2}-2 a b-b=0 \\\\ \quad & x(2 a+b)+y(a+1)-2 a^{2}-2 a b-2 b+b=0 \\\\ \quad & x(2 a+b)+y(a+1)-2\left(a^{2}+a b+b\right)+b=0 \\\\ \therefore\ & x(2 a+b)+y(a+1)-2(0)+b=0 \\\\ \therefore\ & x(2 a+b)+y(a+1)+b=0 \end{aligned}$



6. Find the coordinates of the points on the curve $x^{2}-y^{2}=3 x y-39$ at which the tangents are (i) parallel (ii) perpendicular to the line $x+y=1$.





Curve: $x^{2}-y^{2}=3 x y-39\\\\ $ Differentiate with respect to $x$. $\begin{aligned} &\\ \quad &2 x-2 y \frac{d y}{d x}=3 x \frac{d y}{d x}+3 y \\\\ \quad &(3 x+2 y) \frac{d y}{d x}=2 x-3 y \\\\ \quad &\frac{d y}{d x}=\frac{2 x-3 y}{3 x+2 y}\\\\ \end{aligned}$ Line: $x+y=1 \Rightarrow y=1-x\\\\ $ $\therefore$ The gradient of line $=-1\\\\ $ (i) tangent // given line $\begin{aligned} &\\ \quad & \frac{2 x-3 y}{3 x+2 y}=-1 \\\\ \quad & 2 x-3 y=-3 x-2 y \\\\ \quad & y=5 x \\\\ \therefore\ & x^{2}-(5 x)^{2}=3 x(5 x)-39 \\\\ &\quad x^{2}-25 x^{2}=15 x^{2}-39 \\\\ \therefore\ & x^{2}=1 \\\\ & x=\pm 1\\\\ \therefore\ & y=\pm 5\\\\ \end{aligned}$ Thus, the points on the curve at which the tangents are parallel to the line $x + y = 1$ are $(-1, -5)$ and $(1, 5)$. (ii) tangent $\perp$ given line $\begin{aligned} &\\ \quad & \frac{2 x-3 y}{3 x+2 y}=1 \\\\ \quad & 2 x-3 y=3 x+2 y \ \quad & x=-5 y \\\\ \therefore\ &(-5 y)^{2}-y^{2}=3(-5 y) y-39 \\\\ \quad & 25 y^{2}-y^{2}=-15 y^{2}-39 \\\\ \quad & y^{2}=-1\\\ \end{aligned}$ Since $y^{2}>0$ for all $y \in \mathbb{R}, y^{2}=-1$ is impossible. Therefore, there is no tangent on the curve which is parallel to the line $x+y=1$.



7. The equation of a curve is $x y(x+y)=2 a^{3}$, where $a$ is a non-zero constant. Show that there is only one point on the curve at which the tangent is parallel to the $x$ axis, and find the coordinates of this point.





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



8. Show that the tangent lines to the curve $x^{2}-x y+y^{2}=3$, at the points where the curve cuts the $x$ axis, are parallel to each other.





Curve $: x^{2}-x y+y^{2}=3\\\\ $ Differentiate with respect to $x$. $\begin{aligned} &\\ 2 x-x \frac{d y}{d x}-y+2 y \frac{d y}{d x}&=0 \\\\ (2 y-x) \frac{d y}{d x}&=y-2 x \\\\ \frac{d y}{d x}&=\frac{y-2 x}{2 y-x}\\\\ \end{aligned}$ When the curve cuts the $x$ axis, $\begin{aligned} &\\ y&=0 \\\\ x^{2}&=3\\\\ x&=\pm \sqrt{3}\\\\ \end{aligned}$ $\therefore$ The curve cuts the $x$-axis at $(-\sqrt{3}, 0)$ and $(\sqrt{3}, 0)$. $\begin{aligned} &\\ \therefore\ &\text{ At }(-\sqrt{3}, 0), \\\\ m_{1}&=\left.\frac{d y}{d x}\right|_{(-\sqrt{3}, 0)}\\\\ &=\frac{0-2(-\sqrt{3})}{2(0)-(-\sqrt{3})}\\\\ &=2\\\\ \therefore\ & \text{ At } (\sqrt{3}, 0),\\\\ m_{2}&=\left.\frac{d y}{d x}\right|_{(\sqrt{3}, 0)}\\\\ &=\frac{0-2(\sqrt{3})}{2(0)-(\sqrt{3})}\\\\ &=2\\\\ \therefore\ m_{1}&=m_{2}\\\\ \end{aligned}$ Hence, The two tangents are parallel.



9. Show that the equation of the tangent line to the curve $\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1$ at $(p, q)$ is $\dfrac{p x}{a^{2}}+\dfrac{q y}{b^{2}}=1$.





$\begin{aligned} \text{Curve }: &\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\\\\ \text{At } (p, q), & \frac{p^{2}}{a^{2}}+\frac{q^{2}}{b^{2}}=1\\\\ \text{Differentiate } & \text{ with respect to } x.\\\\ \frac{2 x}{a^{2}}+\frac{2 y}{b^{2}} \cdot \frac{d y}{d x} &=0 \\\\ \frac{d y}{d x}& =-\frac{b^{2} x}{a^{2} y}\\\\ \text{At } (p, q), &\\\\ \left.\frac{d y}{d x}\right|_{(p, q)}&=-\frac{b^{2} p}{a^{2} q}\\\\ \text{The equation } & \text{ of tangent at } (p, q) \text{ is}\\\\ y-q&=-\frac{b^{2} p}{a^{2} q}(x-p)\\\\ \text{Multiplying } & \text{ both sides with } \frac{q}{b^{2}},\\\\ \frac{q y}{b^{2}}-\frac{q^{2}}{b^{2}}&=-\frac{p x}{a^{2}}+\frac{p^{2}}{a^{2}} \\\\ \frac{p x}{a^{2}}+\frac{q y}{b^{2}}&=\frac{p^{2}}{a^{2}}+\frac{q^{2}}{b^{2}} \\\\ \therefore \frac{p x}{a^{2}}+\frac{q y}{b^{2}}&=1 \quad \left[\because \frac{p^{2}}{a^{2}}+\frac{q^{2}}{b^{2}}=1\right] \end{aligned}$



10. Show that the equation of the tangent line to the curve $\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1$ at $(m, n)$ is $\dfrac{m x}{a^{2}}-\dfrac{n y}{b^{2}}=1 .$





$\begin{aligned} \text{Curve }: \qquad & \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\\\\ \text{At } (m, n),\qquad & \frac{m^{2}}{a^{2}}-\frac{n^{2}}{b^{2}}=1\\\\ \text{Differentiate } & \text{ with respect to } x.\\\\ \frac{2 x}{a^{2}}-\frac{2 y}{b^{2}} \cdot \frac{d y}{d x} &=0 \\\\ \frac{d y}{d x} &=\frac{b^{2} x}{a^{2} y} \\\\ \text { At } & (m, n),\\\\ \left.\frac{d y}{d x}\right|_{(m, n)} &=\frac{b^{2} m}{a^{2} n}\\\\ \text{The equation } & \text{ of tangent at } (m, n) \text{ is}\\\\ y-n &=\frac{b^{2} m}{a^{2} n}(x-m)\\\\ \text{Multiplying } & \text{ both sides with } \frac{n}{b^{2}},\\\\ \frac{n y}{b^{2}}-\frac{n^{2}}{b^{2}} &=\frac{m x}{a^{2}}-\frac{m^{2}}{a^{2}} \\\\ \frac{m x}{a^{2}}-\frac{n y}{b^{2}} &=\frac{m^{2}}{a^{2}}-\frac{n^{2}}{b^{2}} \\\\ \therefore \frac{m x}{a^{2}}-\frac{n y}{b^{2}} &=1 \quad\left[\because \frac{m^{2}}{a^{2}}-\frac{n^{2}}{b^{2}}=1\right] \end{aligned}$



11. Show that the tangents to each curve $2 x^{2}+y^{2}=6$ and $y^{2}=4 x$ at the point of intersection of the two curves are perpendicular to each other.





Curve 1: $2 x^{2}+y^{2}=6\\\\ $ Differentiate with respect to $x$. $\begin{aligned} &\\ 4 x+2 y \frac{d y}{d x}=0\\\\ \frac{d y}{d x}=-\frac{2 x}{y}\\\\ \end{aligned}$ Curve 2: $y^{2}=4 x\\\\ $ Differentiate with respect to $x$. $\begin{aligned} &\\ 2 y \frac{d y}{d x}=4 \\\\ \frac{d y}{d x}=\frac{2}{y}\\\\ \end{aligned}$ At the point of intersection of two curves, $\begin{aligned} &\\ 2 x^{2}+4 x &=6 \\\\ x^{2}+2 x-3 &=0 \\\\ \therefore\ (x+3)(x-1)&=0 \\\\ \therefore\ x =-3 \text { or } x&=1\\\\ \end{aligned}$ When $x=-3, y^{2}=4(-3)=-12 \notin \mathrm{R} \text{(reject)}\\\\ $ When $x=1, y^{2}=4(1)=2\\\\ $ Thus, the point of intersection of the two curves is $(1,2)\\\\ $. At $(1,2)$, gradient of Curve 1 $=m_{1}=-\frac{2}{2}=-1\\\\ $ At $(1,2)$, gradient of Curve 2 $=m_{2}=\frac{2}{2}=1\\\\ $ $\therefore m_{1} m_{2}=-1(1)=-1\\\\ $ Therefore, the tangents to each curve $2 x^{2}+y^{2}=6$ and $y^{2}=4 x$ at the point of intersection of the two curves are perpendicular to each other.



12. Let $l$ be any tangent line to the curve $\sqrt{x}+\sqrt{y}=\sqrt{c}$ where $c$ is a non-zero constant. If $l$ cuts the $x$-axis at $(a, 0)$ and the $y$-axis at $(0, b)$, show that $a+b=c$.





Curve: $\sqrt{x}+\sqrt{y}=\sqrt{c}$ Differentiate with respect to $x$. $\begin{aligned} &\\ \frac{1}{2 \sqrt{x}}+\frac{1}{2 \sqrt{y}} \cdot \frac{d y}{d x}&=0 \\\\ \frac{d y}{d x}&=-\frac{\sqrt{y}}{\sqrt{x}}\\\\ \end{aligned}$ Let $\left(x_{1}, y_{1}\right)$ be any point on the curve. $\begin{aligned} &\\ \therefore \sqrt{x_{1}}+\sqrt{y_{1}}&=\sqrt{c} \\\\ \left.\frac{d y}{d x}\right|_{\left(x_{1}, y_{1}\right)}&=-\frac{\sqrt{y_{1}}}{\sqrt{x_{1}}}\\\\ \end{aligned}$ The equation of tangent at $\left(x_{1}, y_{1}\right)$ is $\begin{aligned} &\\ y-y_{1}=-\frac{\sqrt{y_{1}}}{\sqrt{x_{1}}}\left(x-x_{1}\right)\\\\ \end{aligned}$ The tangent cuts $x$-axis at $(a, 0)$ and $y$-axis at $(0, b)$. $\begin{aligned} &\\ \therefore 0-y_{1}&=-\frac{\sqrt{y_{1}}}{\sqrt{x_{1}}}\left(a-x_{1}\right) \\\\ a&=x_{1}+\sqrt{x_{1} y_{1}} \\\\ \therefore b-y_{1}&=-\frac{\sqrt{y_{1}}}{\sqrt{x_{1}}}\left(0-x_{1}\right)\\\\ b&=y_{1}+\sqrt{x_{1} y_{1}} \\\\ \therefore\ a+b&=x_{1}+2 \sqrt{x_{1} y_{1}}+y_{1} \\\\ a+b&=\left(\sqrt{x_{1}}\right)^{2}+2 \sqrt{x_{1} y_{1}}+\left(\sqrt{y_{1}}\right)^{2} \\\\ a+b&=\left(\sqrt{x_{1}}+\sqrt{y_{1}}\right)^{2} \\\\ a+b&=(\sqrt{c})^{2} \\\\ \therefore\ a+b&=c \end{aligned}$



13. Find the normals to the curve $x y+2 x-y=0$ that are parallel to the line $2 x+y=0$.





$\begin{aligned} \text{Curve }: x y+2 x-y=0\\\\ \text{Differentiate with respect to } x.\\\\ x \frac{d y}{d x}+y+2-\frac{d y}{d x}&=0 \\\\ (1-x) \frac{d y}{d x}&=y+2 \\\\ \frac{d y}{d x}&=\frac{y+2}{1-x} \\\\ \therefore \text { Gradient of tangent }&=\frac{2+y}{1-x} \\\\ \therefore \text { Gradient of normal }&=-\frac{1}{\frac{d y}{d x}} \\\\ &=\frac{x-1}{y+2}\\\\ \text{Line }: 2 x+y&=0\\\\ \therefore\ y&=-2 x\\\\ \therefore\ \text{ Gradient of line } &=-2\\\\ \end{aligned}$ Since normals $\parallel$ given line, $\begin{aligned} \frac{x-1}{y+2}&=-2 \\\\ \therefore\ x&=-2 y-3 \end{aligned}$ Substituting $x=-2, y-3$ in curve equation, $\begin{aligned} &\\ (-2 y-3) y+2(-2 y-3)-y&=0 \\\\ \therefore\ y^{2}+4 y+3&=0 \\\\ \therefore\ (y+3)(y+1)&=0 \\\\ \therefore\ y=-3 \text { or } y&=-1\\\\ \text{When } y=-3,\quad x&=3\\\\ \text{When } y=-1,\quad x&=-1\\\\ \end{aligned}$ The equation of normal line $\begin{aligned} &\\ &(-1,-1) \text { is } y+1=-2(x+1) \\\\ &\therefore \quad 2 x+y+3=0\\\\ \end{aligned}$ The equation of normal line at $(3,-3)$ is $\begin{aligned} &\\ y+3&=-2(x-3) \\\\ \therefore\ 2 x+y-3&=0\\\\ \end{aligned}$



14. If $f(x)+x^{2}[f(x)]^{3}=10$ and $f(1)=2$, find $f^{\prime}(1)$.





$\begin{aligned} f(x)+x^{2}[f(x)]^{3} &=10 \\\\ \text { Differentiate with respect to } & x . \\\\ f^{\prime}(x)+3 x^{2}[f(x)]^{2} \cdot f^{\prime}(x)+2 x[f(x)]^{3} &=0 \\\\ f(1) &=2 \quad \text { (given) } \\\\ \text { When } x &=1 \\\\ \therefore\ f^{\prime}(1)+3(1)^{2}[f(1)]^{2} \cdot f^{\prime}(1)+2(1)[f(1)]^{3} &=0 \\\\ \therefore\ f^{\prime}(1)+3(1)^{2}[2]^{2} \cdot f^{\prime}(1)+2(1)[2]^{3} &=0 \\\\ \therefore\ f^{\prime}(1)+12 f^{\prime}(1)+16 &=0 \\\\ \therefore\ 13 f^{\prime}(1) &=-16 \\\\ \therefore\ f^{\prime}(1) &=-\frac{16}{13} \end{aligned}$



15. If $L$ is any normal line to the curve $x^{2}+y^{2}=1$, show that $L$ passes through the origin.





Curve: $x^{2}+y^{2}=1$ Differentiate with respect to $x$. $\begin{aligned} &\\ 2 x+2 y \frac{d y}{d x}&=0 \\\\ \frac{d y}{d x}&=-\frac{x}{y} \\\\ \therefore \text { Gradient of tangent }&=-\frac{x}{y}\\\\ \therefore \text { Gradient of normal }&=\frac{y}{x}\\\\ \end{aligned}$ Let $(a, b)$ be any point on the curve. $\begin{aligned} &\\\\ \therefore a^{2}+b^{2}&=1\\\\ b^{2}&=1-a^{2}\\\\ \end{aligned}$ Gradient of normal at $(a, b)=\dfrac{b}{a}$ Equation of normal at $(a, b)$ is $\begin{aligned} &\\ y-b&=\frac{b}{a}(x-a) \\\\ \therefore\ L: y&=\frac{b}{a} x\\\\ \text{When } x=0, y&=0\\\\ \end{aligned}$ $\therefore\ L$ passes through the origin.



16. Where does the normal line to the curve $x^{2}-x y+y^{2}=3$ at the point $(-1,1)$ intersect the curve a second time?





Curve: $x^{2}-x y+y^{2}=3\\\\ $ Differentiate with respect to $x$. $\begin{aligned} &\\ 2 x-x \frac{d y}{d x}-y+2 y \frac{d y}{d x}&=0 \\\\ \frac{d y}{d x}&=\frac{2 x-y}{x-2 y} \\\\ \text { At } (-1,1),& \\\\ \text { Gradient of tangent } &=\frac{d y}{d x} \\\\ &=\frac{2(-1)-1}{-1-2(1)} \\\\ &=1\\\\ \therefore \text { Gradient of normal }&=-1\\\\ \end{aligned}$ Let $(a, b)$ be a point on the curve that the normal intersect the curve a second time. $\begin{aligned} &\\ \therefore \frac{b-1}{a+1}=-1 \\\\ b=-a\\\\ \end{aligned}$ Since $(a, b)$ lies on the curve, $a^{2}-a b+b^{2}=3$. $\begin{aligned} &\\ \therefore a^{2}-a(-a)+(-a)^{2}&=3 \\\\ \therefore 3 a^{2}&=3 \\\\ a^{2}&=1 \\\\ \therefore a&=\pm 1\\\\ \text{ When } a=1, b=-1\\\\ \end{aligned}$ $\therefore$ The second point of intersection of normal and the curve is $(1,-1)$.

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