Question 13 Marks
Find the rate of change of demand $( x )$ of a commodity with respect to its price $( y )$ if $y =\frac{5 x+7}{2 x-13}$
Answer$
\begin{gathered}
\quad y=\frac{5 x+7}{2 x-13} \\
\therefore \frac{d y}{d x}=\frac{d}{d x}\left(\frac{5 x+7}{2 x-13}\right) \\
=\frac{(2 x-13) \cdot \frac{d}{d x}(5 x+7)-(5 x+7) \cdot \frac{d}{d x}(2 x-13)}{(2 x-13)^2} \\
=\frac{(2 x-13) \cdot(5 \times 1+0)-(5 x+7) \cdot(2 \times 1-0)}{(2 x-13)^2} \\
\\
=\frac{10 x-65-10 x-14}{(2 x-13)^2}=\frac{-79}{(2 x-13)^2}
\end{gathered}
$
$
=\frac{10 x-65-10 x-14}{(2 x-13)^2}=\frac{-79}{(2 x-13)^2}
$
Hence, rate of change of demand $(x)$ w.r.t. price $(y)$
$
=\frac{d x}{d y}=\frac{1}{\left(\frac{x y}{d x}\right)}=-\frac{(2 x-13)^2}{79}
$
View full question & answer→Question 23 Marks
If $y =\sqrt[5]{\left(3 x^2+8 x+5\right)^4}$, find $\frac{d y}{d x}$
Answer$
\text { Given : } y=\sqrt[5]{\left(3 x^2+8 x+5\right)^4}
$
Let $u=3 x^2+8 x+5$
Then $y=\sqrt[5]{u^4}=u^{\frac{4}{5}}$
$
\begin{aligned}
\therefore \frac{d y}{d u} & =\frac{d}{d u}\left(u^{\frac{4}{5}}\right)=\frac{4}{5} u^{\frac{4}{5}-1} \\
& =\frac{4}{5}\left(3 x^2+8 x+5\right)^{-\frac{1}{5}}
\end{aligned}
$
$
\text { and } \begin{aligned}
\frac{d u}{d x} & =\frac{d}{d x}\left(3 x^2+8 x+5\right) \\
& =3 \frac{d}{d x}\left(x^2\right)+8 \frac{d}{d x}(x)+\frac{d}{d x}(5) \\
& =3 \times 2 x+8 \times 1+0=6 x+8
\end{aligned}
$
$
\begin{aligned}
\therefore \frac{d y}{d x} & =\frac{d y}{d u} \cdot \frac{d u}{d x} \\
& =\frac{4}{5}\left(3 x^2+8 x+5\right)^{-\frac{1}{5}} \cdot(6 x+8) .
\end{aligned}
$
View full question & answer→Question 33 Marks
If $y = x ^3+3 x y^2+3 x ^2 y$, find $\frac{d y}{d x}$
Answer$
y=x^3+3 x y^2+3 x^2 y
$
Differentiating both sides w.r.t. $x$, we get
$
\frac{d y}{d x}=3 x^2+3\left[x \cdot \frac{d}{d x}\left(y^2\right)+y^2 \cdot \frac{d}{d x}(x)\right]+
$
$3\left[x^2 \cdot \frac{d y}{d x}+y \cdot \frac{d}{d x}\left(x^2\right)\right]$
$
\begin{aligned}
& \therefore \frac{d y}{d x}=3 x^2+3\left[x \times 2 y \frac{d y}{d x}+y^2 \times 1\right]+3\left[x^2 \cdot \frac{d y}{d x}+y \times 2 x\right] \\
& \therefore \frac{d y}{d x}=3 x^2+6 x y \frac{d y}{d x}+3 y^2+3 x^2 \frac{d y}{d x}+6 x y \\
& \therefore\left(1-6 x y-3 x^2\right) \frac{d y}{d x}=3 x^2+3 y^2+6 x y \\
& \therefore \frac{d y}{d x}=\frac{3 x^2+3 y^2+6 x y}{1-6 x y-3 x^2} \\
& =\frac{-3\left(x^2+y^2+2 x y\right)}{6 x y+3 x^2-1} .
\end{aligned}
$
View full question & answer→Question 43 Marks
If $y=\left(6 x^3-3 x^2-9 x\right)^{10}$, find $\frac{d y}{d x}$
AnswerGiven $y=\left(6 x^3-3 x^2-9 x\right)^{10}$
Let $u=6 x^3-3 x^2-9 x$
Then $y=u^{10}$
$
\begin{aligned}
& \begin{aligned}
\therefore \frac{d y}{d u} & =\frac{d}{d u}\left(u^{10}\right)=10 u^9 \\
& =10\left(6 x^3-3 x^2-9 x\right)^9
\end{aligned} \\
& \text { and } \frac{d u}{d x}=\frac{d}{d x}\left(6 x^3-3 x^2-9 x\right)
\end{aligned}
$
$
\begin{aligned}
& =6 \frac{d}{d x}\left(x^3\right)-3 \frac{d}{d x}\left(x^2\right)-9 \frac{d}{d x}(x) \\
& =6 \times 3 x^2-3 \times 2 x-9 \times 1 \\
& =18 x^2-6 x-9 \\
\therefore \frac{d y}{d x}= & \frac{d y}{d u} \cdot \frac{d u}{d x} \\
& =10\left(6 x^2-3 x^2-9 x\right)^9 \cdot\left(18 x^2-6 x-9\right) .
\end{aligned}
$
View full question & answer→Question 53 Marks
Find $\frac{d y}{d x}$ if, :
$
x=\sqrt{1+u^2}, y=\log \left(1+u^2\right)
$
Answer$
x =\sqrt{1+u^2}, y =\log \left(1+ u ^2\right)
$
Differentiating $x$ and $y$ w.r.t. $u$, we get,
$
\begin{aligned}
& \begin{aligned}
\frac{d x}{d u}=\frac{d}{d u} & \left(\sqrt{1+u^2}\right)=\frac{1}{2 \sqrt{1+u^2}} \cdot \frac{d}{d u}\left(1+u^2\right) \\
= & \frac{1}{2 \sqrt{1+u^2}} \times(0+2 u)=\frac{u}{\sqrt{1+u^2}}
\end{aligned} \\
& \text { and } \frac{d y}{d u}=\frac{d}{d u}\left[\log \left(1+u^2\right)\right] \\
& =\frac{1}{1+u^2} \cdot \frac{d}{d u}\left(1+u^2\right) \\
& =\frac{1}{1+u^2} \times(0+2 u)=\frac{2 u}{1+u^2} \\
& \begin{aligned}
\therefore \frac{d y}{d x}= & \frac{(d y / d u)}{(d x / d u)}=\frac{\left(\frac{2 u}{1+u^2}\right)}{\left(\frac{u}{\sqrt{1+u^2}}\right)} \\
& =\frac{2 u}{1+u^2} \times \frac{\sqrt{1+u^2}}{u}=\frac{2}{\sqrt{1+u^2}}
\end{aligned}
\end{aligned}
$
View full question & answer→Question 63 Marks
Solve the following : If $e ^{ x }+ e ^{ y }= e ^{( x + y )}$, then show that $\frac{d y}{d x}=-e^{y-x}$.
Answer$
e ^{ x }+ e ^{ y }= e ^{( x + y )}
$
Differentiating both sides w.r.t. $x$, we get
$
\begin{aligned}
& e^x+e^y \cdot \frac{d y}{d x}=e^{(x+y)} \cdot \frac{d}{d x}(x+y) \\
& \therefore e^x+e^y \cdot \frac{d y}{d x}=e^{(x+y)} \cdot\left(1+\frac{d y}{d x}\right) \\
& \therefore e^x+e^y \cdot \frac{d y}{d x}=e^{(x+y)}+e^{(x+y)} \cdot \frac{d y}{d x}
\end{aligned}
$
$
\begin{aligned}
& \therefore\left[e^y-e^{(x+y)}\right] \frac{d y}{d x}=e^{(x+y)}-e^x \\
& \therefore\left(e^y-e^x-e^y\right) \frac{d y}{d x}=e^x+e^y-e^x \\
& \therefore-e^x \cdot \frac{d y}{d x}=e^y \\
& \therefore \frac{d y}{d x}=-\frac{e^y}{e^x}=-e^{y-x} .
\end{aligned}
$
View full question & answer→Question 73 Marks
Solve the following : If $\log (x+y)=\log (x y)+a$, then show that $\frac{d y}{d x}=\frac{-y^2}{x^2}$
Answer$
\begin{aligned}
& \log (x+y)=\log (x y)+a \\
& \therefore \log (x+y)=\log x+\log y+a
\end{aligned}
$
Differentiating both sides w.r.t. $x$, we get
$
\begin{aligned}
& \frac{1}{x+y} \cdot \frac{d}{d x}(x+y)=\frac{1}{x}+\frac{1}{y} \cdot \frac{d y}{d x}+0 \\
& \therefore \frac{1}{x+y} \cdot\left(1+\frac{d y}{d x}\right)=\frac{1}{x}+\frac{1}{y} \cdot \frac{d y}{d x} \\
& \therefore \frac{1}{x+y}+\frac{1}{x+y} \cdot \frac{d y}{d x}=\frac{1}{x}+\frac{1}{y} \frac{d y}{d x} \\
& \therefore\left(\frac{1}{x+y}-\frac{1}{y}\right) \frac{d y}{d x}=\frac{1}{x}-\frac{1}{x+y} \\
& \therefore\left[\frac{y-x-y}{y(x+y)}\right] \frac{d y}{d x}=\frac{x+y-x}{x(x+y)} \\
& \therefore\left[\frac{-x}{y(x+y)}\right] \frac{d y}{d x}=\frac{y}{x(x+y)} \\
& \therefore-\frac{x \cdot \frac{d y}{y}}{d x}=\frac{y}{x} \\
& \therefore \frac{d y}{d x}=-\frac{y^2}{x^2} .
\end{aligned}
$
View full question & answer→Question 83 Marks
Find $\frac{d y}{d x}$ if, :
$
x y=\log (x y)
$
Answer$
\begin{aligned}
& x y=\log (x y) \\
& \therefore x y=\log x+\log y
\end{aligned}
$
Differentiating both sides w.r.t. $x_r$ we get
$
\begin{aligned}
& x y=\log (x y) \\
& \therefore x y=\log x+\log y
\end{aligned}
$
Differentiating both sides w.r.t. $x$, we get
$
\begin{aligned}
& x \cdot \frac{d y}{d x}+y \cdot \frac{d}{d x}(x)=\frac{1}{x}+\frac{1}{y} \cdot \frac{d y}{d x} \\
& \therefore x \cdot \frac{d y}{d x}+y \times 1=\frac{1}{x}+\frac{1}{y} \cdot \frac{d y}{d x} \\
& \therefore\left(x-\frac{1}{y}\right) \frac{d y}{d x}=\frac{1}{x}-y \\
& \therefore\left(\frac{x y-1}{y}\right) \frac{d y}{d x}=\frac{1-x y}{x}=\frac{-(x y-1)}{x} \\
& \therefore \frac{1}{y} \frac{d y}{d x}=-\frac{1}{x} \\
& \therefore \frac{d y}{d x}=-\frac{y}{x} .
\end{aligned}
$
View full question & answer→Question 93 Marks
Find $\frac{d y}{d x}$ if, :
$
x^y=e^{(x-y)}
$
Answer$
\begin{aligned}
& x ^{ y }= e ^{( x - y )} \\
& \therefore \log x ^{ y }=\log e ^{( x - y )} \\
& \therefore y \log x =( x - y ) \log e \\
& \therefore y \log x = x - y \ldots \ldots[\because \log e =1] \\
& \therefore y + y \log x = x \\
& \therefore y (1+\log x )= x \\
& \therefore y =\frac{x}{1+\log x} \\
& \therefore \frac{d y}{d x}=\frac{d}{d x}\left(\frac{x}{1+\log x}\right) \\
& \qquad \frac{(1+\log x) \cdot \frac{d}{d x}(x)-x \frac{d}{d x}(1+\log x)}{(1+\log x)^2}
\end{aligned}
$
$
\begin{aligned}
& =\frac{(1+\log x) \cdot 1-x\left(0+\frac{1}{x}\right)}{(1+\log x)^2} \\
& =\frac{1+\log x-1}{(1+\log x)^2} \\
& =\frac{\log x}{(1+\log x)^2} .
\end{aligned}
$
View full question & answer→Question 103 Marks
Find $\frac{d y}{d x}$ if, :
$
y=x^x+a^x
$
Answer$
y=x^x+a^x
$
Let $u=x^x$
Then $\log u=\log x^x=x \cdot \log x$
Differentiating both sides w.r.t. $x$, we get
$
\begin{aligned}
\frac{1}{u} \cdot \frac{d u}{d x} & =\frac{d}{d x}(x \cdot \log x) \\
& =x \cdot \frac{d}{d x}(\log x)+(\log x) \cdot \frac{d}{d x}(x) \\
& =x \times \frac{1}{x}+(\log x) \times 1 \\
\therefore \frac{d u}{d x} & =u(1+\log x)=x^x(1+\log x)
\end{aligned}
$
Now, $y=u+a^x \quad$
$
\begin{aligned}
\therefore \frac{d y}{d x} & =\frac{d u}{d x}+\frac{d}{d x}\left(a^x\right) \\
& =x^x(1+\log x)+a^x \cdot \log a
\end{aligned}
$
View full question & answer→Question 113 Marks
Find $\frac{d y}{d x}$ if, :
$
y=(2 x+5)^x
$
Answer$
\begin{aligned}
& y=(2 x+5)^x \\
& \therefore \log y=\log (2 x+5)^x=x \log (2 x+5)
\end{aligned}
$
Differentiating both sides w.r.t. $x$, we get
$
\begin{aligned}
\frac{1}{y} \cdot \frac{d y}{d x} & =\frac{d}{d x}[x \log (2 x+5)] \\
& =x \frac{d}{d x}[\log (2 x+5)]+[\log (2 x+5)] \cdot \frac{d}{d x}(x) \\
& =x \times \frac{1}{2 x+5} \cdot \frac{d}{d x}(2 x+5)+[\log (2 x+5)] \times 1 \\
& =\frac{x }{2 x+5} \times(2 \times 1+0)+\log (2 x+5) \\
\therefore \frac{d y}{d x} & =y\left[\frac{2 x}{2 x+5}+\log (2 x+5)\right] \\
& =(2 x+5)^x\left[\log (2 x+5)+\frac{2 x}{2 x+5}\right] .
\end{aligned}
$
View full question & answer→Question 123 Marks
Find $\frac{d y}{d x}$ if, :
$
y=\left(1+\frac{1}{x}\right)^x
$
Answer$
\begin{aligned}
& y=\left(1+\frac{1}{x}\right)^x \\
& \therefore \log y=\log \left(1+\frac{1}{x}\right)^x=x \log \left(1+\frac{1}{x}\right)
\end{aligned}
$
Differentiating both sides w.r.t. $x$, we get
$
\begin{aligned}
\frac{1}{y} \cdot \frac{d y}{d x} & =\frac{d}{d x}\left[x \log \left(1+\frac{1}{x}\right)\right] \\
& =x \frac{d}{d x}\left[\log \left(1+\frac{1}{x}\right)\right]+\left[\log \left(1+\frac{1}{x}\right)\right] \cdot \frac{d}{d x}(x) \\
& =x \times \frac{1}{1+\frac{1}{x} \cdot \frac{d}{d x}\left(1+\frac{1}{x}\right)+\left[\log \left(1+\frac{1}{x}\right)\right] \times 1} \\
& =x \times \frac{x}{x+1} \times\left(0-\frac{1}{x^2}\right)+\log \left(1+\frac{1}{x}\right) \\
\therefore \frac{d y}{d x} & =y\left[\frac{-1}{x+1}+\log \left(1+\frac{1}{x}\right)\right] \\
& =\left(1+\frac{1}{x}\right)^x\left[\log \left(1+\frac{1}{x}\right)-\frac{1}{1+x}\right]
\end{aligned}
$
View full question & answer→Question 133 Marks
Find $\frac{d y}{d x}$ if, :
$
y=e^{x^x}
$
Answer$
\begin{aligned}
& y=e^{x^x} \\
& \therefore \log y=\log e^{x^x}=x^x \log e \\
& \therefore \log y=x^x \quad \ldots[\because \log e=1]
\end{aligned}
$
Differentiating both sides w.r.t. $x$, we get
$
\begin{aligned}
& \frac{1}{y} \cdot \frac{d y}{d x}=\frac{d}{d x}\left(x^x\right) \\
& \therefore \frac{d y}{d x}=y \cdot \frac{d}{d x}\left(x^x\right)=e^{x^x} \cdot \frac{d}{d x}\left(x^x\right)
\end{aligned}
$
Let $u=x^x $
Then $\log u=\log x^x=x \log x$
Differentiating both sides w.r.t. $x$, we get
$
\begin{aligned}
& \begin{aligned}
\frac{1}{u} \cdot \frac{d u}{d x} & =\frac{d}{d x}(x \log x) \\
& =x \cdot \frac{d}{d x}(\log x)+(\log x) \cdot \frac{d}{d x}(x) \\
& =x \times \frac{1}{x}+(\log x) \times 1
\end{aligned} \\
& \therefore \frac{d u}{d x}=u(1+\log x) \\
& \therefore \frac{d}{d x}\left(x^x\right)=x^x(1+\log x) \\
& \therefore \text { from }(1), \\
& \frac{d y}{d x}=e^{x^x} \cdot x^x(1+\log x) .
\end{aligned}
$
View full question & answer→Question 143 Marks
Find $\frac{d y}{d x}$ if, :
$y=x^{x^{2 x}}$
Answer$
\begin{aligned}
& y=x^{x^{2 x}} \\
& \therefore \log y=\log x^{x^{2 x}}=x^{2 x} \cdot \log x
\end{aligned}
$
Differentiating both sides w.r.t. $x$, we get
$
\begin{aligned}
\frac{1}{y} \cdot \frac{d y}{d x} & =\frac{d}{d x}\left(x^{2 x} \cdot \log x\right) \\
& =x^{2 x} \cdot \frac{d}{d x}(\log x)+(\log x) \cdot \frac{d}{d x}\left(x^{2 x}\right) \\
& =x^{2 x} \times \frac{1}{x}+(\log x) \cdot \frac{d}{d x}\left(x^{2 x}\right) \\
\therefore \frac{d y}{d x} & =y\left[\frac{x^{2 x}}{x}+(\log x) \cdot \frac{d}{d x}\left(x^{2 x}\right)\right] \\
& =x^{x^{2 x}}\left[\frac{x^{2 x}}{x}+(\log x) \cdot \frac{d}{d x}\left(x^{2 x}\right)\right]
\end{aligned}
$
Let $u=x^{2 x}$
Then $\log u=\log x^{2 x}=2 x \log x$ Differentiating both sides w.r.t. $x$, we get
$
\begin{aligned}
\frac{1}{u} \cdot \frac{d u}{d x} & =2 \frac{d}{d x}(x \log x) \\
& =2\left[x \frac{d}{d x}(\log x)+(\log x) \cdot \frac{d}{d x}(x)\right] \\
& =2\left[x \times \frac{1}{x}+(\log x) \times 1\right]
\end{aligned}
$
$
\begin{aligned}
& \therefore \frac{d u}{d x}=2 u(1+\log x) \\
& \therefore \frac{d}{d x}\left(x^{2 x}\right)=2 x^{2 x}(1+\log x)
\end{aligned}
$
$\therefore$ from (1),
$
\begin{aligned}
\frac{d y}{d x} & =x^{2^{2 x}}\left[\frac{x^{2 x}}{x}+(\log x) \times 2 x^{2 x}(1+\log x)\right] \\
& =x^{2^{2 x}} \cdot x^{2 x} \cdot \log x\left[2(1+\log x)+\frac{1}{x \log x}\right]
\end{aligned}
$
View full question & answer→Question 153 Marks
Find the marginal demand of a commodity where demand is x and price is y : $
y=\frac{5 x+9}{2 x-10}
$
Answer$
\begin{aligned}
& \text { Given : } y=\frac{5 x+9}{2 x-10} \\
& \therefore \frac{d y}{d x}=\frac{d}{d x}\left(\frac{5 x+9}{2 x-10}\right) \\
& =\frac{(2 x-10) \cdot \frac{d}{d x}(5 x+9)-(5 x+9) \cdot \frac{d}{d x}(2 x-10)}{(2 x-10)^2} \\
& =\frac{(2 x-10)(5 \times 1+0)-(5 x+9)(2 \times 1-0)}{(2 x-10)^2} \\
& =\frac{10 x-50-10 x-18}{(2 x-10)^2}=\frac{-68}{(2 x-10)^2} \\
\end{aligned}
$
By the derivative of inverse function,
$
\frac{d x}{d y}=\frac{1}{\left(\frac{d y}{d x}\right)}=-\frac{(2 x-10)^2}{68}
$
Hence, marginal demand $=\frac{d x}{d y}=\frac{-(2 x-10)^2}{68}$
View full question & answer→Question 163 Marks
Find the marginal demand of a commodity where demand is x and price is y : $
y=\frac{x+2}{x^2+1}
$
AnswerGiven : $y=\frac{x+2}{x^2+1}$
$
\begin{aligned}
\therefore \frac{d y}{d x} & =\frac{d}{d x}\left(\frac{x+2}{x^2+1}\right) \\
& =\frac{\left(x^2+1\right) \cdot \frac{d}{d x}(x+2)-(x+2) \cdot \frac{d}{d x}\left(x^2+1\right)}{\left(x^2+1\right)^2} \\
& =\frac{\left(x^2+1\right)(1+0)-(x+2)(2 x+0)}{\left(x^2+1\right)^2} \\
& =\frac{x^2+1-2 x^2-4 x}{\left(x^2+1\right)^2}=\frac{1-4 x-x^2}{\left(x^2+1\right)^2}
\end{aligned}
$
By the derivative of inverse function, $\frac{d x}{d y}=\frac{1}{\left(\frac{d y}{d x}\right)}=\frac{\left(x^2+1\right)^2}{1-4 x-x^2}$
Hence, marginal demand $=\frac{d x}{d y}$
$
=\frac{\left(x^2+1\right)^2}{1-4 x-x^2}
$
View full question & answer→Question 173 Marks
Find the marginal demand of a commodity where demand is x and price is y : $
y=x e^{-x}+7
$
AnswerGiven : $y=x e^{-x}+7$
$
\begin{aligned}
\therefore \frac{d y}{d x} & =\frac{d}{d x}\left(x e^{-x}+7\right) \\
& =\frac{d}{d x}\left(x e^{-x}\right)+\frac{d}{d x}(7) \\
& =x \cdot \frac{d}{d x}\left(e^{-x}\right)+e^{-x} \cdot \frac{d}{d x}(x)+0
\end{aligned}
$
$
\begin{aligned}
& =x \times e^{-x} \cdot \frac{d}{d x}(-x)+e^{-x} \times 1 \\
& =x e^{-x}(-1)+e^{-x} \\
& =e^{-x}(-x+1)=\frac{1-x}{e^x}
\end{aligned}
$
By the derivative of inverse function,
$
\frac{d x}{d y}=\frac{1}{\left(\frac{d y}{d x}\right)}=\frac{e^x}{1-x}
$
Hence, marginal demand $=\frac{d x}{d y}=\frac{e^x}{1-x}$.
View full question & answer→Question 183 Marks
Find the rate of change of demand (x) of a commodity with respect to price (y) if :
$y=25 x+\log \left(1+x^2\right)$
AnswerGiven $y=25 x+\log \left(1+x^2\right)$
$
\begin{aligned}
\therefore \frac{d y}{d x} & =\frac{d}{d x}\left[25 x+\log \left(1+x^2\right)\right] \\
& =25 \frac{d}{d x}(x)+\frac{d}{d x}\left[\log \left(1+x^2\right)\right] \\
& =25 \times 1+\frac{1}{1+x^2} \cdot \frac{d}{d x}\left(1+x^2\right) \\
& =25+\frac{1}{1+x^2} \times(0+2 x) \\
& =25+\frac{2 x}{1+x^2}=\frac{25+25 x^2+2 x}{1+x^2} \\
& =\frac{25 x^2+2 x+25}{1+x^2}
\end{aligned}
$
By derivative of inverse function,
$
\frac{d x}{d y}=\frac{1}{\left(\frac{d y}{d x}\right)}=\frac{1+x^2}{25 x^2+2 x+25}
$
Hence, the rate of change of demand $( x )$ with respect to price $( y )$
$
\frac{d x}{d y}=\frac{1+x^2}{25 x^2+2 x+25}
$
View full question & answer→Question 193 Marks
Find the rate of change of demand (x) of a commodity with respect to price (y) if :
$
y=18 x+\log (x-4)
$
AnswerGiven $y=18 x+\log (x-4)$
$
\begin{aligned}
\therefore \frac{d y}{d x} & =\frac{d}{d x}[18 x+\log (x-4)] \\
& =18 \frac{d}{d x}(x)+\frac{d}{d x}[\log (x-4)] \\
& =18 \times 1+\frac{1}{x-4} \cdot \frac{d}{d x}(x-4)
\end{aligned}
$
$
\begin{aligned}
& =18+\frac{1}{x-4} \times(1-0) \\
& =18+\frac{1}{x-4}=\frac{18 x-72+1}{x-4} \\
& =\frac{18 x-71}{x-4}
\end{aligned}
$
By derivative of inverse function
$
\frac{d x}{d y}=\frac{1}{\left(\frac{d y}{d x}\right)}=\frac{x-4}{18 x-71}
$
Hence, the rate of change of demand ( $x$ ) with respect to price (y) $=\frac{d x}{d y}=\frac{x-4}{18 x-71}$
View full question & answer→Question 203 Marks
Find $\frac{d y}{d x}$ if, :
$
y=5^{(x+\log x)}
$
AnswerGiven : $y=5^{(x+\log x)}$
Let $u=x+\log x$
Then $y=5^u$
$
\begin{aligned}
& \therefore \frac{d y}{d u}=\frac{d}{d u}\left(5^u\right)=5^u \cdot \log 5 \\
& =5^{(x+\log x)} \cdot \log 5 \\
& \text { and } \frac{d u}{d x}=\frac{d}{d x}(x+\log x) \\
& =1+\frac{1}{x} \\
& \therefore \frac{d y}{d x}=\frac{d y}{d u} \cdot \frac{d u}{d x} \\
& =5^{(x+\log x)} \cdot \log 5 \cdot\left(1+\frac{1}{x}\right) . \\
\end{aligned}
$
View full question & answer→Question 213 Marks
Find $\frac{d y}{d x}$ if, :
$
y=a^{(1+\log x)}
$
AnswerGiven : $y=a^{(1+\log x)}$
Let $u=1+\log x$
Then $y=a^u$
$
\begin{aligned}
\therefore \frac{d y}{d u} & =\frac{d}{d u}\left(a^u\right)=a^u \cdot \log a \\
& =a^{(1+\log x)} \cdot \log a
\end{aligned}
$
and $\frac{d u}{d x}=\frac{d}{d x}(1+\log x)$
$
=0+\frac{1}{x}=\frac{1}{x}
$
$
\begin{aligned}
\therefore \frac{d y}{d x} & =\frac{d y}{d u} \cdot \frac{d u}{d x} \\
& =a^{(1+\log x)} \cdot \log a \cdot \frac{1}{x} .
\end{aligned}
$
View full question & answer→Question 223 Marks
Find $\frac{d y}{d x}$ if, :
$
y=e^{5 x^2-2 x+4}
$
AnswerGiven : $y=e^{5 x^2-2 x+4}$
Let $u=5 x^2-2 x+4$
Then $y=e^u$
$
\begin{aligned}
\therefore \frac{d y}{d u} & =\frac{d}{d u}\left(e^u\right)=e^u \\
& =e^{5 x^2-2 x+4}
\end{aligned}
$
and $\frac{d u}{d x}=\frac{d}{d x}\left(5 x^2-2 x+4\right)$
$
\begin{aligned}
& =5 \frac{d}{d x}\left(x^2\right)-2 \frac{d}{d x}(x)+\frac{d}{d x}(4) \\
& =5 \times 2 x-2 \times 1+0=10 x-2 \\
\therefore \frac{d y}{d x} & =\frac{d y}{d u}-\frac{d u}{d x} \\
& =e^{5 x^2-2 x+4} \times(10 x-2) \\
& =(10 x-2) e^{5 x^2-2 x+4} .
\end{aligned}
$
View full question & answer→Question 233 Marks
Find $\frac{d y}{d x}$ if, :
$
y=\log \left(a x^2+b x+c\right)
$
Answer$
\text { Given } y=\log \left(a x^2+b x+c\right)
$
Let $u = ax x ^2+ bx + c$
Then $y=\log u$
$
\begin{aligned}
\therefore \frac{d y}{d u} & =\frac{d}{d u}(\log u)=\frac{1}{u} \\
& =\frac{1}{a x^2+b x+c}
\end{aligned}
$
and $\frac{d u}{d x}=\frac{d}{d x}(a x+b x+c)$
$
\begin{aligned}
& =a \frac{d}{d x}\left(x^2\right)+b \frac{d}{d x}(x)+\frac{d}{d x}(c) \\
& =a \times 2 x+b \times 1 \times 0=2 a x+b \\
\therefore \frac{d y}{d x} & =\frac{d y}{d u} \cdot \frac{d u}{d x}=\frac{1}{a x^2+b x+c} \times(2 a x+b) \\
& =\frac{2 a x+b}{a x^2+b x+c} .
\end{aligned}
$
View full question & answer→Question 243 Marks
Find $\frac{d y}{d x}$ if, :
$
y=\log \left(10 x^4+5 x^3-3 x^2+2\right)
$
AnswerGiven $y=\log \left(10 x^4+5 x^3-3 x^2+2\right)$
Let $u=10 x^4+5 x^3-3 x^2+2$
Then $y =\log u$
$
\begin{aligned}
& \therefore \frac{d y}{d u}=\frac{d}{d u}(\log u)=\frac{1}{u} \\
& =\frac{1}{10 x^4+5 x^3-3 x^2+2} \\
& \text { and } \frac{d u}{d x}=\frac{d}{d x}\left(10 x^4+5 x^3-3 x^2+2\right) \\
& =10 \frac{d}{d x}\left(x^4\right)+5 \frac{d}{d x}\left(x^3\right)-3 \frac{d}{d x}\left(x^2\right)+\frac{d}{d x}(2) \\
& =10 \times 4 x^3+5 \times 3 x^2-3 \times 2 x+0 \\
& =40 x^3+15 x^2-6 x \\
& \therefore \frac{d y}{d x}=\frac{d y}{d u} \cdot \frac{d u}{d x} \\
& =\frac{1}{10 x^4+5 x^3-3 x^2+2} \times\left(40 x^3+15 x^2-6 x\right) \\
& =\frac{40 x^3+15 x^2-6 x}{10 x^4+5 x^3-3 x^2+2} \text {. } \\
\end{aligned}
$
View full question & answer→