Question 12 Marks
Find $\frac{d y}{d x}$ if $y = x ^{ x }$.
Answer$
\begin{aligned}
& y=x^x \\
& \therefore \log y=\log x^x=x \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}(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 \\
& \therefore \frac{d y}{d x}=y(1+\log x) \\
& =x^x(1+\log x) . \\
\end{aligned}
$
View full question & answer→Question 22 Marks
Find the rate of change of demand $(x)$ of a commodity with respect to its price $( y )$ if $y =25+30 x - x ^2$.
Answer$
\begin{aligned}
& y=25+30 x-x^2 \\
\therefore \frac{d y}{d x} & =\frac{d}{d x}\left(25+30 x-x^2\right) \\
& =\frac{d}{d x}(25)+30 \frac{d}{d x}(x)-\frac{d}{d x}\left(x^2\right) \\
& =0+30 \times 1-2 x \\
= & 30-2 x
\end{aligned}
$
Hence, the rate of change of demand $(x)$ w.r.t. price $(y)$
$
=\frac{d x}{d y}=\frac{1}{\left(\frac{d y}{d x}\right)}=\frac{1}{30-2 x}
$
View full question & answer→Question 32 Marks
If $y=[\log (\log (\log x))]^2$, find $\frac{d y}{d x}$
Answer$
\begin{aligned}
& \quad y=[\log (\log (\log x))]^2 \\
& \therefore \frac{d y}{d x}=\frac{d}{d x} \cdot[\log (\log (\log x))]^2 \\
& =2[\log (\log (\log x))] \cdot \frac{d}{d x}[\log (\log (\log x))] \\
& =2 \log [\log (\log x)] \times \frac{1}{\log (\log x)} \cdot \frac{d}{d x}[\log (\log x)] \\
& =\frac{2 \log [\log (\log x)]}{\log (\log x)} \times \frac{1}{\log x} \cdot \frac{d}{d x}(\log x) \\
& =\frac{2 \log [\log (\log x)]}{\log (\log x)} \times \frac{1}{\log x} \times \frac{1}{x} \\
& =\frac{2 \log [\log (\log x)]}{x \cdot \log x \cdot \log (\log x)} .
\end{aligned}
$
View full question & answer→Question 42 Marks
Find $\frac{d^2 y}{d x^2}$, if $y = x ^2 \cdot e ^{ x }$
Answer$
y=x^2 \cdot e^x
$
Differentiating w.r.t. $x$, we get
$
\begin{aligned}
\frac{d y}{d x} & =\frac{d}{d x}\left(x^2 e^x\right)=x^2 \frac{d}{d x}\left(e^x\right)+e^x \cdot \frac{d}{d x}\left(x^2\right) \\
& =x^2 \cdot e^x+e^x \times 2 x=e^x\left(x^2+2 x\right)
\end{aligned}
$
Differentiating again w.r.t. $x$, we get
$
\begin{aligned}
& \begin{aligned}
\frac{d^2 y}{d x^2} & =\frac{d}{d x}\left[e^x\left(x^2+2 x\right)\right] \\
& =e^x \cdot \frac{d}{d x}\left(x^2+2 x\right)+\left(x^2+2 x\right) \cdot \frac{d}{d x}\left(e^x\right) \\
& =e^x(2 x+2)+\left(x^2+2 x\right) \cdot e^x
\end{aligned} \\
& = e ^{ x }\left(2 x +2+ x ^2+2 x \right) \\
& = e ^{ x }\left( x ^2+4 x +2\right) .
\end{aligned}
$
View full question & answer→Question 52 Marks
Find $\frac{d^2 y}{d x^2}$, if $y =2 at , x = at t ^2$.
Answer$
x=a t^2, y=2 a t
$
Differentiating $x$ and $y$ w.r.t. $t$, we get
$
\begin{aligned}
& \frac{d x}{d t}=\frac{d}{d t}\left(a t^2\right)=a \frac{d}{d t}\left(t^2\right) \\
& =a \times 2 t=2 a t \\
& \text { and } \frac{d y}{d t}=\frac{d}{d t}(2 a t)=2 a \frac{d}{d t}(t) \\
& =2 a \times 1=2 a \\
& \therefore \frac{d y}{d x}=\frac{(d y / d t)}{(d x / d t)}=\frac{2 a}{2 a t}=\frac{1}{t} \\
& \therefore \frac{d^2 y}{d x^2}=\frac{d}{d x}\left(\frac{1}{t}\right)=\frac{d}{d t}\left(\frac{1}{t}\right) \cdot \frac{d t}{d x} \\
& =-\frac{1}{t^2} \times \frac{1}{\left(\frac{d x}{d t}\right)}=-\frac{1}{t^2} \times \frac{1}{2 a t} \\
& =-\frac{1}{2 a t^3} \text {. } \\
\end{aligned}
$
View full question & answer→Question 62 Marks
Differentiate $e ^{(4 x+5)}$ with resepct to $10^{4 x}$.
AnswerLet $u=e^{(4 x+5)}$ and $v=10^{4 x}$
Then we want to find $\frac{d u}{d v}$
Differentiating $u$ and $v$ w.r.t. $x$, we get
$
\begin{aligned}
\frac{d u}{d x} & =\frac{d}{d x}\left[e^{(4 x+5)}\right]=e^{(4 x+5)} \cdot \frac{d}{d x}(4 x+5) \\
& =e^{(4 x+5)} \times(4 \times 1+0)=4 e^{(4 x+5)}
\end{aligned}
$
$
\begin{aligned}
& \text { and } \begin{aligned}
\frac{d v}{d x} & =\frac{d}{d x}\left(10^{4 x}\right)=10^{4 x} \cdot \log 10 \cdot \frac{d}{d x}(4 x) \\
& =10^{4 x} \cdot(\log 10) \times 4=4 \cdot 10^{4 x} \cdot \log 10 \\
\therefore \frac{d u}{d v} & =\frac{(d u / d x)}{(d v / d x)}=\frac{4 e^{(4 x+5)}}{4 \cdot 10^{4 x} \cdot \log 10}=\frac{e^{(4 x+5)}}{10^{4 x} \cdot \log 10} .
\end{aligned}
\end{aligned}
$
View full question & answer→Question 72 Marks
Differentiate $\log \left(1+x^2\right)$ with respect to $a^x$.
AnswerLet $u =\log \left(1+ x ^2\right)$ and $v = a ^{ x }$
Then we want to find $\frac{d u}{d v}$
Differentiating $u$ and $v$ w.r.t. $x$, we get
$
\begin{aligned}
\frac{d u}{d x} & =\frac{d}{d x}\left[\log \left(1+x^2\right)\right]=\frac{1}{1+x^2} \cdot \frac{d}{d x}\left(1+x^2\right) \\
& =\frac{1}{1+x^2} \times(0+2 x)=\frac{2 x}{1+x^2}
\end{aligned}
$
and $\frac{d v}{d x}=\frac{d}{d x}\left(a^x\right)=a^x \cdot \log a$
$
\therefore \frac{d u}{d v}=\frac{(d u / d x)}{(d v / d x)}=\frac{\left(\frac{2 x}{1+x^2}\right)}{a^x \cdot \log a}=\frac{2 x}{\left(1+x^2\right) \cdot a^x \log a} \text {. }
$
View full question & answer→Question 82 Marks
Find $\frac{d y}{d x}$ if $x = e ^{3 t }, y =e^{\sqrt{t}}$.
Answer$
x = e ^{3 t }, y =e^{\sqrt{t}}
$
Differentiating $x$ and $y$ w.r.t. $t$, we get
$
\begin{aligned}
\frac{d x}{d t} & =\frac{d}{d t}\left(e^{3 t}\right)=e^{3 t} \cdot \frac{d}{d t}(3 t) \\
& =e^{3 t} \times 3=3 e^{3 t}
\end{aligned}
$
$
\text { and } \frac{d y}{d t}=\frac{d}{d t}(e \sqrt{t})=e \sqrt{t} \cdot \frac{d}{d t}(\sqrt{t})
$
$
\begin{aligned}
& =e \sqrt{t} \times \frac{1}{2 \sqrt{t}}=\frac{e \sqrt{t}}{2 \sqrt{t}} \\
\therefore \frac{d y}{d x} & =\frac{(d y / d t)}{(d x / d t)}=\frac{\left(\frac{e \sqrt{t}}{2 \sqrt{t}}\right)}{3 e^{3 t}} \\
& =\frac{1}{6 \sqrt{t}} \cdot e^{(\sqrt{t}-3 t)} .
\end{aligned}
$
View full question & answer→Question 92 Marks
Find $\frac{d y}{d x}$ if $x =5 t ^2, y =10 t$
Answer$
x=5 t^2, y=10 t
$
Differentiating $x$ and $y$ w.r.t. $t$, we get
$
\begin{aligned}
& \frac{d x}{d t}=5 \frac{d}{d t}\left(t^2\right)=5 \times 2 t=10 t \\
& \text { and } \frac{d y}{d t}=10 \frac{d}{d t}(t)=10 \times 1=10 \\
& \therefore \frac{d y}{d x}=\frac{(d y / d t)}{(d x / d t)}=\frac{10}{10 t}=\frac{1}{t}.
\end{aligned}
$
View full question & answer→Question 102 Marks
If $x^3 y^3=x^2-y^2$, find $\frac{d y}{d x}$.
Answer$
x^3 y^3=x^2-y^2
$
Differentiating both sides w.r.t. $x$, we get
$
\begin{aligned}
& x^3 \cdot \frac{d}{d x}\left(y^3\right)+y^3 \cdot \frac{d}{d x}\left(x^3\right)=2 x-2 y \frac{d y}{d x} \\
& \therefore x^3 \times 3 y^2 \frac{d y}{d x}+y^3 \times 3 x^2=2 x-2 y \frac{d y}{d x} \\
& \therefore\left(3 x^2 y^2+2 y\right) \frac{d y}{d x}=2 x-3 x^2 y^3
\end{aligned}
$
$
\begin{aligned}
& \therefore y\left(2+3 x^3 y\right) \frac{d y}{d x}=x\left(2-3 x y^3\right) \\
& \therefore \frac{d y}{d x}=\frac{x\left(2-3 x y^3\right)}{y\left(2+3 x^3 y\right)} .
\end{aligned}
$
View full question & answer→Question 112 Marks
If $x^3+y^2+x y=7$, find $\frac{d y}{d x}$
Answer$
x^3+y^2+x y=7
$
Differentiating both sides w.r.t. $x$, we get
$
\begin{aligned}
& 3 x^2+2 y \cdot \frac{d y}{d x}+x \frac{d y}{d x}+y \cdot \frac{d}{d x}(x)=0 \\
& \therefore 3 x^2+2 y \frac{d y}{d x}+x \frac{d y}{d x}+y \times 1=0
\end{aligned}
$
$
\begin{aligned}
& \therefore(2 y+x) \frac{d y}{d x}=-3 x^2-y \\
& \therefore \frac{d y}{d x}=\frac{-\left(y+3 x^2\right)}{2 y+x} .
\end{aligned}
$
View full question & answer→Question 122 Marks
Solve the following : If $x =a\left(1-\frac{1}{t}\right), y =a\left(1+\frac{1}{t}\right)$, then show that $\frac{d y}{d x}=-1$
Answer$
x=a\left(1-\frac{1}{t}\right), y=a\left(1+\frac{1}{t}\right)
$
Differentiating $x$ and $y$ w.r.t. $t$, we get
$
\begin{aligned}
& \frac{d x}{d t}=a \frac{d}{d t}\left(1-\frac{1}{t}\right)=a\left[0-(-1) t^{-2}\right]=\frac{a}{t^2} \\
& \text { and } \frac{d y}{d t}=a \frac{d}{d t}\left(1+\frac{1}{t}\right)=a\left[0+(-1) t^{-2}\right]=-\frac{a}{t^2} \\
& \therefore \frac{d y}{d x}=\frac{(d y / d t)}{(d x / d t)}=\frac{\left(-\frac{a}{t^2}\right)}{\left(\frac{a}{t^2}\right)}=-1 .
\end{aligned}
$
View full question & answer→Question 132 Marks
Find $\frac{d y}{d x}$ if, :
$
x=e^{3 t}, y=e^{(4 t+5)}
$
Answer$
x=e^{3 t}, y=e^{(4 t+5)}
$
Differentiating $x$ and $y$ w.r.t. $t$, we get
$
\begin{aligned}
\frac{d x}{d t} & =\frac{d}{d t}\left(e^{3 t}\right)=e^{3 t} \cdot \frac{d}{d t}(3 t) \\
& =e^{3 t} \times 3 \times 1=3 e^{3 t}
\end{aligned}
$
$
\text { and } \begin{aligned}
\frac{d y}{d t} & =\frac{d}{d t}\left[e^{(4 t+5)}\right]=e^{(4 t+5)} \cdot \frac{d}{d t}(4 t+5) \\
& =e^{(4 t+5)} \times(4 \times 1+0)=4 e ^{(4 t+5)}
\end{aligned}
$
$
\begin{aligned}
\therefore \frac{d y}{d x} & =\frac{(d y / d t)}{(d x / d t)}=\frac{4 e^{(4 t+5)}}{3 e^{3 t}} \\
& =\frac{4}{3} e^{4 t+5-3 t}=\frac{4}{3} e^{t+5}
\end{aligned}
$
View full question & answer→Question 142 Marks
Find $\frac{d y}{d x}$ if, :
$
x=2 a t^2, y=a t^4
$
Answer$
x=2 a t^2
$
Differentiating both sides w.r.t. $t$, we get
$
\begin{aligned}
& \frac{d x}{d t}=4 a t \\
& y=a t^4
\end{aligned}
$
Differentiating both sides w.r.t. t, we get
$
\begin{aligned}
& \frac{d y}{d t}=4 a t^3 \\
& \therefore \frac{d y}{d x}=\frac{\left(\frac{d y}{d t}\right)}{\left(\frac{d y}{d t}\right)}=\frac{4 a t^3}{4 a t}=t^2
\end{aligned}
$
View full question & answer→Question 152 Marks
Find $\frac{d y}{d x}$ if, :
$
x = a t ^2, y =2 at
$
Answer$
x = at { }^2, y =2 at
$
Differentiating $x$ and $y$ w.r.t. $t$, we get
$
\frac{d x}{d t}=a \frac{d}{d t}\left(t^2\right)=a \times 2 t=2 a t
$
$
\begin{aligned}
& \text { and } \frac{d y}{d t}=2 a \frac{d}{d t}(t)=2 a \times 1=2 a \\
& \therefore \frac{d y}{d x}=\frac{(d y / d t)}{(d x / d t)}=\frac{2 a}{2 a t}=\frac{1}{t} .
\end{aligned}
$
View full question & answer→Question 162 Marks
Find $\frac{d y}{d x}$ if, :
$
y \cdot e^x+x \cdot e^y=1
$
Answer$
y \cdot e^x+x \cdot e^y=1
$
Differentiating both sides w.r.t. $x_r$ we get
$
\begin{aligned}
& \frac{d}{d x}\left(y e^x\right)+\frac{d}{d x}\left(x e^y\right)=0 \\
& \therefore y \cdot \frac{d}{d x}\left(e^x\right)+e^x \cdot \frac{d y}{d x}+x \cdot \frac{d}{d x}\left(e^y\right)+e^y \cdot \frac{d}{d x}(x)=0 \\
& \therefore y \cdot e^x+e^x \cdot \frac{d y}{d x}+x \cdot e^y \cdot \frac{d y}{d x}+e^y \times 1=0 \\
& \therefore\left(e^x+x e^y\right) \frac{d y}{d x}=-e^y-y e^x \\
& \therefore \frac{d y}{d x}=-\left(\frac{e^y+y e^x}{e^x+x e^y}\right) .
\end{aligned}
$
View full question & answer→Question 172 Marks
Find $\frac{d y}{d x}$ if, :
$
x^3+x^2 y+x y^2+y^3=81
$
Answer$
x^3+x^2 y+x y^2+y^3=81
$
Differentiating both sides w.r.t. $x_1$ we get
$
\begin{array}{r}
3 x^2+\left[x^2 \frac{d y}{d x}+y \cdot \frac{d}{d y}\left(x^2\right)\right]+\left[x \cdot \frac{d}{d x}\left(y^2\right)+y^2 \cdot \frac{d}{d x}(x)\right]+ \\
3 y^2 \frac{d y}{d x}=0
\end{array}
$
$
\therefore 3 x^2+x^2 \frac{d y}{d x}+y \times 2 x+x \times 2 y \cdot \frac{d y}{d x}+y^2 \times 1+
$
$
3 y^2 \frac{d y}{d x}=0
$
$
\begin{aligned}
& \therefore x^2 \frac{d y}{d x}+2 x y \frac{d y}{d x}+3 y^2 \frac{d y}{d x}=-3 x^2-2 x y-y^2 \\
& \therefore\left(x^2+2 x y+3 y^2\right) \frac{d y}{d x}=-\left(3 x^2+2 x y+y^2\right) \\
& \therefore \frac{d y}{d x}=-\left(\frac{3 x^2+2 x y+y^2}{x^2+2 x y+3 y^2}\right) .
\end{aligned}
$
View full question & answer→Question 182 Marks
Find $\frac{d y}{d x}$ if, :
$
x^3+y^3+4 x^3 y=0
$
Answer$
x^3+y^3+4 x^3 y=0
$
Differentiating both sides w.r.t. $x_r$ we get
$
\begin{aligned}
& 3 x^2+3 y^2 \frac{d y}{d x}+4\left[x^3 \frac{d y}{d x}+y \frac{d}{d x}\left(x^3\right)\right]=0 \\
& \therefore 3 x^2+3 y^2 \frac{d y}{d x}+4 x^3 \frac{d y}{d x}+4 y \times 3 x^2=0 \\
& \therefore 3 y^2 \frac{d y}{d x}+4 x^3 \frac{d y}{d x}=-3 x^2-12 x^2 y \\
& \therefore\left(3 y^2+4 x^3\right) \frac{d y}{d x}=-3 x^2(1+4 y) \\
& \therefore \frac{d y}{d x}=-\frac{3 x^2(1+4 y)}{3 y^2+4 x^3} .
\end{aligned}
$
View full question & answer→Question 192 Marks
Find $\frac{d y}{d x}$ if, :
$
y=x^{e^x}
$
Answer$
\begin{aligned}
& y=x^{e^x} \\
& \therefore \log y=\log x^{e^x}=e^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(e^x \cdot \log x\right) \\
& =e^x \cdot \frac{d}{d x}(\log x)+(\log x) \cdot \frac{d}{d x}\left(e^x\right) \\
& =e^x \cdot \frac{1}{x}+(\log x)\left(e^x\right) \\
\therefore \frac{d y}{d x} & =y\left[\frac{e^x}{x}+e^x \cdot \log x\right] \\
& =x^{e^x} \cdot e^x\left[\frac{1}{x}+\log x\right]
\end{aligned}
$
View full question & answer→Question 202 Marks
Find the rate of change of demand (x) of a commodity with respect to price (y) if :
$
y=12+10 x+25 x^2
$
AnswerGiven $y=12+10 x+25 x^2$
$
\begin{aligned}
\therefore \frac{d y}{d x} & =\frac{d}{d x}\left(12+10 x+25 x^2\right) \\
& =\frac{d}{d x}(12)+10 \frac{d}{d x}(x)+25 \frac{d}{d x}\left(x^2\right) \\
& =0+10 \times 1+25 \times 2 x=10+50 x
\end{aligned}
$
By derivative of inverse function,
$
\frac{d x}{d y}=\frac{1}{\left(\frac{d y}{d x}\right)}=\frac{1}{10+50 x}
$
Hence, the rate of change of demand $( x )$ with respect to price $( y )$ $=\frac{d x}{d y}=\frac{1}{10+50 x}$
View full question & answer→Question 212 Marks
Find $\frac{d y}{d x}$ if, :
$
y=\log (\log x)
$
AnswerGiven $y=\log (\log x)$
Let $u =\log x$
Then $y =\log u$
$
\therefore \frac{d y}{d u}=\frac{d}{d u}(\log u)
$
$
\begin{aligned}
& =\frac{1}{u}=\frac{1}{\log x} \\
& \text { and } \frac{d u}{d x}=\frac{d}{d x}(\log x)=\frac{1}{x} \\
& \therefore \frac{d y}{d x}=\frac{d y}{d u} \cdot \frac{d u}{d x}=\frac{1}{\log x} \times \frac{1}{x} \\
& =\frac{1}{x \log x} \text {. } \\
\end{aligned}
$
View full question & answer→Question 222 Marks
Find $\frac{d y}{d x}$ if, :
$
y=\left(5 x^3-4 x^2-8 x\right)^9
$
AnswerGiven : $y=\left(5 x^3-4 x^2-8 x\right)^9$
Let $u=5 x^3-4 x^2-8 x$
Then $y=u^9$
$
\begin{aligned}
\therefore \frac{d y}{d u} & =\frac{d}{d u}\left(u^9\right)=9 u^8 \\
& =9\left(5 x^3-4 x^2-8 x\right)^8
\end{aligned}
$
and $\frac{d u}{d x}=\frac{d}{d x}\left(5 x^3-4 x^2-8 x\right)$
$
\begin{aligned}
& =5 \frac{d}{d x}\left(x^3\right)-4 \frac{d}{d x}\left(x^2\right)-8 \frac{d}{d x}(x) \\
& =5 \times 3 x^2-4 \times 2 x-8 \times 1 \\
& =15 x^2-8 x-8
\end{aligned}
$
$
\begin{aligned}
\therefore \frac{d y}{d x} & =\frac{d y}{d u} \cdot \frac{d u}{d x}\\
& =9\left(5 x^3-4 x^2-8 x\right)^8\left(15 x^2-8 x-8\right) .
\end{aligned}
$
View full question & answer→Question 232 Marks
Find $\frac{d y}{d x}$ if, :
$
y=\sqrt[3]{a^2+x^2}
$
AnswerGiven : $y=\sqrt[3]{a^2+x^2}$
Let $u=a^2+x^2$
Then $y=\sqrt[3]{u}$
$
\begin{aligned}
\therefore \frac{d y}{d u}=\frac{d}{d u}\left(u^{\frac{1}{3}}\right) & =\frac{1}{3} u^{-\frac{2}{3}} \\
& =\frac{1}{3}\left(a^2+x^2\right)^{-\frac{2}{3}}
\end{aligned}
$
$
\text { and } \begin{aligned}
\frac{d u}{d x} & =\frac{d}{d x}\left(a^2+x^2\right) \\
& =0+2 x=2 x
\end{aligned}
$
$
\begin{aligned}
\therefore \frac{d y}{d x} & =\frac{d y}{d u} \cdot \frac{d u}{d x} \\
& =\frac{1}{3}\left(a^2+x^2\right)^{-\frac{2}{3}} \cdot 2 x \\
& =\frac{2 x}{3}\left(a^2+x^2\right)^{-\frac{2}{3}}
\end{aligned}
$
View full question & answer→Question 242 Marks
Find $\frac{d y}{d x}$ if, :
$
y=\sqrt{x+\frac{1}{x}}
$
AnswerGiven : $y=\sqrt{x+\frac{1}{x}}$
Let $u=x+\frac{1}{x}$
Then $y=\sqrt{u}$
$
\therefore \frac{d y}{d u}=\frac{d}{d u}\left(u^{\frac{1}{2}}\right)=\frac{1}{2} u^{-\frac{1}{2}}
$
$
=\frac{1}{2 \sqrt{u}}=\frac{1}{2 \sqrt{x+\frac{1}{x}}}
$
$
\text { and } \begin{aligned}
\frac{d u}{d x} & =\frac{d}{d x}\left(x+\frac{1}{x}\right) \\
& =\frac{d}{d x}(x)+\frac{d}{d x}\left(x^{-1}\right) \\
& =1+(-1) x^{-2}=1-\frac{1}{x^2}
\end{aligned}
$
$
\begin{aligned}
\therefore \frac{d y}{d x} & =\frac{d y}{d u} \cdot \frac{d u}{d x}=\frac{1}{2 \sqrt{x+\frac{1}{x}}} \cdot\left(1-\frac{1}{x^2}\right) \\
& =\frac{1}{2}\left(x+\frac{1}{x}\right)^{-\frac{1}{2}}\left(1-\frac{1}{x^2}\right) .
\end{aligned}
$
View full question & answer→