Solve the Following Question.(5 Marks)

Question 15 Marks

Find $\frac{d y}{d x}$ if $y = x ^{ x }+(7 x -1)^{ x }$

Answer

$
y=x^x+(7 x-1)^x
$
Let $u=x^x$ and $v=(7 x-1)^x$
Then $y=u+v$
$
\therefore \frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}
$
Take $u=x^x$
$
\therefore \log u=\log x^x=x \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 \log x) \\
& =x \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}
$
Also, $v=(7 x-1)^x$
$
\begin{aligned}
& \therefore \log v=\log (7 x-1)^x=x \log (7 x-1) \\
& \begin{aligned}
\frac{1}{v} \cdot \frac{d v}{d x} & =\frac{d}{d x}[x \log (7 x-1)] \\
& =x \frac{d}{d x}[\log (7 x-1)]+[\log (7 x-1)] \cdot \frac{d}{d x}(x) \\
& =x \times \frac{1}{7 x-1} \cdot \frac{d}{d x}(7 x-1)+[\log (7 x-1)] \times 1
\end{aligned}
\end{aligned}
$
$
=\frac{x}{7 x-1} \times(7 \times 1-0)+\log (7 x-1)
$
$
\begin{aligned}
\therefore \frac{d v}{d x} & =v\left[\frac{7 x}{7 x-1}+\log (7 x-1)\right] \\
& =(7 x-1)^x\left[\frac{7 x}{7 x-1}+\log (7 x-1)\right]
\end{aligned}
$
From (1), (2) and (3), we get
$
\frac{d y}{d x}=x^x(1+\log x)+(7 x-1)^x\left[\frac{7 x}{7 x-1}+\log (7 x-1)\right] .
$

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Question 25 Marks

Find $\frac{d y}{d x}$, if $y =\sqrt{\frac{(3 x-4)^3}{(x+1)^4(x+2)}}$

Answer

$
\begin{aligned}
& y=\sqrt{\frac{(3 x-4)^3}{(x+1)^4(x+2)}} \\
& =\frac{(3 x-4)^{\frac{3}{2}}}{(x+1)^{\frac{4}{2}} \cdot(x+2)^{\frac{1}{2}}}
\end{aligned}
$
Taking logarithm of both sides, we get
$
\begin{aligned}
& \log y=\log \left[\frac{(3 x-4)^{\frac{3}{2}}}{(x+1)^{\frac{4}{2}} \cdot(x+2)^{\frac{1}{2}}}\right] \\
& =\log (3 x-4)^{\frac{3}{2}}-\left[\log (x+1)^2+\log (x+2)^{\frac{1}{2}}\right] \\
& =\frac{3}{2} \log (3 x-4)-2 \log (x+1)-\frac{1}{2} \log (x+2)
\end{aligned}
$
Differentiating both sides w.r.t. $x$, we get
$
\begin{aligned}
& \frac{1}{y} \cdot \frac{d y}{d x}=\frac{3}{2} \cdot \frac{d}{d x}[\log (3 x-4)]-2 \frac{d}{d x}[\log (x+1)]-\frac{1}{2} \cdot \frac{d}{d x}[\log (x+2)] \\
& =\frac{3}{2} \cdot \frac{1}{3 x-4} \cdot \frac{d}{d x}(3 x-4)-2 \cdot \frac{1}{x+1} \cdot \frac{d}{d x}(x+1)-\frac{1}{2} \cdot \frac{1}{x+2} \cdot \frac{d}{d x}(x+2) \\
& \therefore \frac{1}{y} \cdot \frac{d y}{d x}=\frac{3}{2(3 x-4)} \times 3-\frac{2}{x+1} \times 1-\frac{1}{2(x+2)} \times 1 \\
& \therefore \frac{1}{y} \cdot \frac{d y}{d x}=\frac{9}{2(3 x-4)}-\frac{2}{x+1}-\frac{1}{2(x+2)} \\
& \therefore \frac{d y}{d x}=\frac{y}{2}\left[\frac{9}{3 x-4}-\frac{4}{x+1}-\frac{1}{x+2}\right] \\
& \therefore \frac{d y}{d x}=\frac{1}{2} \sqrt{\frac{(3 x-4)^3}{(x+1)^4(x+2)}}\left[\frac{9}{3 x-4}-\frac{4}{x+1}-\frac{1}{x+2}\right]
\end{aligned}
$

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Question 35 Marks

If $a x^2+2 h x y+b y^2=0$, then show that $\frac{d^2 y}{d x^2}=0$.

Answer

$
\begin{aligned}
& a x^2+2 h x y+b y^2=0 \ldots \ldots . .(1) \\
& \therefore a x^2+h x y+h x y+b y^2=0 \\
& \therefore x(a x+h y)+y(h x+b y)=0 \\
& \therefore x(a x+h y)=-y(h x+b y) \\
& \therefore \frac{a x+h y}{h x+b y}=-\frac{y}{x} \ldots \ldots . .(2)
\end{aligned}
$
Differentiating (1) w.r.t. $x$, we get
$
\begin{aligned}
& a \times 2 x+2 h\left[x \frac{d y}{d x}+y \cdot \frac{d}{d x}(x)\right]+b \times 2 y \frac{d y}{d x}=0 \\
& \therefore 2 a x+2 h x \frac{d y}{d x}+2 h y \times 1+2 b y \frac{d y}{d x}=0 \\
& \therefore(2 h x+2 b y) \frac{d y}{d x}=-2 a x-2 h y \\
& \therefore \frac{d y}{d x}=\frac{-2(a x+h y)}{2(h x+b y)}=-\left(\frac{a x+h y}{h x+b y}\right) \\
& \therefore \frac{d y}{d x}=\frac{y}{x} \\
& \therefore \frac{d^2 y}{d x^2}=\frac{d}{d x}\left(\frac{y}{x}\right) \\
& =\frac{x \frac{d y}{d x}-y \cdot \frac{d}{d x}(x)}{x^2} \\
& =\frac{x\left(\frac{y}{x}\right)-y \times 1}{x^2} \\
& \ldots\left[\because \frac{d y}{d x}=\frac{y}{x}\right] \\
& =\frac{y-y}{x^2}=\frac{0}{x^2} \\
\end{aligned}
$
$\frac{d^2 y}{d x^2}=0$.

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Question 45 Marks

If $x^2+6 x y+y^2=10$, then show that $\frac{d^2 y}{d x^2}=\frac{80}{(3 x+y)^3}$.

Answer

$
x^2+6 x y+y^2=10
$
Differentiating both sides w.r.t. a, we get
$
\begin{aligned}
& 2 x+6\left[x \frac{d y}{d x}+y \cdot \frac{d}{d x}(x)\right]+2 y \frac{d y}{d x}=0 \\
& \therefore 2 x+6 x \frac{d y}{d x}+6 y \times 1+2 y \frac{d y}{d x}=0 \\
& \therefore(6 x+2 y) \frac{d y}{d x}=-2 x-6 y \\
& \therefore \frac{d y}{d x}=\frac{-2(x+3 y)}{2(3 x+y)}=-\left(\frac{x+3 y}{3 x+y}\right) \\
& \therefore \frac{d^2 y}{d x^2}=-\frac{d}{d x}\left(\frac{x+3 y}{3 x+y}\right) \\
& =-\left[\frac{(3 x+y) \cdot \frac{d}{d x}(x+3 y)-(x+3 y) \cdot \frac{d}{d x}(3 x+y)}{(3 x+y)^2}\right] \\
& =-\left[\frac{(3 x+y)\left(1+3 \frac{d y}{d x}\right)-(x+3 y)\left(3+\frac{d y}{d x}\right)}{(3 x+y)^2}\right] \\
& =\frac{1}{(3 x+y)^2}\left[-(3 x+y)\left\{1-\frac{3(x+3 y)}{3 x+y}\right\}+\right. \\
& \left.(x+3 y)\left(3-\frac{x+3 y}{3 x+y}\right)\right] \\
& =\frac{1}{(3 x+y)^2}\left[-(3 x+y)\left(\frac{3 x+y-3 x-9 y)}{3 x+y}\right)+\right. \\
& \left.(x+3 y)\left(\frac{9 x+3 y-x-3 y}{3 x+y}\right)\right] \\
\end{aligned}
$
$
\begin{aligned}
& =\frac{1}{(3 x+y)^2}\left[8 y+\frac{(x+3 y)(8 x)}{3 x+y}\right] \\
& =\frac{1}{(3 x+y)^2}\left[\frac{8 y(3 x+y)+(x+3 y) 8 x}{(3 x+y)}\right] \\
& =\frac{24 x y+8 y^2+8 x^2+24 x y}{(3 x+y)^3} \\
& =\frac{8 x^2+48 x y+8 y^2}{(3 x+y)^3}=\frac{8\left(x^2+6 x y+y^2\right)}{(3 x+y)^3} \\
& =\frac{8(10)}{(3 x+y)^3}
\end{aligned}
$
$
\therefore \frac{d^2 y}{d x^2}=\frac{80}{(3 x+y)^3} \text {. }
$

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Question 55 Marks

If $x ^7 \cdot y ^9=( x + y )^{16}$, then show that $\frac{d y}{d x}=\frac{y}{x}$.

Answer

$
x^7 \cdot y^9=(x+y)^{16}
$
Taking logarithm of both sides, we get
$
\begin{aligned}
& \log x^7 \cdot y^9=\log (x+y)^{16} \\
& \therefore \log x^7+\log y^9=16 \log (x+y) \\
& \therefore 7 \log x+9 \log y=16 \log (x+y)
\end{aligned}
$
Differentiating both sides w.r.t. $x$, we get
$
\begin{aligned}
& 7\left(\frac{1}{x}\right)+9\left(\frac{1}{y}\right) \frac{d y}{d x}=16\left(\frac{1}{x+y}\right) \frac{d}{d x}(x+y) \\
& \therefore \frac{7}{x}+\frac{9}{y} \frac{d y}{d x}=\frac{16}{x+y}\left(1+\frac{d y}{d x}\right) \\
& \therefore \frac{7}{x}+\frac{9}{y} \frac{d y}{d x}=\frac{16}{x+y}+\frac{16}{x+y} \frac{d y}{d x} \\
& \therefore \frac{9}{y} \frac{d y}{d x}-\frac{16}{x+y} \frac{d y}{d x}=\frac{16}{x+y}-\frac{7}{x} \\
& \therefore\left(\frac{9}{y}-\frac{16}{x+y}\right) \frac{d y}{d x}=\frac{16}{x+y}-\frac{7}{x} \\
& \therefore\left[\frac{9 x+9 y-16 y}{y(x+y)}\right] \frac{d y}{d x}=\frac{16 x-7 x-7 y}{x(x+y)} \\
& \therefore\left[\frac{9 x-7 y}{y(x+y)}\right] \frac{d y}{d x}=\frac{9 x-7 y}{x(x+y)} \\
& \therefore \frac{d y}{d x}=\frac{9 x-7 y}{x(x+y)} \times \frac{y(x+y)}{9 x-7 y}
\end{aligned}
$
$\frac{dy}{dx}=$$\frac{x}{y}$

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Question 65 Marks

Find $\frac{d y}{d x}$ if, :
$y = (\log x)^x + x^{\log x}$

Answer

$y = (\log x)^x + x^{\log x}$
Let $u=(\log x)^x$ and $v=x^{\log x}$
Then $y=u+v$
$\therefore \frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}$
Take $u=(\log x)^x \quad$
$\therefore \log u=\log (\log x)^x=x \log (\log x)$
Differentiating both sides w.r.t. $x$, we get
$\frac{1}{u} \cdot \frac{d u}{d x} =\frac{d}{d x}[x \log (\log x)]$
$=x \cdot \frac{d}{d x}[\log (\log x)]+[\log (\log x)] \cdot \frac{d}{d x}(x)$
$=x \times \frac{1}{\log x} \cdot \frac{d}{d x}(\log x)+[\log (\log x)] \times 1$
$=x \times \frac{1}{\log x} \times \frac{1}{x}+\log (\log x)$
\therefore \frac{d u}{d x} $=u\left[\frac{1}{\log x}+\log (\log x)\right]$
$=(\log x)^x\left[\frac{1}{\log x}+\log (\log x)\right]$
Also, $v=x^{\log x}$
$\therefore \log v=\log x^{\log x}=\log x \cdot \log x=(\log x)^2$
Differentiating both sides w.r.t. $x$, we get
$\frac{1}{v} \cdot \frac{d v}{d x} =\frac{d}{d x}(\log x)^2$
$=2(\log x) \cdot \frac{d}{d x}(\log x)$
$=2 \log x \times \frac{1}{x}$
$\therefore \frac{d v}{d x} =v\left[\frac{2 \log x}{x}\right]$
$=x^{\log x\left[\frac{2 \log x}{x}\right]}$
From (1), (2) and (3), we get
$\frac{d y}{d x}=(\log x)^x\left[\frac{1}{\log x}+\log (\log x)\right]+x^{\log x}\left[\frac{2 \log x}{x}\right]$

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Maths (commerce) Solve the Following Question.(5 Marks)

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