Solve the Following Question.(4 Marks)

Question 14 Marks

Find $\frac{d y}{d x}$ if $y =2^{x^x}$.

Answer

Given : $y=2^{x^x}$
Let $u=x^x$
Then $y=2^u$
$
\begin{aligned}
\therefore \frac{d y}{d u} & =\frac{d}{d u}\left(2^u\right)=2^u \cdot \log 2 \\
& =2^{x^x} \cdot \log 2
\end{aligned}
$
Now, $u=x^x$
$
\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) \\
\therefore \frac{d y}{d x} & =\frac{d y}{d u} \cdot \frac{d u}{d x} \\
& =2^{x^x} \cdot \log 2 \cdot x^x(1+\log x) \\
& =2^{x^x} \cdot x^x(\log 2)(1+\log x) .
\end{aligned}
$
... [By (1) and (2)]

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Question 24 Marks

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

Answer

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

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Question 34 Marks

Solve the following : If $x = t \cdot \log t , y = t ^{ t }$, then show that $\frac{d y}{d x}- y =0$.

Answer

$
x=t \log t
$
Differentiating w.r.t. $t$, we get
$
\begin{aligned}
\frac{d x}{d t} & =\frac{d}{d t}(t \cdot \log t) \\
& =t \frac{d}{d t}(\log t)+(\log t) \cdot \frac{d}{d t}(t) \\
& =t \times \frac{1}{t}+(\log t) \times 1=1+\log t
\end{aligned}
$
Also, $y=t^t$
$
\therefore \log y=\log t^t=t \log t \text {. }
$
Differentiating both sides w.r.t. $t$, we get
$
\begin{aligned}
& \frac{1}{y} \cdot \frac{d y}{d t}=\frac{d}{d t}(t \log t) \\
& =t \cdot \frac{d}{d t}(\log t)+(\log t) \cdot \frac{d}{d t}(t) \\
& =t \times \frac{1}{t}+(\log t) \times 1 \\
& \therefore \frac{d y}{d t}=y(1+\log t) \\
& \therefore \frac{d y}{d x}=\frac{(d y / d t)}{(d x / d t)}=\frac{y(1+\log t)}{1+\log t}=y \\
& \therefore \frac{d y}{d x}-y=0 .
\end{aligned}
$

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Question 44 Marks

Solve the following : If $x =\frac{4 t}{1+t^2}, y =3\left(\frac{1-t^2}{1+t^2}\right)$, then show that $\frac{d y}{d x}=-\frac{9 x}{4 y}$

Answer

$
x=\frac{4 t}{1+t^2}, \quad y=3\left(\frac{1-t^2}{1+t^2}\right)
$
Differentiating $x$ and $y$ w.r.t. $t$, we get
$
\begin{aligned}
\frac{d x}{d t}=\frac{d}{d t}\left(\frac{4 t}{1+t^2}\right) & =\frac{\left(1+t^2\right) \cdot \frac{d}{d t}(4 t)-4 t \cdot \frac{d}{d t}\left(1+t^2\right)}{\left(1+t^2\right)^2} \\
& =\frac{\left(1+t^2\right)(4)-4 t(0+2 t)}{\left(1+t^2\right)^2} \\
& =\frac{4+4 t^2-8 t^2}{\left(1+t^2\right)^2}=\frac{4-4 t^2}{\left(1+t^2\right)^2} \\
& =\frac{4\left(1-t^2\right)}{\left(1+t^2\right)^2}
\end{aligned}
$
$
\text { and } \begin{aligned}
\frac{d y}{d t} & =3 \frac{d}{d t}\left(\frac{1-t^2}{1+t^2}\right) \\
& =3\left[\frac{\left(1+t^2\right) \cdot \frac{d}{d t}\left(1-t^2\right)-\left(1-t^2\right) \cdot \frac{d}{d t}\left(1+t^2\right)}{\left(1+t^2\right)^2}\right] \\
& =3\left[\frac{\left(1+t^2\right)(0-2 t)-\left(1-t^2\right)(0+2 t)}{\left(1+t^2\right)^2}\right] \\
& =3\left[\frac{-2 t-2 t^3-2 t+2 t^3}{\left(1+t^2\right)^2}\right] \\
& =\frac{-12 t}{\left(1+t^2\right)^2}
\end{aligned}
$
$
\therefore \frac{d y}{d x}=\frac{(d y / d t)}{(d x / d t)}=\frac{\left[\frac{-12 t}{\left(1+t^2\right)^2}\right]}{\left[\frac{4\left(1-t^2\right)}{\left(1+t^2\right)^2}\right]}
$
$\therefore \frac{d y}{d x}=\frac{-3 t}{1-t^2}$
$
\frac{-9 x}{4 y}=\frac{-9}{4} \cdot \frac{\left(\frac{4 t}{1+t^2}\right)}{3\left(\frac{1-t^2}{1+t^2}\right)}=\frac{-3 t}{1-t^2}
$
From (1) and (2)
$
\frac{d y}{d x}=-\frac{9 x}{4 y}
$

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Question 54 Marks

Find $\frac{d y}{d x}$ if, :
$
x =\left(u+\frac{1}{u}\right)^2, y =(2)^{\left(u+\frac{1}{u}\right)}
$

Answer

$
x =\left(u+\frac{1}{u}\right)^2, y =(2)^{\left(u+\frac{1}{u}\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(u+\frac{1}{u}\right)^2=2\left(u+\frac{1}{u}\right) \cdot \frac{d}{d u}\left(u+\frac{1}{u}\right) \\
& =2\left(u+\frac{1}{u}\right)\left(1-\frac{1}{u^2}\right)
\end{aligned} \\
& \text { and } \frac{d y}{d u}=\frac{d}{d u}\left[2^{\left(u+\frac{1}{u}\right)}\right]
\end{aligned}
$
$
\begin{aligned}
& =2^{\left(u+\frac{1}{u}\right)} \cdot \log 2 \cdot \frac{d}{d u}\left(u+\frac{1}{u}\right) \\
& =2^{\left(u+\frac{1}{u}\right)} \cdot \log 2 \cdot\left(1-\frac{1}{u^2}\right)
\end{aligned}
$
$
\begin{aligned}
\therefore \frac{d y}{d x}=\frac{(d y / d u)}{(d x / d u)} & =\frac{2^{\left(u+\frac{1}{u}\right)} \cdot \log 2 \cdot\left(1-\frac{1}{u^2}\right)}{2\left(u+\frac{1}{u}\right)\left(1-\frac{1}{u^2}\right)} \\
& =\frac{2^{\left(u+\frac{1}{u}\right)} \cdot \log 2}{2\left(u+\frac{1}{u}\right)} \\
& =\frac{y \log 2}{2 \sqrt{x}}
\end{aligned}
$

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Question 64 Marks

Solve the following : If $x ^5 \cdot y ^7=( x + y )^{12}$, then show that $\frac{d y}{d x}=\frac{y}{x}$

Answer

$
\begin{aligned}
& x^5 \cdot y^7=(x+y)^{12} \\
& \therefore \log \left(x^5 \cdot y^7\right)=\log (x+y)^{12} \\
& \therefore \log x^5+\log y^7=\log (x+y)^{12} \\
& \therefore 5 \log x+7 \log y=12 \log (x+y)
\end{aligned}
$
Differentiating both sides w.r.t. $x$, we get
$
\begin{aligned}
& 5 \times \frac{1}{x}+7 \times \frac{1}{y} \cdot \frac{d y}{d x}=12 \times \frac{1}{x+y} \cdot \frac{d}{d x}(x+y) \\
& \therefore \frac{5}{x}+\frac{7}{y} \cdot \frac{d y}{d x}=\frac{12}{x+y}\left(1+\frac{d y}{d x}\right) \\
& \therefore \frac{5}{x}+\frac{7}{y} \frac{d y}{d x}=\frac{12}{x+y}+\frac{12}{x+y} \cdot \frac{d y}{d x}
\end{aligned}
$
$
\begin{aligned}
& \therefore\left(\frac{7}{y}-\frac{12}{x+y}\right) \frac{d y}{d x}=\frac{12}{x+y}-\frac{5}{x} \\
& \therefore\left[\frac{7 x+7 y-12 y}{y(x+y)}\right] \frac{d y}{d x}=\frac{12 x-5 x-5 y}{x(x+y)} \\
& \therefore\left[\frac{7 x-5 y}{y(x+y)}\right] \frac{d y}{d x}=\frac{7 x-5 y}{x(x+y)} \\
& \therefore \frac{1}{y} \cdot \frac{d y}{d x}=\frac{1}{x}
\end{aligned}
$
$
\therefore \frac{d y}{d x}=\frac{y}{x}
$

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Question 74 Marks

Find $\frac{d y}{d x}$ if, :
$
y=10^{x^x}+10^{x^{10}}+10^{10^x}
$

Answer

$
y=10^{x^x}+10^{x^{10}}+10^{10^x}
$
Let $u=x^x$
Then $\log u=\log x^x=x \log x$ 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}
$
Now, $y=10^u+10^{x^{10}}+10^{10^x}$
$
\begin{aligned}
\therefore \frac{d y}{d x} & =\frac{d}{d x}\left(10^u\right)+\frac{d}{d x}\left(10^{x^{10}}\right)+\frac{d}{d x}\left(10^{10^x}\right) \\
& =10^u \cdot \log 10 \cdot \frac{d u}{d x}+10^{x^{10}} \cdot \log 10 \cdot \frac{d}{d x}\left(x^{10}\right)+ \\
\end{aligned}
$
$
\begin{array}{r}
=10^{x^x} \cdot \log 10 \cdot x^x(1+\log x)+10^{x^{10}} \cdot \log 10 \times 10^{x^9}+ \\
10^{10^x} \cdot \log 10 \times 10^x \cdot \log 10 \ldots[\text { By (1)] }
\end{array}
$
$
\begin{aligned}
& \therefore \frac{d y}{d x}=10^{x^x} \cdot x^x \cdot(\log 10)(1+\log x)+ \\
& \quad 10^{x^{10}}\left(10^{x^9}\right) \log 10+10^{10^x} \cdot 10^x \cdot(\log 10)^2 .
\end{aligned}
$

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Question 84 Marks

Find $\frac{d y}{d x}$ if, :
$
y=\sqrt[3]{\frac{(3 x-1)}{(2 x+3)(5-x)^2}}
$

Answer

$
\begin{aligned}
& y=\sqrt[3]{\frac{3 x-1}{(2 x+3)(5-x)^2}} \\
& \begin{aligned}
\therefore \log y & =\log \left[\frac{3 x-1}{(2 x+3)(5-x)^2}\right]^{\frac{1}{3}} \\
& =\frac{1}{3} \log \left[\frac{3 x-1}{(2 x+3)(5-x)^2}\right]
\end{aligned}
\end{aligned}
$
$
\begin{aligned}
& =\frac{1}{3}\left[\log (3 x-1)-\log (2 x+3)-\log (5-x)^2\right] \\
& =\frac{1}{3} \log (3 x-1)-\frac{1}{3} \log (2 x+3)-\frac{2}{3} \log (5-x)
\end{aligned}
$
Differentiating both sides w.r.t. $x$, we get
$
\begin{aligned}
\frac{1}{y} \cdot \frac{d y}{d x} & =\frac{1}{3} \frac{d}{d x}[\log (3 x-1)]-\frac{1}{3} \frac{d}{d x}[\log (2 x+3)]- \\
& \frac{2}{3} \frac{d}{d x}[\log (5-x)] \\
& =\frac{1}{3} \times \frac{1}{3 x-1} \cdot \frac{d}{d x}(3 x-1)-\frac{1}{3} \times \frac{1}{2 x+3} \cdot \frac{d}{d x}(2 x+3)- \\
& =\frac{1}{3(3 x-1)} \times(3 \times 1-0)-\frac{1}{3(2 x+3)} \times(2 \times 1+0)- \\
\therefore \frac{d y}{d x} & =y\left[\frac{3}{3(3 x-1)}-\frac{1}{3(2 x+3)}+\frac{2}{3(5-x)}\right] \\
= & \frac{1}{3} \cdot \sqrt[3]{\frac{3 x-1}{(2 x+3)(5-x)^2}}\left[\frac{3}{3 x-1}-\frac{2}{2 x+3}+\frac{2}{5-x}\right] .
\end{aligned}
$

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Maths (commerce) Solve the Following Question.(4 Marks)

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