Question 12 Marks
Solve the following: : The supply $S$ for a commodity at price $P$ is given by $S=P^2+9 P-2$. Find the marginal supply when the price is $7 /-$.
AnswerGiven, $S=P^2+9 P-2$
$
\begin{aligned}
\text { Marginal supply } & =\frac{d S}{d P} \\
& =\frac{d}{d P}\left(P^2+9 P-2\right) \\
& =\frac{d}{d P}\left(P^2\right)+9 \frac{d}{d P}(P)-\frac{d}{d P}(2) \\
& =2 P+9(1)-0 \\
& =2 P+9
\end{aligned}
$
When $\mathrm{P}=7$,
$
\begin{aligned}
\text { Marginal supply } & =\left(\frac{\mathrm{dS}}{\mathrm{dP}}\right)_{p-7}=2(7)+9 \\
& =14+9=23
\end{aligned}
$
$\therefore$ The marginal supply is 23 , at $\mathrm{P}=7$.
View full question & answer→Question 22 Marks
Solve the following: : If for a commodity; the price demand relation is given as $D=\left(\frac{P+5}{P-1}\right)$. Find the marginal demand when price is ₹ 2 /
View full question & answer→Question 32 Marks
Solve the following: : The demand $D$ for a price $P$ is given as $D=\frac{27}{P}$, find the rate of change of demand when the price is ₹ 3 /-.
View full question & answer→Question 42 Marks
Solve the following: : The total cost of producing $x$ items is given by $C=x^2+4 x+4$. Find the average cost and the marginal cost. What is the marginal cost when $x=7$ ?
View full question & answer→Question 52 Marks
Solve the following: : The supply $S$ of electric bulbs at price $P$ is given by $S=2 p^3+5$. Find the marginal supply when the price is ₹ $5 /-$. Interpret the result.
View full question & answer→Question 62 Marks
Find $\frac{d y}{d x}$ if : $y=x \log x\left(x^2+1\right)$
Answer$
y=x \log x\left(x^2+1\right)
$
Differentiating w.r.t. $x$, we get
$
\begin{aligned}
\frac{\mathrm{d} y}{\mathrm{~d} x}= & \frac{\mathrm{d}}{\mathrm{d} x}(x)(\log x)\left(x^2+1\right) \\
= & (x)(\log x) \frac{\mathrm{d}}{\mathrm{d} x}\left(x^2+1\right) \\
& \quad-\left(x^2+1\right) \frac{\mathrm{d}}{\mathrm{d} x}((x)(\log x)) \\
= & (x \log x)(2 x+0) \text { } \\
& \quad \quad+\left(x^2+1\right)\left[x \frac{\mathrm{d}}{\mathrm{d} x}(\log x)+(\log x) \frac{\mathrm{d}}{\mathrm{d} x}(x)\right] \\
= & 2 x^2 \log x+\left(x^2+1\right)\left[x \times \frac{1}{x}+(\log x)(1)\right] \\
= & 2 x^2 \log x+\left(x^2+1\right)(1+\log x) \\
= & 2 x^2 \log x+\left(x^2+1\right)+\left(x^2+1\right) \log x
\end{aligned}
$
View full question & answer→Question 72 Marks
Find $\frac{d y}{d x}$ if : $y=\frac{e^x}{\log x}$
Answer$
y=\frac{\mathrm{e}^x}{\log x}
$
Differentiating w.r.t. $x$, we get
$
\begin{aligned}
\frac{\mathrm{d} y}{\mathrm{~d} x} & =\frac{\mathrm{d}}{\mathrm{d} x}\left(\frac{\mathrm{e}^x}{\log x}\right)_{\text { }} \\
& =\frac{(\log x) \frac{\mathrm{d}}{\mathrm{d} x}\left(\mathrm{e}^x\right)-\left(\mathrm{e}^x\right) \frac{\mathrm{d}}{\mathrm{d} x}(\log x)}{(\log x)^2} \\
& =\frac{(\log x) \mathrm{e}^x-\mathrm{e}^x\left(\frac{1}{x}\right)}{(\log x)^2} \\
& =\frac{\mathrm{e}^x\left(\log x-\frac{1}{x}\right)}{(\log x)^2}
\end{aligned}
$
View full question & answer→Question 82 Marks
Find $\frac{d y}{d x}$ if : $\mathrm{y}=\frac{(\log x+1)}{x}$
Answer$
y=\frac{(\log x+1)}{x}
$
Differentiating w.r.t. $x$, we get
$
\begin{aligned}
\frac{\mathrm{d} y}{\mathrm{~d} x} & =\frac{\mathrm{d}}{\mathrm{d} x}\left[\frac{\log x+1}{x}\right] \\
& =\frac{x \frac{\mathrm{d}}{\mathrm{d} x}(\log x+1)-(\log x+1) \frac{\mathrm{d}}{\mathrm{d} x}(x)}{x^2} \\
& =\frac{x\left(\frac{1}{x}+0\right)-(\log x+1)(1)}{x^2} \\
& =\frac{1-\log x-1}{x^2} \text { } \\
& =\frac{-\log x}{x^2}
\end{aligned}
$
View full question & answer→Question 92 Marks
Find $\frac{d y}{d x}$ if : $\mathrm{y}=\frac{1+x}{2+x}$
Answer$
y=\frac{1+x}{2+x}
$
Differentiating w.r.t. $x$, we get
$
\begin{aligned}
\frac{\mathrm{d} y}{\mathrm{~d} x} & =\frac{\mathrm{d}}{\mathrm{d} x}\left(\frac{1+x}{2+x}\right) \text { } \\
& =\frac{(2+x) \frac{\mathrm{d}}{\mathrm{d} x}(1+x)-(1+x) \frac{\mathrm{d}}{\mathrm{d} x}(2+x)}{(2+x)^2} \\
& =\frac{(2+x)(0+1)-(1+x)(0+1)}{(2+x)^2 \text { }} \\
\frac{\mathrm{d} y}{\mathrm{~d} x} & =\frac{(2+x)-(1+x)}{(2+x)^2}=\frac{2+x-1-x}{(2+x)^2}=\frac{1}{(2+x)^2}
\end{aligned}
$
View full question & answer→Question 102 Marks
Find $\frac{d y}{d x}$ if : $y=x^3-2 x^2+\sqrt{x}+1$
Answer$
y=x^3-2 x^2+\sqrt{x}+1
$
Differentiating w.r.t. $x$, we get
$
\begin{aligned}
\frac{\mathrm{d} y}{\mathrm{~d} x} & =\frac{\mathrm{d}}{\mathrm{d} x}\left(x^3-2 x^2+\sqrt{x}+1\right) \\
& =\frac{\mathrm{d}}{\mathrm{d} x}\left(x^3\right)-2 \frac{\mathrm{d}}{\mathrm{d} x}\left(x^2\right)+\frac{\mathrm{d}}{\mathrm{d} x}(\sqrt{x})+\frac{\mathrm{d}}{\mathrm{d} x}(1) \\
& =3 x^2-2(2 x)+\frac{\mathrm{d}}{\mathrm{d} x}\left(x^{\frac{1}{2}}\right)+0 \\
& =3 x^2-4 x+\frac{1}{2} x^{\frac{1}{2}-1} \text { } \\
& =3 x^2-4 x+\frac{1}{2} x^{\frac{-1}{2}} \\
\frac{\mathrm{d} y}{\mathrm{~d} x} & =3 x^2-4 x+\frac{1}{2 \sqrt{x}}
\end{aligned}
$
View full question & answer→Question 112 Marks
Find $\frac{d y}{d x}$ if : $\mathrm{y}=\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)^2$
Answer$
\begin{aligned}
y & =\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)^2 \\
\therefore \quad y & =x+2+\frac{1}{x} \text { }
\end{aligned}
$
Differentiating w.r.t. $x$, we get
$
\begin{aligned}
& \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\mathrm{d}}{\mathrm{d} x}\left(x+2+\frac{1}{x}\right) \\
= & \frac{\mathrm{d}}{\mathrm{d} x}(x)+\frac{\mathrm{d}}{\mathrm{d} x}(2)+\frac{\mathrm{d}}{\mathrm{d} x}\left(\frac{1}{x}\right) \\
= & 1+0+\frac{\mathrm{d}}{\mathrm{d} x}\left(x^{-1}\right) \\
= & 1+(-1) x^{-2} \\
= & 1-\frac{1}{x^2} \text { }
\end{aligned}
$
View full question & answer→Question 122 Marks
Find $\frac{d y}{d x}$ if : $y=(\sqrt{x}+1)^2$
Answer$
\begin{aligned}
& y=(\sqrt{x}+1)^2 \\
\therefore \quad y & =x+2 \sqrt{x}+1
\end{aligned}
$
Differentiating w.r.t. $x$, we get
$
\begin{aligned}
& \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\mathrm{d}}{\mathrm{d} x}(x+2 \sqrt{x}+1) \\
& =\frac{\mathrm{d}}{\mathrm{d} x}(x)+2 \frac{\mathrm{d}}{\mathrm{d} x}(\sqrt{x})+\frac{\mathrm{d}}{\mathrm{d} x}(1) \\
& =1+2\left(\frac{1}{2 \sqrt{x}}\right)+0 \\
& \frac{\mathrm{d} y}{\mathrm{~d} x}=1+\frac{1 \text { }}{\sqrt{x}} \\
&
\end{aligned}
$
View full question & answer→Question 132 Marks
Find $\frac{d y}{d x}$ if : $y=x^2+\frac{1}{x^2}$
Answer$
\begin{aligned}
y & =x^2+\frac{1}{x^2} \\
\therefore \quad y & =x^2+x^{-2}
\end{aligned}
$
Differentiating w.r.t. $x$, we get
$
\begin{aligned}
\frac{\mathrm{d} y}{\mathrm{~d} x} & =\frac{\mathrm{d}}{\mathrm{d} x}\left(x^2+x^{-2}\right) \\
& =\frac{\mathrm{d}}{\mathrm{d} x}\left(x^2\right)+\frac{\mathrm{d}}{\mathrm{d} x}\left(x^{-2}\right) \\
& =2 x-2 x^{-3} \\
& =2 x-\frac{2}{x^3}
\end{aligned}
$
View full question & answer→Question 142 Marks
Solve the following examples: : The supply $S$ for a commodity at price $P$ is given by $S=P^2+9 P-2$. Find the marginal supply when the price is 7 .
View full question & answer→Question 152 Marks
Solve the following examples: : The total cost of producing $x$ units is given by $C=10 e^{2 x}$, find its marginal cost and average cost when $x=2$.
View full question & answer→Question 162 Marks
Solve the following examples: : The total cost of ' $t$ ' toy cars is given by $C=5(2 t)+17$. Find the marginal cost and average cost at $\mathrm{t}=3$.
View full question & answer→Question 172 Marks
Solve the following examples: : The demand $D$ for a price $P$ is given as $D=\frac{27}{P}$, find the rate of change of demand when the price is 3 .
View full question & answer→Question 182 Marks
Differentiate the following functions w.r.t. x. : $\frac{\left(2 e^x-1\right)}{\left(2 e^x+1\right)}$
AnswerLet $y=\frac{2 \mathrm{e}^x-1}{2 \mathrm{e}^x+1}$
Differentiating w.r.t. $x$, we get
$
\begin{aligned}
& \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\mathrm{d}}{\mathrm{d} x}\left(\frac{2 \mathrm{e}^x-1}{2 \mathrm{e}^x+1}\right) \text { } \\
& =\frac{\left(2 \mathrm{e}^x+1\right) \frac{\mathrm{d}}{\mathrm{d} x}\left(2 \mathrm{e}^x-1\right)-\left(2 \mathrm{e}^x-1\right) \frac{\mathrm{d}}{\mathrm{d} x}\left(2 \mathrm{e}^x+1\right)}{\left(2 \mathrm{e}^x+1\right)^2} \\
& =\frac{\left(2 \mathrm{e}^x+1\right)\left(2 \mathrm{e}^x-0\right)-\left(2 \mathrm{e}^x-1\right)\left(2 \mathrm{e}^x+0\right)}{\left(2 \mathrm{e}^x+1\right)^2} \\
& =\frac{\left(2 \mathrm{e}^x+1\right)\left(2 \mathrm{e}^x\right)-\left(2 \mathrm{e}^x-1\right)\left(2 \mathrm{e}^x\right)}{\left(2 \mathrm{e}^x+1\right)^2} \\
& =\frac{2 \mathrm{e}^x\left(2 \mathrm{e}^x+1-2 \mathrm{e}^x+1\right)}{\left(2 \mathrm{e}^x+1\right)^2} \\
& =\frac{2 \mathrm{e}^x(2)}{\left(2 \mathrm{e}^x+1\right)^2} \\
& =\frac{4 \mathrm{e}^x}{\left(2 \mathrm{e}^x+1\right)^2}
\end{aligned}
$
View full question & answer→Question 192 Marks
Differentiate the following functions w.r.t. x. : $\frac{e^x}{e^x+1}$
Answer$
y=\frac{\mathrm{e}^x}{\mathrm{e}^x+1}
$
Differentiating w.r.t. $x$, we get
$
\begin{aligned}
& \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\mathrm{d}}{\mathrm{d} x}\left(\frac{\mathrm{e}^x}{\mathrm{e}^x+1}\right) \text { } \\
&=\frac{\left(\mathrm{e}^x+1\right) \frac{\mathrm{d}}{\mathrm{d} x}\left(\mathrm{e}^x\right)-\mathrm{e}^x \frac{\mathrm{d}}{\mathrm{d} x}\left(\mathrm{e}^x+1\right)}{\left(\mathrm{e}^x+1\right)^2} \\
&=\frac{\left(\mathrm{e}^x+1\right) \mathrm{e}^x-\mathrm{e}^x\left(\mathrm{e}^x+0\right)}{\left(\mathrm{e}^x+1\right)^2} \\
&=\frac{\mathrm{e}^x\left(\mathrm{e}^x+1-\mathrm{e}^x\right)}{\left(\mathrm{e}^x+1\right)^2} \\
&= \frac{\mathrm{e}^x}{\left(\mathrm{e}^x+1\right)^2}
\end{aligned}
$
View full question & answer→Question 202 Marks
Differentiate the following functions w.r.t. x. : \begin{equation}
\frac{1}{e^x+1}
\end{equation}
AnswerLet $y=\frac{1}{\mathrm{e}^x+1}$
Differentiating w.r.t. $x$, we get
$
\begin{aligned}
\frac{\mathrm{d} y}{\mathrm{~d} x} & =\frac{\mathrm{d}}{\mathrm{d} x}\left(\frac{1}{\mathrm{e}^x+1}\right) \\
& =\frac{\left(\mathrm{e}^x+1\right) \frac{\mathrm{d}}{\mathrm{d} x}(1)-(1) \frac{\mathrm{d}}{\mathrm{d} x}\left(\mathrm{e}^x+1\right)}{\left(\mathrm{e}^x+1\right)^2} \\
& =\frac{\left(\mathrm{e}^x+1\right)(0)-(1)\left(\mathrm{e}^x+0\right)}{\left(\mathrm{e}^x+1\right)^2} \\
& =\frac{\mathrm{e}^x+1-\mathrm{e}^x}{\left(\mathrm{e}^x+1\right)^2} \\
& =\frac{1}{\left(\mathrm{e}^x+1\right)^2}
\end{aligned}
$
View full question & answer→Question 212 Marks
Differentiate the following functions w.r.t. x. : \begin{equation}
\frac{x^2+1}{x}
\end{equation}
AnswerLet $y=\frac{x^2+1}{x}$
Differentiating w.r.t. $x$, we get
$
\begin{aligned}
\frac{\mathrm{d} y}{\mathrm{~d} x} & =\frac{\mathrm{d}}{\mathrm{d} x}\left(\frac{x^2+1}{x}\right) \\
& =\frac{x \frac{\mathrm{d}}{\mathrm{d} x}\left(x^2+1\right)-\left(x^2+1\right) \frac{\mathrm{d}}{\mathrm{d} x}(x)}{x^2} \\
& =\frac{x(2 x+0)-\left(x^2+1\right)(1)}{x^2} \\
& =\frac{2 x^2-x^2-1}{x^2} \\
\frac{\mathrm{d} y}{\mathrm{~d} x} & =\frac{x^2-1}{x^2}
\end{aligned}
$
View full question & answer→Question 222 Marks
Differentiate the following functions w.r.t. x. : \begin{equation}
\frac{x}{x+1}
\end{equation}
AnswerLet $y=\frac{x}{x+1}$
Differentiating w.r.t. $x$, we get
$
\begin{aligned}
\frac{\mathrm{d} y}{\mathrm{~d} x} & =\frac{\mathrm{d}}{\mathrm{d} x}\left(\frac{x}{x+1}\right) \text { } \\
& =\frac{(x+1) \frac{\mathrm{d}}{\mathrm{d} x}(x)-x \frac{\mathrm{d}}{\mathrm{d} x}(x+1)}{(x+1)^2} \\
& =\frac{(x+1)(1)-x(1+0)}{(x+1)^2} \\
& =\frac{x+1-x}{(x+1)^2} \\
& =\frac{1}{(x+1)^2}
\end{aligned}
$
View full question & answer→Question 232 Marks
Differentiate the following w.r.t. x. : \begin{equation}
\mathrm{e}^{\mathrm{x}} \log \mathrm{x}
\end{equation}
AnswerLet $y=e^x \log x$
Differentiating w.r.t. $x$, we get
$
\begin{aligned}
\frac{\mathrm{d} y}{\mathrm{~d} x} & =\frac{\mathrm{d}}{\mathrm{d} x}\left(\mathrm{e}^x \log x\right) \\
& =\mathrm{e}^x \frac{\mathrm{d}}{\mathrm{d} x}(\log x)+(\log x) \frac{\mathrm{d}}{\mathrm{d} x}\left(\mathrm{e}^x\right) \\
& =\mathrm{e}^x\left(\frac{1}{x}\right)+(\log x)\left(\mathrm{e}^x\right) \\
& =\mathrm{e}^x\left(\frac{1}{x}+\log x\right)
\end{aligned}
$
View full question & answer→Question 242 Marks
Differentiate the following w.r.t. x. : \begin{equation}
x^{\frac{5}{2}} e^x
\end{equation}
AnswerLet $\mathrm{y}=x^{\frac{5}{2}} e^x$
Differentiating w.r.t. $x$, we get
$
\begin{aligned}
& \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\mathrm{d}}{\mathrm{d} x}\left(\begin{array}{c}
x^{\frac{5}{2}} \mathrm{e}^x \\
\text { MaharashtraBoardSolutions.in }
\end{array}\right. \\
& =x^{\frac{5}{2}} \frac{\mathrm{d}}{\mathrm{d} x}\left(\mathrm{e}^x\right)+\mathrm{e}^x \frac{\mathrm{d}}{\mathrm{d} x}\left(x^{\frac{5}{2}}\right) \\
& =x^{\frac{5}{2}}\left(\mathrm{e}^x\right)+\mathrm{e}^x\left(\frac{5}{2} x^{\frac{3}{2}}\right) \\
& =\mathrm{e}^x\left(x^{\frac{5}{2}}+\frac{5}{2} x^{\frac{3}{2}}\right) \\
&
\end{aligned}
$
View full question & answer→Question 252 Marks
Differentiate the following w.r.t. x. : $\sqrt{x}\left(x^2+1\right)^2$
Answer$
\text { Let } \begin{aligned}
\mathrm{y} & =\sqrt{x}\left(x^2+1\right)^2 \\
y & =x^{\frac{1}{2}}\left(x^4+2 x^2+1\right) \\
y & =x^{\frac{9}{2}}+2 x^{\frac{5}{2}}+x^{\frac{1}{2}}
\end{aligned}
$
Differentiating w.r.t. $x$, we get
$
\begin{aligned}
\frac{\mathrm{d} y}{\mathrm{~d} x} & =\frac{\mathrm{d}}{\mathrm{d} x}\left(x^{\frac{9}{2}}+2 x^{\frac{5}{2}}+x^{\frac{1}{2}}\right) \\
& =\frac{\mathrm{d}}{\mathrm{d} x} x^{\frac{9}{2}}+2 \frac{\mathrm{d}}{\mathrm{d} x} x^{\frac{5}{2}}+\frac{\mathrm{d}}{\mathrm{d} x} \sqrt{x} \\
& =\frac{9}{2} x^{\frac{9}{2}-1}+2 \times \frac{5}{2} x^{\frac{5}{2}-1}+\frac{1}{2 \sqrt{x}} \\
& =\frac{9}{2} x^{\frac{7}{2}}+5 x^{\frac{3}{2}}+\frac{1}{2 \sqrt{x}}
\end{aligned}
$
View full question & answer→Question 262 Marks
Differentiate the following w.r.t. x. : $\frac{2}{7} x^{\frac{7}{2}}+\frac{5}{2} x^{\frac{2}{5}}$
AnswerLet $\mathrm{y}=\frac{2}{7} x^{\frac{7}{2}}+\frac{5}{2} x^{\frac{2}{5}}$
Differentiating w.r.t. $x$, we get
$
\begin{aligned}
\frac{\mathrm{d} y}{\mathrm{~d} x} & =\frac{\mathrm{d}}{\mathrm{d} x}\left(\frac{2}{7} x^{\frac{7}{2}}+\frac{5}{2} x^{\frac{2}{5}}\right) \\
& =\frac{2}{7} \frac{\mathrm{d}}{\mathrm{d} x} x^{\frac{7}{2}}+\frac{5}{2} \frac{\mathrm{d}}{\mathrm{d} x} x^{\frac{2}{5}} \\
& =\frac{2}{7} \times \frac{7}{2} x^{\frac{7}{2}-1}+\frac{5}{2} \times \frac{2}{5} x^{\frac{2}{5}-1} \\
& =x^{\frac{5}{2}}+x^{\frac{-3}{5}}
\end{aligned}
$
View full question & answer→Question 272 Marks
Differentiate the following w.r.t. x. : $x^{\frac{5}{2}}+5 x^{\frac{7}{5}}$
Answer$
x^{\frac{5}{2}}+5 x^{\frac{7}{5}}
$
Solution:
Let $\mathrm{y}=x^{\frac{5}{2}}+5 x^{\frac{7}{5}}$
Differentiating w.r.t. $x$, we get
$
\begin{aligned}
\frac{\mathrm{d} y}{\mathrm{~d} x} & =\frac{\mathrm{d}}{\mathrm{d} x}\left(\begin{array}{c}
x^{\frac{5}{2}}+5 x^{\frac{7}{5}}
\end{array}\right) \\
& =\frac{\mathrm{d}}{\mathrm{d} x} x^{\frac{5}{2}}+5 \frac{\mathrm{d}}{\mathrm{d} x} x^{\frac{7}{5}} \\
& =\frac{5}{2} x^{\frac{5}{2}-1}+5 \frac{7}{5} x^{\frac{7}{5}-1} \\
& =\frac{5}{2} x^{\frac{3}{2}}+7 x^{\frac{2}{5}}
\end{aligned}
$
View full question & answer→Question 282 Marks
Differentiate the following w.r.t. x. : $x \sqrt{x}+\log x-e^x$
Answer$
\begin{aligned}
& \text { Let } \mathrm{y}=\mathrm{x} \sqrt{ } \mathrm{x}+\log \mathrm{x}-\mathrm{e}^{\mathrm{x}} \\
& =x^{\frac{3}{2}}+\log x-e^x
\end{aligned}
$
Differentiating w.r.t. $x$, we get
$
\begin{aligned}
& \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\mathrm{d}}{\mathrm{d} x}\left(\begin{array}{c}
x^{\frac{3}{2}}+\log x-\mathrm{e}^x \\
\text { }
\end{array}\right. \\
& =\frac{\mathrm{d}}{\mathrm{d} x} x^{\frac{3}{2}}+\frac{\mathrm{d}}{\mathrm{d} x} \log x-\frac{\mathrm{d}}{\mathrm{d} x} \mathrm{e}^x \\
& =\frac{3}{2} x^{\frac{3}{2}-1}+\frac{1}{x}-\mathrm{e}^x \\
& =\frac{3}{2} x^{\frac{1}{2}}+\frac{1}{x}-\mathrm{e}^x \\
& =\frac{3}{2} \sqrt{x}+\frac{1}{x}-\mathrm{e}^x \\
&
\end{aligned}
$
View full question & answer→