Question 12 Marks
If $y=\left(2 x^{2}+3\right)(3 x-2)$ then find derivative of $y$ with respect to $x$.
Answer$y =\left(2 x^{2}+3\right)(3 x-2)$
$\text { Take, } u=2 x^{2}+3 \text { and } v=3 x-2 .$
$\therefore \frac{d u}{d x} =4 x \text { and } \frac{d v}{d x}=3 $
$ \text { Now, } y=u \cdot v . $
$\therefore \frac{d y}{d x} =u \frac{d v}{d x}+v \frac{d u}{d x}$
$ =\left(2 x^{2}+3\right)(3)+(3 x-2)(4 x)$
$ =6 x^{2}+9+12 x^{2}-8 x$
$ =18 x^{2}-8 x+9 $
Note : Illustration $9$ can also be solved using working rule $1$ by simplifying $y$ i.e. multiplying two terms of $y$
View full question & answer→Question 22 Marks
Find $\frac{d y}{d x}$ for $y=x^{4}-3 x^{2}+2 x-3$
Answer$ y =x^{4}-3 x^{2}+2 x-3$
$\therefore \frac{d y}{d x} =\frac{d}{d x}\left(x^{4}-3 x^{2}+2 x-3\right)$
$ =\frac{d}{d x}\left(x^{4}\right)-\frac{d}{d x}\left(3 x^{2}\right)+\frac{d}{d x}(2 x)-\frac{d}{d x}(3)$
$ =\frac{d}{d x}\left(x^{4}\right)-3 \frac{d}{d x}\left(x^{2}\right)+2 \frac{d}{d x}(x)-\frac{d}{d x}(3)$
$ =4 x^{3}-3(2 x)+2(1)-(0)$
$ =4 x^{3}-6 x+2 $
View full question & answer→Question 32 Marks
Obtain derivative of $f(x)=k \quad(k$ is constant) with the help of definition.
AnswerHere, $f(x)=k$
$ \therefore f(x+h)=k $
Now, $f^{\prime}(x)=\lim _{h \rightarrow 0} \underset{h}{f(x+h)-f(x)}$
$ =\lim _{h \rightarrow 0} \frac{k-k}{h}$
$ =\lim _{h \rightarrow 0} \frac{0}{h}$
$ =0 $
Hence, if $f(x)=k$ then $f^{\prime}(x)=0$
View full question & answer→Question 42 Marks
The cost function of a commodity is $C=5 x^{2}+6 x+2000$, where $x$ is the number of units produced. Find marginal cost when production is $\mathbf{5 0}$ units.
AnswerCost function $C=5 x^{2}+6 x+2000$
$\therefore \quad$ Marginal Cost $\frac{d C}{d x}=10 x+6$
When $x=50$ then
$
\text { Marginal Cost } \frac{d C}{d x}=10(50)+6
$
$
=506
$
Interpretation : The cost of producing the 51st unit is approximately ₹ 506 .
View full question & answer→Question 52 Marks
If the demand function of pizza is $p=150-4 x$ then find the marginal revenue when demand is of $3$ pizzas and interpret it.
AnswerHere, demand function $p=150-4 x$
Now, revenue function $R=p \cdot x$
$ =(150-4 x) x$
$\therefore R =150 x-4 x^{2} $
Marginal revenue $\frac{d R}{d x}=150-8 x$
When demand of pizza is $x=3$ then
Marginal revenue $\frac{d R}{d x}=150-8(3)$
Interpretation : Revenue for selling the $4 th$ pizza is approximately $₹ 126$.
View full question & answer→Question 62 Marks
If $f(x)=4 x^{3}+2 x^{2}+7 x+9$ then for which value of $x, f^{\prime \prime}(x)=52$ ?
Answer$ f(x)=4 x^{3}+2 x^{2}+7 x+9$
$\therefore f^{\prime}(x)=12 x^{2}+4 x+7$
$\therefore f^{\prime \prime}(x)=24 x+4 $
Now, $f^{\prime \prime}(x)=52$
$ \therefore 24 x+4=52$
$ \therefore 24 x=48$
$ \therefore x=2 $
View full question & answer→Question 72 Marks
Obtain $\frac{d y}{d x}$ for $y=3 x^{4}-2 x^{3}+x^{2}-8 x+7$. Also obtain its value at $x=1$.
Answer$ y=3 x^{4}-2 x^{3}+x^{2}-8 x+7$
$\therefore \frac{d y}{d x} =12 x^{3}-6 x^{2}+2 x-8$
$\therefore \frac{d^{2} y}{d x^{2}} =\frac{d}{d x}\left[\frac{d y}{d x}\right]$
$ =\frac{d}{d x}\left[12 x^{3}-6 x^{2}+2 x-8\right]$
$ =36 x^{2}-12 x+2 $
Putting $x=1$,
$ \frac{d^{2} y}{d x^{2}} =36(1)^{2}-12(1)+2$
$ =36-12+2$
$ =26 $
View full question & answer→Question 82 Marks
If $f(x)=x^{2}-x+3$ then for which value of $x, f^{\prime}(x)=0$ ?
Answer$ \begin{array}{r} \text { Here, } f(x)=x^{2}-x+3$
$\therefore f^{\prime}(x)=2 x-1+0 $
Now, $f^{\prime}(x)=0$ is given
$ \therefore 2 x-1=0$
$ \therefore 2 x=1$
$ \therefore x=\frac{1}{2} $
View full question & answer→Question 92 Marks
If $f(x)=3 x^{2}+2 x+1$ then find $f^{\prime}(x)$ and hence obtain $f^{\prime}(1)$.
AnswerHere, $ f(x)=3 x^{2}+2 x+1$
$ \therefore f^{\prime}(x) =6 x+2$
$\therefore f^{\prime}(1) =6(1)+2$
$ =8 $
View full question & answer→Question 102 Marks
If $y=2+3 x+4 x^{2}+\frac{5}{6-7 x}$ then find $\frac{d y}{d x}$.
Answer$ y =2+3 x+4 x^{2}+\frac{5}{6-7 x}$
$\therefore \frac{d y}{d x} =\frac{d}{d x}\left[2+3 x+4 x^{2}+\frac{5}{6-7 x}\right]$
$ =0+3(1)+4(2 x)+\frac{d}{d x}\left(\frac{5}{6-7 x}\right)$
$ =3+8 x+\frac{(6-7 x)(0)-5(-7)}{(6-7 x)^{2}} \quad[\because \text { Division rule }]$
$ =3+8 x+\frac{35}{(6-7 x)^{2}} $
View full question & answer→Question 112 Marks
Find $\frac{d y}{d x}, y=\sqrt{x^{2}+3}$.
Answer$ y=\sqrt{x^{2}+3} $
Taking $u=x^{2}+3, y=\sqrt{u}$
$ \therefore \frac{d u}{d x}=2 x \text { and } \frac{d y}{d u}=\frac{1}{2 \sqrt{u}} \text {. } $
Now, $\frac{d y}{d x}=\frac{d y}{d u} \times \frac{d u}{d x}$
$ =\left(\frac{1}{2 \sqrt{u}}\right)(2 x)$
$ =\frac{x}{\sqrt{u}} $
Putting value of $u$,
$ \frac{d y}{d x}=\frac{x}{\sqrt{x^{2}+3}} $
View full question & answer→Question 122 Marks
Differentiate $y=(3 x+7)^{8}$ with respect to $x$.
Answer$ y=(3 x+7)^{8} $
Taking $u=3 x+7, y=u^{8}$
$ \therefore \frac{d u}{d x}=3 \text { and } \frac{d y}{d u}=8 u^{7} $
Now, $\frac{d y}{d x}=\frac{d y}{d u} \times \frac{d u}{d x}$
$ =\left(8 u^{7}\right)(3)$
$ =24 u^{7} $
Putting value of $u$,
$ \frac{d y}{d x}=24(3 x+7)^{7} $
View full question & answer→Question 132 Marks
If $y=\frac{3}{4 x+5}$ then differentiate $y$ with respect to $x$.
Answer$ y=\frac{3}{4 x+5}$
$\text { Take } u=3 \text { and } v=4 x+5 .$
$ \therefore \frac{d u}{d x}=0 \text { and } \frac{d v}{d x}=4 $
Now, $ y=\frac{u}{v}$
$ \therefore \frac{d y}{d x} =\frac{v \frac{d u}{d x}-u \frac{d v}{d x}}{v^{2}}$
$ =\frac{(4 x+5)(0)-3(4)}{(4 x+5)^{2}}$
$ =\frac{0-12}{(4 x+5)^{2}}$
$ =\frac{-12}{(4 x+5)^{2}} $
View full question & answer→Question 142 Marks
Obtain the derivative of $f(x)=x$ with the help of definition.
AnswerHere, $f(x)=x$
$ \therefore f(x+h)=x+h $
Now, $f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$
$ =\lim _{h \rightarrow 0} \frac{(x+h)-x}{h}$
$ =\lim _{h \rightarrow 0} \frac{h}{h}$
$ =1 \quad(\because h \neq 0) $
Hence, if $f(x)=x$ then $f^{\prime}(x)=1$.
View full question & answer→Question 152 Marks
Write the chain rule of differentiation.
Answer
| If $y$ is a differentiable function of $u$ and $u$ is a differentiable function of $x,$ then $\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$ |
| This rule is known as Chain Rule, Which used in differentiation of composite function |
View full question & answer→Question 162 Marks
Find $f”(x)$ if $f(x) = \sqrt[4]{x}$.
Answer
| Here, $f(x) = \sqrt[4]{x} = x^{\frac{1}{4}}$ |
| $f'(x) = \frac{1}{4} \mathrm{x}^{1 / 4} 4^{-1}=\frac{1}{4} \mathrm{x}^{-3 / 4}$ |
| Now, $f’’(x) = \frac{1}{4}\left(-\frac{3}{4}\right) x^{\frac{-3}{4}-1}$ |
| $= \frac{-3}{16} x^{-7 / 4}=\frac{-3}{16 x^{7 / 4}}$ |
| Hence, if $f(x) = x^{\frac{1}{4}}$ Then $f’’(x) =\frac{-3}{16 x^7 / 4}$ |
View full question & answer→Question 172 Marks
State the condition for production cost function $c$ to be minimum.
AnswerThe condition for minimizing the production cost function $c$ are as follows : $(i)\ \frac{dc}{dx} = 0$ and $(ii)\ \frac{\mathrm{d}^2 \mathrm{C}}{\mathrm{dx}^2} > 0$
View full question & answer→Question 182 Marks
What are the conditions for profit function $p$ to be maximum$?$
AnswerThe condition for minimizing the profit $p$ are as follows : $(i)\ \frac{dp}{dx} = 0$ and $(ii)\ \frac{\mathrm{d}^2 \mathrm{p}}{\mathrm{dx}^2} < 0$
View full question & answer→Question 192 Marks
Define elasticity of demand.
AnswerThe ratio of percentage change in the demand of a commodity due to percentage change in the Price is called elasticity of deman(D) i.e.
Elasticity of demand $=\frac { Percentage\ Change\ in\ Demand } { Percentage\ Change\ in\ Price }$
View full question & answer→Question 202 Marks
Explain marginal cost and give its formula.
AnswerThe change in cost due to small change in production is called marginal cost. Marginal cost can be obtained by talking the derivative cost function with respect to x. Thus, when the production is x then Marginal cost $=\frac{dc}{dx}$
View full question & answer→Question 212 Marks
State necessary and sufficient conditions for a function to be maximum at $x=a.$
AnswerThe necessary and sufficient conditions for a function to be maximum at $x = a$ are follows: $(i)\ f’ (a) = 0$ and $(ii)\ f” (a)<0$
View full question & answer→Question 222 Marks
Obtain marginal cost if the production cost function is $C=0.0012 x^2-0.18 x+25$.
Answer$C=0.0012 x^2-0.18 x+25$
Now, marginal cost
$=\frac{d \mathrm{C}}{d x}$
$=\frac{d}{d x}\left[0.0012 x^2-0.18 x+25\right]$
$=0.0012(2 x)-0.18(1)+0$
$=0.0024 x-0.18$
Hence, $C=0.0012 x^2-0.18 x+25$ then marginal cost obtained is $0.0024 x-0.18$
View full question & answer→Question 232 Marks
State the division rule of derivative.
Answer
| if $u$ and $v$ are differentiable function of $x$ and if $Y=\frac{u}{v}; v\neq 0$ then $\frac{d y}{d x}=\frac{v\left(\frac{d u}{d x}\right)-u\left(\frac{d v}{d x}\right)}{v^2}$. |
View full question & answer→Question 242 Marks
Find $\frac{d^2 y}{d x^2}$ if $y =\sqrt{x}+\frac{1}{\sqrt{x}}$
Answer
| Here, $y = \sqrt{x}+\frac{1}{\sqrt{x}}$ |
| $=x^{\frac{1}{2}}+x^{-\frac{1}{2}}$ |
| $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{2} \mathrm{x}^{-\frac{1}{2}}-\frac{1}{2} \mathrm{x}^{-\frac{3}{2}}$ |
| Now,$\frac{d^2 y}{d x^2}=\frac{d}{d x}\left[\frac{d y}{d x}\right]$ |
| $=\frac{d}{d x}\left[\frac{1}{2} x^{-\frac{1}{2}}-\frac{1}{2} x^{-\frac{3}{2}}\right]$ |
| $=-\frac{1}{4} x^{-\frac{3}{2}}+\frac{3}{4} x^{-\frac{5}{2}}$ |
| $=\frac{-1}{4 x^{\frac{3}{2}}}+\frac{3}{4 x^{\frac{5}{2}}}$ |
| $\therefore\ \frac{d^2 y}{d x^2}=\frac{-1}{4 x^{\frac{3}{2}}}+\frac{3}{4 x^{\frac{5}{2}}}$ |
View full question & answer→Question 252 Marks
Find ${\frac{d^2 y}{d x^2}}$if $ y=2 x^3+5 x^2-3+\frac{4}{x^2}-\frac{5}{x^3}$
AnswerHere $y=2 x^3+5 x^2-3+\frac{4}{x^2}-\frac{5}{x^3}$
$=2 x^3+5 x^2-3+4 x^{-2}-5 x^{-3}$
$\therefore \frac{d y}{d x}=6 x^2+10 x^2-0-8 x^{-3}-15 x^{-4}$
Now,$\frac{d^2 y}{d x^2} \quad=\frac{d}{d x}\left[\frac{d y}{d x}\right]$
$=\frac{d}{d x}\left[6 x^2+10 x^2-8 x^{-3}-15 x^{-4}\right]$
$=12 x+10+24 x^{-} 60 x^{-5}$
$=12 x+10+\frac{24}{x^4}-\frac{60}{x^5}$
$\therefore \frac{ d ^2 y}{d x^2}=12 x+10+\frac{24}{x^4}-\frac{60}{x^5}$
View full question & answer→Question 262 Marks
If $f(x) = 3x^2 + 3$ then for which value of $x, f’(x) = f(x)?$
Answer
| Here, $f(x) = 3x^2 + 3$ |
| $\therefore f'(x) = 6x + 0 = 6x$ |
| Now, $f’(x) = f(x)$ |
| $6x = 3x^2 + 3$ |
| $\therefore 3x^2 – 6x + 3 = 0$ |
| $\therefore x^2 – 2x + 1 = 0$ |
| $\therefore (x - 1)^2 = 0$ |
| $\therefore (x - 1) (x - 1) = 0$ |
| $\therefore x = 1$ |
|
| Hence, if $x = 1$, then $f’(x) = f(x).$ |
View full question & answer→Question 272 Marks
$f(x) = x^3 + 5x^2 + 3x + 7,$ prove that $f’(2) = 35.$
Answer
| $f(x) = x^3 + 5x^2 + 3x + 7$ |
|
| $\therefore f'(x) = 3x^2 + 10x + 3$ |
| Now, putting $x = 2$ |
| $f'(2)$ |
$= 3(2)^2 + 10(2) + 3$ |
| |
$= 3(4) + 20 + 3$ |
| |
$= 12 + 20 + 3$ |
| |
$= 35$ |
| Hence, if $f(x) x^3 + 5x^2 + 3x + 7$ then $f’(2) = 35$ |
|
| $f(x) = x^3 + 5x^2 + 3x + 7$ |
View full question & answer→Question 282 Marks
Determine whether the function $y = 12 + 4x – 7x^2,$ is increasing or decreasing at $x = 2.$
Answer
| Here, $y = 12 + 4x – 7x^2$ |
| $ = 4 – 14x$ |
| At $x = 2$ |
| |
| $= 4 -28$ |
| $= -24 < 0$ |
| $∴$ Function is decreasing $x = 2.$ |
View full question & answer→Question 292 Marks
When can it be said that a function is decreasing at a point ?
AnswerSuppose, $y = f (x)$ is a function and $h$ is a very small positive number.
$\rightarrow $ If at a point $x = a. $
$f(a + h) < f(a)$ and $f(a) < f(a – h),$
then at $x = a,$
function $f(x)$ is called decreasing function.
For such decreasing function $f'(a) < 0.$
View full question & answer→Question 302 Marks
What is the maximum value of function?
Answer
- If $h$ is small positive number and if $f(a) > f(a + h)$ and also $f(a) > f(a – h)$ then $f(x)$ is said to be maximum at $x = a.$
- Maximum value do not mean largest value of function.
- The function is maximum of $x = a.$
- Only mean that the value of function is maximum in a small interval around $x = a.$
View full question & answer→Question 312 Marks
Find Marginal revenue if revenue function is $90x - \frac{x^2}{2}$.
Answer
| Revenue function $R = 90x - \frac{x^2}{2}$ |
| Marginal revenue $\frac{dr}{dx} = 90 – \frac{2x}{2}$ |
| $= 90 - x$ |
| Hence, if $R = 90x – \frac{x^2}{2}$ then $\frac{dr}{dx} = 90 -x$ |
View full question & answer→Question 322 Marks
Find $f^{\prime \prime}(0)$ if $f(x)=x^4-4 x^3+3 x^2+x+1$
AnswerHere, if $f^{\prime}(x)=x^4-4 x^3+3 x^2+x+1$
$f^{\prime}(x)=4 x^3+12 x^2+6 x+1$
Now, $f^{\prime}(x)=12 x^2+24 x+6$
Therefore, $f^{\prime}(0)=12(0)^2 24(0)+6$
$=0-0+6$
Hence, $f^{\prime}(x)=x^4-4 x^3+3 x^2+x+1$ then $f^{\prime \prime}(0)=6$.
View full question & answer→Question 332 Marks
Answer
| Let $f: A\rightarrow R$ and $a \in A,$ where $A$ is an open interval of $R.$ if $\lim _{h \rightarrow 0} \frac{f(a+h)-f(a)}{h}$ exists, then limit of a function $4$ is called |
| derivative at $x = (A)$ it is denoted by $f’ (a).$ Thus, $ f’(a) =\lim _{h \rightarrow 0} \frac{f(a+h)-f(a)}{h}$ . |
View full question & answer→Question 342 Marks
Obtain the derivatives of the following function with the help of definition: $f(x)=10$
AnswerHere, $f(x) = 10$
$\therefore f(x + h) = 10$
Hence, $f(x) = 10$ then $f’(x) = 0$ View full question & answer→Question 352 Marks
Obtain the derivatives of the following function with the help of definition: $f ( x )=x^{7}$
Answer
Here, $f(x) = x^7$
$\therefore f(x + h) = (x + h)^7$
Take, $x + h = t,$ when $h → 0, t → x$ and $h = t – x$
Hence, $f(x) = x^7$ then $f'(x) = 7x^6$ View full question & answer→Question 362 Marks
Obtain the derivatives of the following function with the help of definition: $f(x)=x^{2}$
Answer
Here, $f(x) = x^2 $
$\therefore f(x + h) = (x + h)^2 = x^2 + 2xh + h^2$
View full question & answer→Question 372 Marks
Obtain the derivatives of the following function with the help of definition: $f(x)=2 x+3$
AnswerHere, $f(x) = 2x + 3$
$\therefore f(x + h) = 2 (x + h) + 3 = 2x + 2h + 3$
View full question & answer→Question 382 Marks
Find the derivative of following functions: $y=e^{n-1}$
View full question & answer→Question 392 Marks
Find the derivative of following functions: $f(x)=\pi^{3}$
View full question & answer→Question 402 Marks
Find the derivative of following functions: $y=3 x^{11}$
View full question & answer→Question 412 Marks
Find the derivative of following functions: $f(x)=\frac{a^{2}+3 a}{9}$
View full question & answer→Question 422 Marks
Find the derivative of following functions: $f(x)=\frac{1}{3} x^{\frac{3}{3}}$
Answer$\frac{1}{5} x^{-\frac{2}{5}}$
View full question & answer→Question 432 Marks
Find the derivative of following functions: $f(x)=\frac{3 k^{2}}{5}$
View full question & answer→Question 442 Marks
If $f(x)=3 x^{2}+3$, then for which value of $x, f^{\prime}(x)=f(x) ?$
Answerfor $x=1, f^{\prime}(x)=f(x)$
View full question & answer→Question 452 Marks
Find the derivative of following functions: $y=e^{n-1}$
View full question & answer→Question 462 Marks
Find the derivative of following functions: $f(x)=\pi^{3}$
View full question & answer→Question 472 Marks
Find the derivative of following functions: $f(x)=\frac{a^{2}+3 a}{9}$
View full question & answer→Question 482 Marks
Find the derivative of following functions: $y=3 x^{11}$
View full question & answer→Question 492 Marks
Find the derivative of following functions: $f(x)=\frac{3 k^{2}}{5}$
View full question & answer→Question 502 Marks
Find the derivative of following functions: $f(x)=\frac{1}{3} x^{\frac{3}{3}}$
Answer$\frac{1}{5} x^{-\frac{2}{5}}$
View full question & answer→