Question
Differentiate sin (2x + 3) w.r.t. x from first principle.

Answer

$y = \sin(2x + 3)$$y = \triangle y = \sin (2x +2\triangle x + 3)$
$\therefore \triangle y \sin (2x + 2\triangle x + 3) -\sin (2x + 3)$
$= 2\cos (2x + 3 + \triangle x) \sin \triangle x$
$\therefore \lim\limits_{\triangle x\rightarrow 0} \frac{\triangle y}{\triangle x} = \lim\limits_{\triangle x \rightarrow 0} 2\cos (2x +3 +\triangle x) \lim\limits_{\triangle x\rightarrow 0} \frac {\sin\triangle x}{\triangle x}$
OR
$\frac{dy}{dx} = \text2\cos (2x + 3) 1 = 2\cos (2x + 3)$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

Find the general solution of $x \frac{d y}{d x}+2 y=x^{2} \log x$
Using properties of determinants, prove the following:
 $\begin{vmatrix}\text{a} & \text{b} & \text{c} \\\text{a}-\text{b} & \text{b}-\text{c} & \text{c}-\text{a}\\\text{b}+\text{c} & \text{c}+\text{a} & \text{a}+\text{b} \end{vmatrix}=\text{a}^3+\text{b}^3+\text{c}^3-3\text{abc}.$
Maximum Z = 15x + 10y
Subject to
$3\text{x}+2\text{y}\leq80$
$2\text{x}+3\text{y}\leq70$
$\text{x},\text{y}\geq0$
A small firm manufactures gold rings and chains. The total number of rings and chains manufactured per day is at most $24$. It takes $1$ hour to make a ring and $30$ minutes to make a chain. The maximum number of hours available per day is $16$. If the profit on a ring is $Rs. 300$ and that on a chain is $Rs. 190,$ find the number of rings and chains that should be manufactured per day, so as to earn the maximum profit. Make it as an $\text{LPP}$ and solve it graphically.
If $\text{A}=\begin{bmatrix}\text{a}&\text{b}\\0&1\end{bmatrix},$ prove that $\text{A}^\text{n}=\begin{bmatrix}\text{a}^\text{n}&\text{b}\Big(\frac{\text{a}^\text{n}-1}{\text{a}-1}\Big)\\0&1\end{bmatrix}$ for every positive integer n.
Find the points of discontinuity, if any of the following function:
$\text{f(x)}=\begin{cases}\frac{\sin\text{x}}{\text{x}},&\text{if }\text{ x}<0\\2\text{x}+3,&\text{ x}\geq0\end{cases}$
Using properties of determinants, solve the following for $x:$
$ \begin{vmatrix} \text{x - 2} & \text{2x - 3 } & \text{3x - 4 } \\ \text{x - 4} & \text{2x - 9} & \text{2x - 16} \\ \text{x -8} & \text{2x - 27} & \text{3x -64} \end{vmatrix}=0$
A firm manufactures two products, each of which must be processed through two departments, 1 and 2. The hourly requirements per unit for each product in each department, the weekly capacities in each department, selling price per unit, labour cost per unit, and raw material cost per unit are summarized as follows:
 
 
Product A
Product B
Weekly capacity
Department 1
3
2
130
Department 2
4
6
260
Selling price per unit
Rs. 25
Rs. 30
 
Labour cost per unit
Rs. 16
Rs. 20
 
Raw material cost per unit
Rs. 4
Rs. 4
 
The problem is to determine the number of units to produce each product so as to maximize total contribution to profit. Formulate this as a LPP.
Solve the following systems of linear equations by cramer's rule:
2x - y = 1,
7x - 2y = -7
Prove that:
$\begin{vmatrix}\text{x}+4&\text{x}&\text{x}\\\text{x}&\text{x}+4&\text{x}\\\text{x}&\text{x}&\text{x}+4\end{vmatrix}=16(3\text{x}+4)$