Question
Differentiate the following from first principle
$\tan\text{2x}$
$\tan\text{2x}$
$\text{f}'(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\tan2(\text{x}+\text{h})-\tan2\text{x}}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin(\text{2x}+\text{2h}-\text{2x})}{\text{h}.\cos(\text{2x}+\text{2h})\cos\text{2x}}\ \Big[\because\tan\text{A}-\tan\text{B}=\frac{\sin\text{A}-\text{B}}{\cos\text{A}.\cos{\text{B}}}\Big]$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin2\text{h}}{\text{h}.\cos(\text{2x}+\text{2h})\cos2\text{x}}$
$=\lim_\limits{\text{h}\rightarrow0}\Big(\frac{\sin2\text{h}}{2\text{h}}\Big)\times\frac{1\times2}{\cos(\text{2h}+\text{2x})\cos2\text{x}}$
$=\frac{2}{\cos2\text{x}.\cos2\text{x}}\ \Big[\because\lim_\limits{\text{h}\rightarrow0}\frac{\sin2\text{h}}{2\text{h}}=1\Big]$
$=2\sec^22\text{x}\ \Big[\because\frac{1}{\cos^2\text{x}}=\sec^2\text{x}\Big]$
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| | Column C1 | | Column C2 |
| a. | In xy-plane. | i. | Ist octant. |
| b. | Point (2, 3, 4) lies in the. | ii. | yz-plane. |
| c. | Locus of the points having x coordinate 0 is. | iii. | z-coordinate is zero. |
| d. | A line is parallel to x-axis if and only. | iv. | z-axis. |
| e. | If x = 0, y = 0 taken together will represent the. | v. | plane parallel to xy-plane. |
| f. | z = c represent the plane. | vi. | if all the points on the line have equal y and z-coordinates. |
| g. | Planes x = a, y = b represent the line. | vii. | from the point on the respective. |
| h. | Coordinates of a point are the distances from the origin to the feet of perpendiculars. | viii. | parallel to z-axis. |
| i. | A ball is the solid region in the space enclosed by a. | ix | disc. |
| j. | Region in the plane enclosed by a circle is known as a. | x. | sphere. |