Download App

Derivatives — MATHS STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 ScienceMATHSDerivatives5 Marks
Question
Differentiate the following from first principle:

$\sqrt{2\text{x}^2-1}$

✓

Answer

We have,
$\text{f(x)}=\sqrt{2\text{x}^2+1}$
$\because\text{f}'\text{(x)}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(\text{x+h})-\text{f}(\text{x})}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\sqrt{2(\text{x+h})^2+1}-\sqrt{2\text{x}^2+1}}{\text{h}}$
Multiplying Numerator and denominator by ${\sqrt{2(\text{x+h})^2+1}+\sqrt{2\text{x}^2+1}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\bigg\{2(\text{x+h})^2+1-(2\text{x}^2+1)\bigg\}}{\text{h}\bigg(\sqrt{2(\text{x+h})^2+1}+\sqrt{2\text{x}^2+1}\bigg)}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{2\text{x}^2+2\text{h}^2+4\text{xh}+1-2\text{x}^2-1}{\text{h}\bigg(\sqrt{2(\text{x+h})^2+1}+\sqrt{2\text{x}^2+1}\bigg)}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{4\text{xh}+2\text{h}^2}{\text{h}\sqrt{2(\text{x+h})^2+1}+\sqrt{2\text{x}^2+1}}$
$=\frac{4\text{x}}{2\sqrt{2\text{x}^2+1}}$
$=\frac{2\text{x}}{\sqrt{2\text{x}^2+1}}$ $$

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

Explore more

More "5 Marks Questions" from Derivatives→Derivatives — all question types→MATHS STD 11 Science question papers→Rajasthan Board STD 11 Science question papers→Rajasthan Board question paper generator→

Similar questions

There are three shares of $A$ in a lottery in which there are $3$ prizes and 9 blanks. In another lottery, $B$ has 2 prizes and 6 blanks. Find the ratio of probabilities of winning of $A$ and $B$.
→
Prove that the following sets of three lines are concurrent:
$\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}=1, \frac{\text{x}}{\text{b}}+\frac{\text{y}}{\text{a}}=1$ and $\text{y}=\text{x}.$
→
Prove that $\sin\text{x}+\sin3\text{x}+...+\sin(\text{2n-1})\text{x}=\frac{\sin ^2\text{nx}}{\sin\text{x}}$ for all $\text{n}\in\text{N}.$
→
Calculate the mean deviation about the median of the following observation:
22, 24, 30, 27, 29, 31, 25, 28, 41, 42
→
Prove that:
$\sin3\text{A}+\sin2\text{A}-\sin\text{A}=4\sin\text{A}\cos\frac{\text{A}}{2}\cos\frac{3\text{A}}{2}$
→
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{3\sin^2\text{x}-2\sin\text{x}^2}{3\text{x}^2}$
→
$\sin^3\text{x}+\sin^3\big(\frac{2\pi}{3}+\text{x}\big)+\sin^3\big(\frac{4\pi}{3}+\text{x}\big)=-\frac{3}{4}\sin3\text{x}$
→
$\sin^3\text{x}+\sin^3\big(\frac{2\pi}{3}+\text{x}\big)+\sin^3\big(\frac{4\pi}{3}+\text{x}\big)=-\frac{3}{4}\sin3\text{x}$
→
Find the vertex, focus, axis, directrix and latus-rectum of the following parabolas:
y2 = 8x + 8y.
→
Prove that:
$\tan20^\circ\tan30^\circ\tan40^\circ\tan80^\circ=1$
→