Download App

CONTINUITY AND DIFFERENTIABILITY — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSCONTINUITY AND DIFFERENTIABILITY5 Marks
Question
Differentiate the following functions from first principles:
$\sin^{-1}(2\text{x}+3)$
✓

Answer

Let $\text{f(x)}=\sin^{-1}(2\text{x}+3)$
$\Rightarrow\ \text{f}(\text{x}+\text{h})=\sin^{-1}(2(\text{x}+\text{h})+3)$
$\Rightarrow\ \text{f}(\text{x}+\text{h})=\sin^{-1}(2\text{x}+2\text{h}+3)$
$\therefore \frac{\text{d}}{\text{dx}}\{\text{f(x)}\}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f(x)}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\sin^{-1}(2\text{x}+2\text{h}+3)-\sin^{-1}(2\text{x}+3)}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\sin^{-1}\Big[(2\text{x}+2\text{h}+3)\sqrt{1+(2\text{x}+3)^2}-(2\text{x}+3)\sqrt{1-(2\text{x}+2\text{h}+3)^2}\Big]}{\text{h}}$
$\Big[\text{Since}, \sin^{-1}\text{x}-\sin^{-1}\text{y}=\sin^{-1}\big[\text{x}\sqrt{1-\text{y}^2}-\text{y}\sqrt{1-\text{x}^2}\big]\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\sin^{-1}\text{z}}{\text{z}}\times\frac{\text{z}}{\text{h}}$
Where, $\text{z}=(2\text{x}+2\text{h}+3)\sqrt{1-(2\text{x}+3)^2}-(2\text{x}+3)\sqrt{1-(2\text{x}+2\text{h}+3)^2}$
$\text{and }\lim\limits_{\text{h}\rightarrow0}\frac{\sin^{-1}\text{h}}{\text{h}}=1$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{z}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{(2\text{x}+2\text{h}+3)\sqrt{1-(2\text{x}+3)^2}-(2\text{x}+3)\sqrt{1-(2\text{x}+2\text{h}+3)^2}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{(2\text{x}+2\text{h}+3)^2-(2\text{x}+3)^2-(2\text{x}+3)^2\big(1-(2\text{x}+2\text{h}+3)^2\big)}{\big\{(2\text{x}+2\text{h}+3)\sqrt{1-(2\text{x}+3)^2}-(2\text{x}+3)\sqrt{1-(2\text{x}+2\text{h}+3)^2}\big\}}$
[Since, rationalizing numerator]
$=\lim\limits_{\text{h}\rightarrow0}\frac{\big[(2\text{x}+3)^2+4\text{h}^2+4\text{h}(2\text{x}+3)\big]\big(1-(2\text{x}+3)^2\big)-(2\text{x}+3)^2\big[1-(2\text{x}+3)^2-4\text{h}^2-4\text{h}(2\text{x}+3)\big]}{\text{h}\big\{(2\text{x}+2\text{h}+3)\sqrt{1-(2\text{x}+3)^2}+(2\text{x}+3)\sqrt{1-(2\text{x}+2\text{h}+3)^2}\big\}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\Big[(2\text{x}+3)^2+4\text{h}^2+4\text{h}(2\text{x}+3)-(2\text{x}+3)^4-4\text{h}^2(2\text{x}+3)^2-4\text{h}(2\text{x}+3)^3 -(2\text{x}+3)^2+(2\text{x}+3)^3+4\text{h}^2(2\text{x}+3)^2+4\text{h}(2\text{x}+3)^3\Big]}{\text{h}\big\{(2\text{x}+2\text{h}+3)\sqrt{1-(2\text{x}+3)^2}+(2\text{x}+3)\sqrt{1-(2\text{x}+2\text{h}+3)^2}\big\}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{4\text{h}\big[\text{h}+(2\text{x}+3)\big]}{\text{h}\Big\{(2\text{x}+2\text{h}+3)\sqrt{1-(2\text{x}+3)^2}+(2\text{x}+3)\sqrt{1-(2\text{x}+2\text{h}+3)^2}\Big\}}$
$=\frac{4\text{h}\big[\text{h}+(2\text{x}+3)\big]}{(2\text{x}+3)\sqrt{1-(2\text{x}+3)^2}+(2\text{x}+3)\sqrt{1-(2\text{x}+3)^2}}$
$=\frac{4(2\text{x}+3)}{2(2\text{x}+3)\sqrt{1-(2\text{x}+3)^2}}$
$=\frac{2}{\sqrt{1-(2\text{x}+3)^2}}$
So,
$\frac{\text{d}}{\text{dx}}\big(\sin^{-1}(2\text{x}+3)\big)=\frac{2}{\sqrt{1-(2\text{x}+3)^2}}$

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 CONTINUITY AND DIFFERENTIABILITY→CONTINUITY AND DIFFERENTIABILITY — all question types→MATHS STD 12 Science question papers→Rajasthan Board STD 12 Science question papers→Rajasthan Board question paper generator→

Similar questions

Find the equation of the plane determined by the intersection of the lines $\frac{\text{x}+3}{3}=\frac{\text{y}}{-2}=\frac{\text{z}-7}{6}$ and $\frac{\text{x}+6}{1}=\frac{\text{y}+5}{-3}=\frac{\text{z}-1}{2}$
→
Find the equation of the plane passing through the intersection of the planes.
$ \vec{r}.(2\hat{\text{i}}+\hat{\text{j}}+3\hat{\text{k})}=7,\text{ }\vec{r}.(2\hat{\text{i}}+5\hat{\text{j}}+3\hat{\text{k}})=9$and the point (2, 1, 3).
→
If $\text{x}=\text{a}\sin\text{t}-\text{b}\cos\text{t},\text{y}=\text{a}\cos\text{t}+\text{b}\sin\text{t},$ Prove that $\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\frac{\text{x}^2+\text{y}^2}{\text{y}^2}$ 
→
Discuss the continuity of the f(x) at the indicated points f(x) = |x - 1| + |x + 1| at x = -1, 1.
→
Reshma wishes to mix two types of food P and Q in such a way that the vitamin contents of the mixture contains at least 8 units of vitamin A and 11 units of vitamin B. Food P costs Rs 60/kg and food Q costs Rs 80/kg. Food P contains 3 units/kg of vitamin A and 5 units/kg of vitamin B while food Q contains 4 units/kg of vitamin A and 2 units/kg of vitamin B. Determine the minimum cost of the mixture.
→
A man owns a field of area 1000 sq.m. He wants to plant fruit trees in it. He has a sum of Rs. 1400 to purchase young trees. He has the choice of two types of trees. Type A requires 10 sq.m of ground per tree and costs Rs. 20 per tree and type B requires 20 sq.m of ground per tree and costs Rs. 25 per tree. When fully grown, type A produces an average of 20kg of fruit which can be sold at a profit of Rs. 2.00 per kg and type B produces an average of 40kg of fruit which can be sold at a profit of Rs. 1.50 per kg. How many of each type should be planted to achieve maximum profit when the trees are fully grown? What is the maximum profit?
→
Prove that in throwing a pair of dice, the occurrence of the number 4 on the first die is independent of the occurrence of 5 on the second die.
→
If O is a point in space, ABC is a triangle and D, E, F are the mid-points of the sides BC, CA and AB respectively of the triangle, prove that $\overrightarrow{\text{OA}}+\overrightarrow{\text{OB}}+\overrightarrow{\text{OC}}=\overrightarrow{\text{OD}}+\overrightarrow{\text{OE}}+\overrightarrow{\text{OF}}$.
→
Find the vector equation (in scalar product form) of the plane containing the line of intersection of the planes x - 3y + 2z - 5 = 0 and 2x - y + 3z - 1 = 0 and passing through (1, -2, 3).
→
Find the point on the parabolas $x^2 = 2y$ which is closest to the point $(0,5).$
→