Download App

Differentiation — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSDifferentiation5 Marks
Question
Differentiate the following functions with respect to x:
$\frac{\text{x}^2+2}{\sqrt{\cos\text{x}}}$
✓

Answer

Let $\text{y}=\frac{\text{x}^2+2}{\sqrt{\cos\text{x}}}$
Differentiate it with respect to x we get,
$\frac{\text{dy}}{\text{dx}}=\frac{\sqrt{\cos\text{x}}\frac{\text{d}}{\text{dx}}(\text{x}^2+2)-(\text{x}^2+2)\frac{\text{d}}{\text{dx}}(\sqrt{\cos\text{x}})}{(\sqrt{\cos\text{x}})^2}$
[Using quotient rule and chain rule]
$=\frac{2\text{x}\sqrt{\cos\text{x}}-(\text{x}^2+2)\Big(-\frac{1}{2}\frac{\sin\text{x}}{\sqrt{\cos\text{x}}}\Big)}{\cos\text{x}}$
$=\frac{2\text{x}\sqrt{\cos\text{x}}+\frac{(\text{x}^2+2)\sin\text{x}}{2\sqrt{\cos\text{x}}}}{\cos\text{x}}$
$=\frac{4\text{x}\cos\text{x}+(\text{x}^2+2)\sin\text{x}}{2(\cos\text{x})^\frac{3}{2}}$
$=\frac{2\text{x}}{\sqrt{\cos\text{x}}}+\frac{1}{2}\frac{(\text{x}^2+2)\sin\text{x}}{(\cos\text{x})^\frac{3}{2}}$
$=\frac{1}{\sqrt{\cos\text{x}}}\Big\{2\text{x}+\frac{1}{2}\frac{(\text{x}^2+2)\sin\text{x}}{\cos\text{x}}\Big\}$
$=\frac{1}{\sqrt{\cos\text{x}}}\Big\{2\text{x}+\frac{(\text{x}^2+2)\tan\text{x}}{2}\Big\}$
So,
$\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}^2+2}{\sqrt{\cos\text{x}}}\Big)=\frac{1}{\sqrt{\cos\text{x}}}\Big\{2\text{x}+\frac{(\text{x}^2+2)\sin\text{x}}{2}\Big\}$

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 Differentiation→Differentiation — all question types→MATHS STD 12 Science question papers→Rajasthan Board STD 12 Science question papers→Rajasthan Board question paper generator→

Similar questions

Prove that the function $\text{f(x)}=\begin{cases}\frac{\sin\text{x}}{\text{x}},&\text{x}<0\\\text{x}+1,&\text{x}\geq0\end{cases}$ is everywhere continuous.
→
Show that the matrix $\text{A}=\begin{bmatrix} 1 & 0 & -2 \\ -2 & -1 & 2 \\ 3 & 4 & 1 \end{bmatrix}$ satisfies the equation,$ \ce{A^3 - A^2 - 3A - I_3 = 0.}$ Hence, find $A^{-1}.$
→
Let $A =(-1,1)$. Then, discuss whether the following functions defined on A are one$-$one, onto or bijective:
i. $f(x)=\frac{x}{2}$
ii. $g(x)=|x|$
iii. $h(x)=x|x|$
iv. $k(x)=x^2$
→
Show that the points A, B, C with position vectors $2\hat{\text{i}} - \hat{\text{j}} + \hat{\text{k}}, \hat{\text{i}} - 3\hat{\text{j}} - 5\hat{\text{k}} \text{ and } 3\hat{\text{i}} - 4\hat{\text{j}} - 4\hat{\text{k}}$ respectively, are the vertices of a right-angled triangle. Hence find the area of the triangle.
→
$\text{Let } \vec{\text a} = \hat{\text{i}} + \hat{\text{j}} + \hat{\text{k}}, \vec{\text{b}} = \hat{\text{i}} \text{ and } \vec{\text{c}} = \text{c}_{1} \hat{\text{i}} + \text{c}_{2} \hat{\text{j}} + \text{c}_{3} \hat{\text{k}}, \text{then}$
  1. Let $c_1 = 1$ and $c_2 = 2$, find $c_3$ which makes $\vec{\text{a}}, \vec{\text{b}} \text{ and }\vec{\text{c}} \text{ coplanar.}$
  2. If $c_2 = –1$ and $c_3 = 1,$ show that no value of $c_1$ can make $\vec{\text{a}}, \vec{\text{b}} \text{ and } \vec{\text{c}} \text{ coplanar}.$
→
A(1, 0, 4) B(0, -11, 1), C(2, -3, 1) are three points and D is the fool of perpendicular from A on BC. Find the coordinates of D.
→
Evaluate the following integrals:
$\int\limits^1_0\sqrt{\frac{1-\text{x}}{1+\text{x}}}\text{ dx}$
→
Find the general solution of (1 + tany) (dx - dy) + 2xdy = 0.
→
Find the intervals in which the following functions are increasing or decreasing. $f(x) = 5 + 36x + 3x^2- 2x^3$
→
If the radius of a sphere is measured as 7m with an error of 0.02m, find the approximate error in calculating its volume.
→