Download App

Differentiation — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDifferentiation5 Marks
Question
Differentiate the following functions with respect to x:
$\frac{\text{x}^2(1-\text{x}^2)}{\cos2\text{x}}$
✓

Answer

Consider $\text{y}=\frac{\text{x}^2+2}{\sqrt{\cos\text{x}}}$
Differentiate it with respect to x and applying the chain and product rule, we get
$\frac{\text{dy}}{\text{dx}}=\frac{\cos2\text{x}\frac{\text{d}}{\text{dx}}\text{x}^2(1-\text{x}^2)^3-\text{x}^2(1-\text{x}^2)^3\frac{\text{d}}{\text{dx}}\cos2\text{x}}{\cos^2 2\text{x}}$
$=\frac{\cos2\text{x}\Big[\text{x}^2\frac{\text{d}}{\text{dx}}(1-\text{x}^2)^3+(1-\text{x}^2)^3\frac{\text{d}}{\text{dx}}\text{x}^2-\text{x}^2(1-\text{x}^2)^3(-2\sin2\text{x})\Big]}{\cos^2 2\text{x}}$
$=\frac{\cos2\text{x}\Big[-6\text{x}^2(1-\text{x}^2)^2+(1-\text{x}^2)^32\text{x}+2\text{x}^2(1-\text{x}^2)^3\sin2\text{x}\Big]}{\cos^2 2\text{x}}$
$=\frac{2\text{x}(1-\text{x}^2)^2}{\cos2\text{x}}-\frac{6\text{x}^3(1-\text{x}^2)^2}{\cos2\text{x}}+\frac{2\text{x}^2(1-\text{x}^2)^3\sin2\text{x}}{\cos^2 2\text{x}}$
$=2\text{x}(1-\text{x}^2)\sec2\text{x}\big\{1-4\text{x}^2+\text{x}(1-\text{x}^2)\tan2\text{x}\big\}$
Therefore,
$\frac{\text{dy}}{\text{dx}}=2\text{x}(1-\text{x}^2)\sec2\text{x}\big\{1-4\text{x}^2+\text{x}(1-\text{x}^2)\tan2\text{x}\big\}$

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 "Solve the Following Question.(5 Marks)" from Differentiation→Differentiation — all question types→Maths STD 12 Science question papers→Maharashtra Board STD 12 Science question papers→Maharashtra Board question paper generator→

Similar questions

For the following pairs of matrices verify that $(AB)^{-1} = B^{-1} A^{-1}$​​​​​​​:$\text{A}=\begin{bmatrix}3 & 2 \\7 & 5 \end{bmatrix}\text{ and B}=\begin{bmatrix}4 & 6 \\3 & 2 \end{bmatrix}$
→
Solve the following differential equation
$\frac{\text{dy}}{\text{dx}}=\frac{1-\cos\text{x}}{1+\cos\text{x}}$
→
Compare the area under the curve $\text{y}=\cos^2\text{x}\text{ and }\text{y}=\sin^2\text{x}$ between $x = 0$ and $\text{x}=\pi.$
→
Evaluate : $\int \frac{1}{2-3 \sin 2 x} \cdot d x$
→
Differentiate $\cos^{-1}(4\text{x}^3-3\text{x})$ with respect to $\tan^{-1}\Big(\frac{\sqrt{1-\text{x}^2}}{\text{x}}\Big),$ if $\frac{1}{2}<\text{x}<1$
→
Let $A=\{1,2,3,4\} ; B=\{3,5,7,9\} ; C=\{7,23,47,79\}$ and $f: A \rightarrow B, g: B \rightarrow C$ be defined as $f(x)=2 x+1$ and $g(x)=x^2-$ 2. Express $(g \circ f)^{-1}$ and $f^{-1} og ^{-1}$ as the sets of ordered pairs and verify that $(g \circ f)^{-1}=f^{-1} \circ g^{-1}$.
→
Show that $\text{f}(\text{x})=\cos\text{x}$ is a decreasing function on $(0,\pi),$ increasing in $(-\pi,0)$ and neither increasing nor decreasing in $(-\pi,\pi).$
→
Find the equations of tangent and normal to the curve at the given point on it : $x=2 \sin ^3 \theta, y=3 \cos ^3 \theta$ at $\theta=\frac{\pi}{4}$
→
Find the angles which the vector $\vec{\text{a}}=\hat{\text{i}}-\hat{\text{j}}+\sqrt{2}\hat{\text{k}}$ makes with the coordinate axes.
→
Find the intervals in which the following functions are increasing or decreasing.
$\text{f}(\text{x})=\frac{3}{10}\text{x}^4-\frac{4}{5}\text{x}^3-3\text{x}^2+\frac{36}{5}\text{x}+11$
→