Download App

Differentiation — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDifferentiation4 Marks
Question
If $\text{x}=\frac{\sin^3\text{t}}{\sqrt{\cos^2\text{t}}},\text{y}=\frac{\cos^3\text{t}}{\sqrt{\cos2\text{t}}},$ find $\frac{\text{dy}}{\text{dx}}$
✓

Answer

We have, $\text{x}=\frac{\sin^{3}\text{t}}{\sqrt{\cos2\text{t}}}$ and $\text{y}=\frac{\cos^{3}\text{t}}{\sqrt{\cos2\text{t}}}$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=\frac{\text{d}}{\text{dt}}\Big[\frac{\sin^{3}\text{t}}{\sqrt{\cos2\text{t}}}\Big] $
$\Rightarrow\frac{\text{dx}}{\text{dt}}=\frac{\sqrt{\cos2\text{t}}\frac{\text{d}}{\text{dt}}(\sin^{3})-\sin^{3}\text{t}\frac{\text{d}}{\text{dt}}\sqrt{\cos2\text{t}}}{\cos2\text{t}}$
[Using quotient rule]
$\Rightarrow\frac{\text{dx}}{\text{dt}}=\frac{\sqrt{\cos2\text{t}}(3\sin^{3}\text{t})\frac{\text{d}}{\text{dt}}(\sin\text{t})-\sin^{3}\times\frac{1}{2\sqrt{\cos2\text{t}}}\frac{\text{d}}{\text{dt}}(\cos2\text{t})}{\cos2\text{t}}$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=\frac{3\sqrt{\cos2\text{t}}(\sin^{2}\text{t}\cos\text{t})-\frac{\sin^{3}\text{t}}{2\sqrt{\cos2\text{t}}}(-2\sin2\text{t})}{\cos2\text{t}}$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=\frac{3\cos2\text{t}\sin^{\text{2}}\text{t}\cos\text{t}+\sin^{3}\text{t}\sin2\text{t}}{\cos2\text{t}\sqrt{\cos2\text{t}}}$
Now, $\frac{\text{dy}}{\text{dt}}=\frac{\text{d}}{\text{dt}}\Big[\frac{\cos^{3}}{\sqrt{\cos2\text{t}}}\Big]$
$\Rightarrow\frac{\text{dy}}{\text{dt}}=\frac{\sqrt{\cos2\text{t}} \frac{\text{d}}{\text{dt}}(\cos^{3}\text{t})-\cos^{3}\text{t}\frac{\text{d}}{\text{dt}}\sqrt{\cos2\text{t}}}{\cos2\text{t}}$
[Using quotient rule]
$\Rightarrow\frac{\text{dy}}{\text{dt}}=\frac{\sqrt{\cos2\text{t}}(3\cos^{2}\text{t})\frac{\text{d}}{\text{dt}}(\cos\text{t})-\cos^{3}\text{t}\times\frac{1}{2\sqrt{\cos2\text{t}}}\frac{\text{d}}{\text{dt}}(\cos2\text{t})}{\cos2\text{t}}$
$\Rightarrow\frac{\text{dy}}{\text{dt}}=\frac{3\sqrt{\cos2\text{t}}\cos^{2}\text{t}-(\sin\text{t})-\frac{\cos^{3}\text{t}}{2\sqrt{\cos2\text{t}}}(-2\sin2\text{t})}{\cos2\text{t}}$
$\Rightarrow\frac{\text{dy}}{\text{dt}}=\frac{-3\cos2\text{t}\cos^{2}\text{t}+\cos^{3}\text{t}\sin2\text{t}}{\cos2\text{t}\sqrt{\cos2\text{t}}}$
$\therefore\frac{\text{dy}}{\text{dt}}=\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{-3\cos2\text{t}\cos^{2}\text{t}\sin\text{t}+\cos^{3}\text{t}\sin2\text{t}}{\cos2\text{t}\sqrt{\cos2\text{t}}}\times\frac{\cos2\text{t}\sqrt{\cos2\text{t}}}{3\cos2\text{t}\sin^{2}\text{t}\cos\text{t}+\sin^{3}\text{t}\sin2\text{t}}$
$\Rightarrow\frac{\text{dy}}{\text{dt}}=\frac{\sin\text{t}\cos{\text{t}}[-3\cos2\text{t}\cos\text{t}+2\cos^{3}\text{t}]}{\sin\text{t}\cos\text{t}[3\cos2\text{t}\sin{\text{t}}+2\sin^{3}\text{t}]}$
$\Rightarrow\frac{\text{dy}}{\text{dt}}=\frac{[-3(2\cos^{2}\text{t}-1)\cos\text{t}+2\cos^{3}\text{t}]}{[3(1-2\sin^{2}\text{t})\sin\text{t}+2\sin^{3}\text{t}]}$
$\begin{bmatrix} \cos2\text{t}=2\cos^2\text{t}-1 \\ \cos2\text{t}=1-2\sin^2\text{t} \end{bmatrix}$
$\Rightarrow\frac{\text{dy}}{\text{dt}}=\frac{-4\cos^{3}\text{t}+3\cos\text{t}}{3\sin\text{t}-4\sin^{3}\text{t}}$
$\Rightarrow\frac{\text{dy}}{\text{dt}}=\frac{-\cos3\text{t}}{\sin3\text{t}}$
$\begin{bmatrix} \cos3\text{t}=4\cos^3\text{t}-3\cos\text{t} \\ \sin3\text{t}=3\sin^2\text{t}-4\sin^3\text{t} \end{bmatrix}$
$\therefore\frac{\text{dy}}{\text{dx}}=-\cos3\text{t}$

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 "4 Marks" from Differentiation→Differentiation — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

If O be the origin and the coordinates of P be (1, 2, -3), then find the equation of the plane passing through P and perpendicular to OP.
→
A furniture trader deals in only two items – chairs and tables. He has ₹ 50,000 to invest and a space to store at most 35 items. A chair costs him ₹ 1000 and a table costs him ₹ 2000 The trader earns a profit of ₹ 150 and ₹ 250 on a chair and table, respectively. Formulate the above problem as an LPP to maximise the profit and solve it graphically.
→
Evaluate the following intregals:
$\int\frac{1}{1-\sin\text{x}+\cos\text{x}}\ \text{dx}$
→
Integrate the following integrals:
$\int\sin2\text{x}\sin4\text{x}\sin6\text{x dx}$
→
An isosceles triangle of vertical angle 2θ is inscribed in a circle of radius a. Show that the area of the triangle is maximum when $\theta = \frac{\pi}{6}.$ 
→
Evaluate the following integrals:
$\int\text{e}^{2\text{x}}\cos^2\text{x }\text{dx}$
→
Evaluate the following integrals:

$\int\frac{\text{x}^3\sin^{-1}\text{x}^2}{\sqrt{1-\text{x}^4}}\text{dx}$

→
Solve the following system of equations by matrix method:
6x - 12y + 25z = 4
4x + 15y - 20z = 3
2x + 18y + 15z = 10
→
There are three categories of students in a class of 60 students:
A : Very hardworking
B : Regular but not so hardworking
C : Careless and irregular 10 students are in category A, 30 in category B and the rest in category C.
It is found that the probability of students of category A, unable to get good marks in the final year examination is 0.002, of category B it is 0.02 and of category C, this probability is 0.20. A student selected at random was found to be one who could not get good marks in the examination. Find the probability that this student is category C.
→
Differentiate the following functions with respect to x:
$\text{x}^{\text{x}\cos\text{x}}+\frac{\text{x}^2+1}{\text{x}^2-1}$
→