Download App

Differentiation — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDifferentiation4 Marks
Question
Find $\frac{\text{dy}}{\text{dx}},$ when
$\text{x}=\frac{2\text{t}}{1+\text{t}^2}\text{ and y}=\frac{1-\text{t}^2}{1+\text{t}^2}$
✓

Answer

Here, $\text{x}=\frac{2\text{t}}{1+\text{t}^{2}}$
Differentiating it with respect to t using quotient rule,
$\frac{\text{dy}}{\text{dx}}=\bigg[\frac{(1+\text{t}^{2})\frac{\text{d}}{\text{dt}}(2\text{t})-2\text{t}\frac{\text{d}}{\text{dt}}(1+\text{t}^{2})}{(1+\text{t}^{2})^{2}}\bigg]$
$=\Big[\frac{(1+\text{t}^{2})(2)-2\text{t}(2\text{t})}{(1+\text{t}^{2})^{2}}\Big]$
$=\Big[\frac{2+2\text{t}^{2}-4\text{t}^{2}}{(1+\text{t}^{2})^{2}}\Big]$
$=\Big[\frac{2-2\text{t}^2}{(1+\text{t}^2)}\Big]$
$\frac{\text{dx}}{\text{dt}}=\frac{2(1-\text{t}^{2})}{(1+\text{t}^{2})^{2}}...(\text{i})$
And, $\text{y}=\frac{2(1-\text{t}^{2})}{(1+\text{t}^{2})^{2}}$
Differentiating it with respect to t using quotient rule,
$\frac{\text{dy}}{\text{dt}}=\bigg[\frac{(1+\text{t}^{2})\frac{\text{d}}{\text{dt}}(1-\text{t}^{2})-(1-\text{t}^{2})\frac{\text{d}}{\text{dt}}(1+\text{t}^{2})}{(1+\text{t}^{2})^{2}}\bigg]$
$=\Big[\frac{(1+\text{t}^{2})(-2\text{t})-(1-\text{t})^{2}(2\text{t})}{(1+\text{t}^{2})^{2}}\Big]$
$=\Big[\frac{-2\text{t}-2\text{t}^{3}-2\text{t}+2\text{t}^{3}}{(1+\text{t}^{2})^{2}}\Big]$
$\frac{\text{dy}}{\text{dx}}=\Big[\frac{-4\text{t}}{(1+\text{t}^{2})^{2}}\Big]...(\text{ii})$
Dividing equation (ii) by (i), 
 $\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{-4\text{t}}{(1-\text{t}^{2})^{2}}\times\frac{(1+\text{t}^{2})^{2}}{2(1+\text{t}^{2})}$
$=\frac{-2\text{t}}{(1-\text{t}^{2})}$
$\frac{\text{dy}}{\text{dx}}=-\frac{\text{x}}{\text{y}} $
$\Big[\text{Since}\frac{\text{x}}{\text{y}}=\frac{2\text{t}}{1+\text{t}^{2}}\times\frac{1+\text{t}^{2}}{1-\text{t}^{2}}=\frac{2\text{t}}{1-\text{t}^{2}}\Big]$

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 "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

Solve the system of linear equation, using matrix method 2x + 3y + 3z = 5; x - 2y + z = - 4; 3x - y - 2z = 3
→
Evaluate the following definite integrals:
$\int_{0}^\limits{\pi}\text{e}^{2\text{x}}\sin\Big(\frac{\pi}{4}+{\text{x}}\Big)\text{dx}$
→
Verify Lagrange's mean value theorem for the following function on the indicated intervals. find a point 'c' in the indicated interval as stated by the Lagrange's mean value theorem.
f(x) = x(x + 4)2 on [0,4]
→
Evaluvate the following intregals:
$\int\frac{1}{1-\cot\text{x}}\text{ dx}$
→
Evaluate the following integrals:

$\int\frac{\text{x}^2+\text{x}-1}{\text{x}^2+\text{x}-6}\text{ dx}$

→
$\text{Evaluate :} \int\limits_0^{\pi} \frac{x \sin x}{1 + \cos^{2} x} dx$
→
If $\begin{vmatrix}\text{p}&\text{q}&\text{c}\\\text{a}&\text{q}&\text{c}\\\text{a}&\text{b}&\text{r} \end{vmatrix}=0,$ find the value of $\frac{\text{p}}{\text{p}-\text{a}}+\frac{\text{q}}{\text{q}-\text{b}}+\frac{\text{r}}{\text{r}-\text{c}},$ $\text{p}\neq\text{a},\text{q}\neq\text{b},\text{r}\neq\text{c}.$
→
Find the equatoion of the passing through the points (1, -1, 2) and (2, -2, 2) and which is perpendicular to the plane 6x - 2y + 2z = 9.
→
Evaluate:

$\int\limits_0^{\pi/2}\frac{\text{x + sin x}}{\text{1 + cosx}}\text{dx}$.

→
Two groups are competing for the positions of the Board of Directors of a Corporation. The probabilities that the first and the second groups will win are 0.6 and 0.4 respectively. Further, if the first group wins, the probability of introducing a new product is 0.7 and the corresponding probability is 0.3 if the second group wins. Find the probability that the new product introduced was by the second group.
→