Download App

Differentiation — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDifferentiation5 Marks
Question
If $\sec\Big(\frac{\text{x}+\text{y}}{\text{x}-\text{y}}\Big)=\text{a}$ prove that $\frac{\text{dx}}{\text{dx}}=\frac{\text{y}}{\text{x}}$
✓

Answer

We have, $\sec\Big(\frac{\text{x}+\text{y}}{\text{x}-\text{y}}\Big)=\text{a}$
$\Rightarrow\frac{\text{x}+\text{y}}{\text{x}-\text{y}}=\sec^{-1}({\text{a}})$
Differentiate with respect to x, we get,
$\Rightarrow\bigg[\frac{(\text{x}-\text{y})\frac{\text{d}}{\text{dx}}(\text{x}+\text{y})-(\text{x}+\text{y})\frac{\text{d}}{\text{dx}}(\text{x}-\text{y})}{(\text{x}-\text{y}}\bigg]=0$
$\Rightarrow(\text{x}-\text{y})\Big(1+\frac{\text{d}}{\text{dx}}\Big)-(\text{x}+\text{y})\Big(1-\frac{\text{d}}{\text{dx}}\Big)=0$
$\Rightarrow(\text{x}-\text{y})+(\text{x}-\text{y})\frac{\text{dy}}{\text{dx}}-(\text{x}+\text{y})+(\text{x}+\text{y})\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\frac{\text{dy}}{\text{dx}}[\text{x}-\text{y}+\text{x}+\text{y}]=\text{x}+\text{y}-\text{x}+\text{y}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}(2\text{x})=2\text{y}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}$

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→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

The odds against a certain event are 5 to 2 and the odds in favour of another event, independent to the former are 6 to 5. Find the probability that,
  1. At least one of the events will occur,
  2. None of the events will occur.
→
If $\text{A}=\begin{bmatrix}1&0&2\\0&2&1\\2&0&3\end{bmatrix},$ then show that A is a root of the polynomial $f(x) = x^3 - 6x^2 + 7x + 2$.
→
cards are drawn successively with replacement from a well shuffled pack of 52 cards. Find the mean and variance of red cards.
→
Solve the following determinant equations:
$\begin{vmatrix}3\text{x}-8&3&3\\3&3\text{x}-8&8\\3&3&3\text{x}-8\end{vmatrix}=0$
→
If $x = \tan \Bigg(\frac{1}{\text{a}}\log \text{y}\Bigg),$show that
$(1+\text{x}^{2})\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}+(\text{2x - a})\frac{\text{dy}}{\text{dx}}=0$.
→
Find the maximum and minimum values of the function $\text{f}(\text{x})=\frac{4}{\text{x}+2}+\text{x}$
→
Evaluate the following:
$\begin{vmatrix}\text{a}+\text{x}&\text{y}&\text{z}\\\text{x}&\text{a}+\text{y}&\text{z}\\\text{x}&\text{y}&\text{a}+\text{z}\end{vmatrix}$
→
Prove that the area the region bounded by $\text{y}=\sqrt{\text{x}}, $and x = 2y + 3 in the first and x-asis.
→
Find the angle between the following pairs of lines:$\frac{\text{x}-5}{1}=\frac{2\text{y}+6}{-2}=\frac{\text{z}-3}{1}$ and $\frac{\text{x}-2}{3}=\frac{\text{y}+1}{4}=\frac{\text{z}-6}{5}$
→
the cartesian equation of a line are $\frac{\text{x}-5}{3}=\frac{\text{y}+4}{7}=\frac{\text{z}-6}{2}.$ Find a vector equation for the line.
→