Download App

Differentiation — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDifferentiation5 Marks
Question
If $\text{y}=\frac{\text{e}^\text{x}-\text{e}^{-\text{x}}}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}},$ prove that $\frac{\text{dy}}{\text{dx}}=1-\text{y}^2$
✓

Answer

Givne, $\text{y}=\frac{\text{e}^{\text{x}}-\text{e}^{-\text{x}}}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}$
Differentiate with respect to x,
$\frac{\text{d}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big(\frac{\text{e}^{\text{x}}-\text{e}^{-\text{x}}}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}\Big)$
$=\Bigg[\frac{\big(\text{e}^{\text{x}}+\text{e}^{-\text{x}}\big)\frac{\text{d}}{\text{dx}}\big(\text{e}^{\text{x}}-\text{e}^{-\text{x}}\big)\frac{\text{d}}{\text{dx}}\big(\text{e}^{\text{x}}+\text{e}^{-\text{x}}\big)}{\big(\text{e}^{\text{x}}+\text{e}^{-\text{x}}\big)^2}\Bigg]$
[Using quotient rule and chain rule]
$=\begin{bmatrix} \frac{\big(\text{e}^{\text{x}}+\text{e}^{-\text{x}}\big)\Big[\text{e}^{\text{x}}-\text{e}^{-\text{x}}\frac{\text{d}}{\text{dx}}(-\text{x})-\big(\text{e}^{\text{x}}-\text{e}^{-\text{x}}\big)\Big(\text{e}^{\text{x}}+\text{e}^{-\text{x}}\frac{\text{d}}{\text{dx}}(-\text{x})\Big)\Big]}{\big(\text{e}^{\text{x}}+\text{e}^{-\text{x}}\big)^2} \end{bmatrix}$
$=\begin{bmatrix} \frac{\big(\text{e}^{\text{x}}+\text{e}^{-\text{x}}\big)\big(\text{e}^{\text{x}}+\text{e}^{-\text{x}}\big)-\big(\text{e}^{\text{x}}-\text{e}^{-\text{x}}\big)\big(\text{e}^{\text{x}}-\text{e}^{-\text{x}}\big)}{\big(\text{e}^{\text{x}}+\text{e}^{-\text{x}}\big)^2} \end{bmatrix}$
$=\bigg[\frac{\text{e}^{2\text{x}}+\text{e}^{-2\text{x}}+2\text{e}^{\text{x}}\times\text{e}^{-\text{x}}-\text{e}^{2\text{x}}-\text{e}^{-2\text{x}}+2\text{e}^{\text{x}}\text{e}^{-\text{x}}}{\big(\text{e}^{\text{x}}+\text{e}^{-\text{x}}\big)^2}\bigg]$
$\frac{\text{dy}}{\text{dx}}\bigg[\frac{4}{(\text{e}^{\text{x}}+\text{e}^{-\text{x}})^2}\bigg]\ .....(\text{i})$
Now,
$1-\text{y}^2=1-\Big(\frac{\text{e}^{\text{x}}-\text{e}^{-\text{x}}}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}\Big)^2$
$=1-\frac{(\text{e}^{\text{x}}-\text{e}^{-\text{x}})^2}{(\text{e}^{\text{x}}+\text{e}^{-\text{x}})^2}$
$=\frac{(\text{e}^{\text{x}}+\text{e}^{-\text{x}})^2-(\text{e}^{\text{x}}-\text{e}^{-\text{x}})^2}{(\text{e}^{\text{x}}+\text{e}^{-\text{x}})^2}$
$=\frac{4}{(\text{e}^{\text{x}}+\text{e}^{-\text{x}})^2}$

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

A manufacturer has employed 5 skilled men and 10 semi-skilled men and makes two models A and B of an article. The making of one item of model A requires 2 hours work by a skilled man and 2 hours work by a semi-skilled man. One item of model B requires 1 hour by a skilled man and 3 hours by a semi-skilled man. No man is expected to work more than 8 hours per day. The manufacturer’s profit on an item of model A is Rs. 15 and on an item of model B is Rs. 10. How many of items of each model should be made per day in order to maximize daily profit? Formulate the above LPP and solve it graphically and find the maximum profit.
→
Write the number of points where f(x) = |x| + |x − 1| is continuous but not differentiable.
→
For the matrix $A = \left[ {\begin{array}{*{20}{c}} 1&1&1 \\ 1&2&{ - 3} \\ 2&{ - 1}&3 \end{array}} \right]$, show that $A^3 - 6A^2 + 5A + 11I = 0.$ Hence find $A^{-1}.$
→
Sketch the region bounded by the curves $y = x^2 + 2, y = x, x = 0$ and $x = 1$. Also find the area of this region.
→
By using properties of determinants, show that:
$\begin{vmatrix}a^2+1&ab&ac\\ab&b^2+1&bc\\ca&cb&c^2+1\end{vmatrix}=1+a^2+b^2+c^2$
→
If $\text{A}=\begin{bmatrix}1&-1&0\\ 2&3&4\\ 0&1&2\end{bmatrix}\text{and }\text{B}=\begin{bmatrix}2&2&-4\\ -4&2&-4\\ 2&-1&5\end{bmatrix}$ are two square matrices, find AB and hence solve the system of linear equations:
x - y = 3, 2x + 3y + 4z = 17, y + 2z = 7
→
Given $\vec{\text{a}}=\frac{1}{7}\big(2\hat{\text{i}}+3\hat{\text{j}}+6\hat{\text{k}}\big),\vec{\text{b}}=\frac{1}{7}\big(3\hat{\text{i}}-6\hat{\text{j}}+2\hat{\text{k}}\big),$$\vec{\text{c}}=\frac{1}{7}\big(6\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}\big),\hat{\text{i}},\hat{\text{j}},\hat{\text{k}}$
being a right handed orthogonal system of unit vector in spece, show that $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ is also another system.
→
The rate of growth of a population is proportional to the number present. If the population of a city doubled in the past 25 year, and the present population is 100000, when will the city have a population of 500000?
→
If $\text{x}=\text{a}\sec\theta,\text{y}=b\tan\theta$ prove that $\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\frac{\text{b}^4}{\text{a}^2\text{y}^3}$
→
Evaluate the following integrals:
$\int\limits^{\frac{\pi}{4}}_{-\frac{\pi}{4}}\frac{\cos^{2}\text{x}}{1+\text{e}^{\text{x}}}\text{ dx}$
→