If y= e^x-e^ -x e^ x +e^ -x , prove that dy dx =1-y^2

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

Similar questions

Find non-zero values of x satisfying the matrix equation:

$\text{x}\begin{bmatrix}2\text{x}&2\\3&\text{x}\end{bmatrix}+2\begin{bmatrix}8&5\text{x}\\4&4\text{x}\end{bmatrix}=2\begin{bmatrix}(\text{x}^2+8)&24\$10)&6\text{x}\end{bmatrix}$

→

Prove that the function f defined by $\text{f(x)}=\begin{cases}\frac{\text{x}}{|\text{x}|+2\text{x}^2},&\text{if x}\neq0\\\text{k},&\text{ if x}=0\end{cases}$ remains discontinuous at x = 0, regardless the choice of k.

→

If $\text{A}=\begin{bmatrix}1&0&-2\\3&-1&0\\-2&1&1\end{bmatrix},\text{B}=\begin{bmatrix}0&5&-4\\-2&1&3\\-1&0&2\end{bmatrix}$ and $\text{C}=\begin{bmatrix}1&5&2\\-1&1&0\\0&-1&1\end{bmatrix},$ verify that A(B - C) = AB - AC.

→

Maximize Z = 50x + 30y
Subject to
$2\text{x}+\text{y}\leq18$
$3\text{x}+2\text{y}\leq34$
$\text{x},\text{y}\geq0$

→

A company sells two different products, A and B. The two products are produced in a common production process, which has a total capacity of 500 man-hours. It takes 5 hours to produce a unit of A and 3 hours to produce a unit of B. The market has been surveyed and company officials feel that the maximum number of unit of A that can be sold is 70 and that for B is 125. If the profit is Rs. 20 per unit for the product A and Rs. 15 per unit for the product B, how many units of each product should be sold to maximize profit?

→

For any a, b, x, y > 0, prove that:
$\frac{2}{3}\tan^{-1}\Big(\frac{3\text{a}\text{b}^2-\text{a}^3}{\text{b}^3-3\text{a}^2\text{b}}\Big)+\frac{2}{3}\tan^{-1}\Big(\frac{3\text{x}\text{y}^2-\text{x}^3}{\text{y}^3-3\text{x}^2\text{y}}\Big)=\tan^{-1}\frac{2\alpha\beta}{\alpha^2-\beta^2}$
where $\alpha=-\text{ax}+\text{by},\beta=\text{bx}+\text{ay}$

→

Evalute the following integrals:
$\int\frac{\sin2\text{x}}{\text{a}^2+\text{b}^2\sin^2\text{x}}\text{dx}$