Question
If $\text{y}=\log\sqrt{\frac{1+\tan\text{x}}{1-\tan\text{x}}},$ prove that $\frac{\text{dy}}{\text{dx}}=\sec2\text{x}$
✓
Answer
We have, $\text{y}=\sqrt{\frac{1+\text{e}^\text{x}}{1-\text{e}^\text{x}}}$
Differentiating with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big(\sqrt{\frac{1+\text{e}^\text{x}}{1-\text{e}^\text{x}}}\Big)$
$=\frac{1}{\sqrt[2]{\Big(\frac{1+\text{e}^\text{x}}{1-\text{e}^\text{x}}\Big)}}\times\frac{\text{d}}{\text{dx}}\Big(\frac{1+\text{e}^\text{x}}{1-\text{e}^\text{x}}\Big)$
[Using chain rule]
$=\frac{1}{2}\times\sqrt{\frac{1-\text{e}^\text{x}}{1+\text{e}^\text{x}}}\Bigg[\frac{\big(1-\text{e}^\text{x}\frac{\text{d}}{\text{dx}}(1+\text{e}^\text{x})-(1+\text{e}^\text{x})\frac{\text{d}}{\text{dx}}(1-\text{e}^\text{x})\big)}{(1-\text{e}^\text{x})^2}\Bigg]$
$=\frac{1}{2}\sqrt{\frac{1-\text{e}^\text{x}}{1+\text{e}^\text{x}}}\bigg[\frac{(1-\text{e}^\text{x})\text{e}^\text{x}+(1+\text{e}^\text{x})\text{e}^\text{x}}{(1-\text{e}^\text{x})^2}\bigg]$
$=\frac{1}{2}\sqrt{\frac{1-\text{e}^\text{x}}{1+\text{e}^\text{x}}}\Big[\frac{2\text{e}^\text{x}}{(1-\text{e}^\text{x})^2}\Big]$
$=\frac{\text{e}^\text{x}}{\sqrt{(1+\text{e}^\text{x})\sqrt{(1+\text{e}^\text{x})}}}\frac{1}{(1-\text{e}^\text{x})}$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{e}^\text{x}}{(1-\text{e}^\text{x})\sqrt{1-\text{e}^{2\text{x}}}}$
Differentiating with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big(\sqrt{\frac{1+\text{e}^\text{x}}{1-\text{e}^\text{x}}}\Big)$
$=\frac{1}{\sqrt[2]{\Big(\frac{1+\text{e}^\text{x}}{1-\text{e}^\text{x}}\Big)}}\times\frac{\text{d}}{\text{dx}}\Big(\frac{1+\text{e}^\text{x}}{1-\text{e}^\text{x}}\Big)$
[Using chain rule]
$=\frac{1}{2}\times\sqrt{\frac{1-\text{e}^\text{x}}{1+\text{e}^\text{x}}}\Bigg[\frac{\big(1-\text{e}^\text{x}\frac{\text{d}}{\text{dx}}(1+\text{e}^\text{x})-(1+\text{e}^\text{x})\frac{\text{d}}{\text{dx}}(1-\text{e}^\text{x})\big)}{(1-\text{e}^\text{x})^2}\Bigg]$
$=\frac{1}{2}\sqrt{\frac{1-\text{e}^\text{x}}{1+\text{e}^\text{x}}}\bigg[\frac{(1-\text{e}^\text{x})\text{e}^\text{x}+(1+\text{e}^\text{x})\text{e}^\text{x}}{(1-\text{e}^\text{x})^2}\bigg]$
$=\frac{1}{2}\sqrt{\frac{1-\text{e}^\text{x}}{1+\text{e}^\text{x}}}\Big[\frac{2\text{e}^\text{x}}{(1-\text{e}^\text{x})^2}\Big]$
$=\frac{\text{e}^\text{x}}{\sqrt{(1+\text{e}^\text{x})\sqrt{(1+\text{e}^\text{x})}}}\frac{1}{(1-\text{e}^\text{x})}$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{e}^\text{x}}{(1-\text{e}^\text{x})\sqrt{1-\text{e}^{2\text{x}}}}$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Show that the following system of linear equations is consistent and also find solution:
x + y + z = 6
x + 2y + 3z = 14
x + 4y + 7z = 30
x + y + z = 6
x + 2y + 3z = 14
x + 4y + 7z = 30
Differentiate the following functions with respect to x:
$\log\sqrt{\frac{1-\cos\text{x}}{1+\cos\text{x}}}$
→$\log\sqrt{\frac{1-\cos\text{x}}{1+\cos\text{x}}}$
Solve the following initial value problems:
$\frac{\text{dy}}{\text{dx}}+\text{y}\cot\text{x}=4\text{x }\text{cosec x},\text{ y}\Big(\frac{\pi}{2}\Big)=0$
→$\frac{\text{dy}}{\text{dx}}+\text{y}\cot\text{x}=4\text{x }\text{cosec x},\text{ y}\Big(\frac{\pi}{2}\Big)=0$
Suppose we have four boxes A, B, C, D containing coloured marbles as given below:
- Box A?
- Box B?
- Box C?
Explain why the experiment of tossing a coin three times is said to have binomial distribution.
Let $\vec{\text{a}}=\hat{\text{i}}+4\hat{\text{j}}+2\hat{\text{k}},\vec{\text{b}}=3\hat{\text{i}}-2\hat{\text{j}}+7\hat{\text{k}}$ and $\vec{\text{c}}=2\hat{\text{i}}-\hat{\text{j}}+4\hat{\text{k}}.$Find a vector $\vec{\text{d}}$ which is perpendicular to both $\vec{\text{a}}$ and $\vec{\text{d}}$ and $\vec{\text{c}}.\vec{\text{d}}=15.$
→A large window has the shape of a rectangle surmounted by an equilateral triangle. If the perimeter of the window is 12 metres find the dimensions of the rectangle will produce the largest area of the window.
A chemical company produces two compounds, A and B. The following table gives the units of ingredients, C and D per kg of compounds A and B as well as minimum requirements of C and D and costs per kg of A and B. Find the quantities of A and B which would give a supply of C and D at a minimum cost.
| | Compound | Minimum requirement | |
| A | B | | |
| Ingredient C | 1 | 2 | 80 |
| Ingredient D | 3 | 1 | 75 |
| Coist (in Rs.) per Kg | 4 | 6 | |
The rate of increase of bacteria in a culture is proportional to the number of bacteria present and it is found that the number doubles in 6 hours. Prove that the bacteria becomes 8 times at the end of 18 hours.
Solve the following differential equation:
$\frac{\text{dy}}{\text{dx}}+\text{y}\cos\text{x}=\sin\text{x}\cos\text{x}$
→$\frac{\text{dy}}{\text{dx}}+\text{y}\cos\text{x}=\sin\text{x}\cos\text{x}$