Question

• Find $\frac{dy}{dx}$ if,

1. $y=\sqrt{tan\sqrt{x}}$
2. $y=tan^{-1}(\frac{cosx}{1+sinx})$
3. $y=log \sqrt{\frac{1+cos3x}{1-cos3x}}$
4. $y=cosec^{-1}(\frac{5}{3cosx+4sinx})$
5. $y=4^{log_2^{sinx}} + 9^{log_3^{cosx}}$
6. $y=sec^{-1}(\frac{1+\frac{e^{2\sqrt{x}}}{\sqrt{x}}}{2})$
7. $y=log(x+\sqrt{x^2+a^2})$.
8. $y=tan^{-1}(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}})$
9. $y=tan^{-1}(\frac{cos5x+sin5x}{cos5x-sin5x})$
10. $y=cot^{-1}(\frac{5-4tanx}{4+5tanx})$

Answer 1

See explanation for derivatives of each function.

Answer 2

Below are the derivatives for each given function:

1. $\frac{dy}{dx}=\frac{\sec^2\sqrt{x}}{4\sqrt{x}\,\sqrt{\tan\sqrt{x}}}$

2. $\frac{dy}{dx}=-\frac{3}{\sin3x}$

3. $\frac{dy}{dx}=0$

4. $\frac{dy}{dx}=\frac{1}{\sqrt{x^2+a^2}}$

5. $\frac{dy}{dx}=5$

6. $\frac{dy}{dx}=-\frac{1}{2}$

7. $\frac{dy}{dx}=1$

8. $\frac{dy}{dx}=\frac{\frac{e^{2\sqrt{x}}}{2x}-\frac{e^{2\sqrt{x}}}{4x^{3/2}}}{\left|\frac{1+\frac{e^{2\sqrt{x}}}{\sqrt{x}}}{2}\right|\sqrt{\left(\frac{1+\frac{e^{2\sqrt{x}}}{\sqrt{x}}}{2}\right)^2-1}}$

9. $\frac{dy}{dx}=\frac{1}{2\sqrt{1-x^2}}$

10. $\frac{dy}{dx}=1$

Detailed Solutions:

(1) Given:

$$y=\sqrt{\tan\sqrt{x}}=(\tan\sqrt{x})^{1/2}$$

Differentiate using the chain‐rule:

$$\frac{dy}{dx}=\frac{1}{2}(\tan\sqrt{x})^{-1/2}\cdot\sec^2\sqrt{x}\cdot\frac{d}{dx}(\sqrt{x}) =\frac{1}{2}(\tan\sqrt{x})^{-1/2}\cdot\sec^2\sqrt{x}\cdot\frac{1}{2\sqrt{x}}$$

Thus,

$$\frac{dy}{dx}=\frac{\sec^2\sqrt{x}}{4\sqrt{x}\,\sqrt{\tan\sqrt{x}}}$$

(2) Given:

$$y=\log\sqrt{\frac{1+\cos3x}{1-\cos3x}}=\frac{1}{2}\ln\left(\frac{1+\cos3x}{1-\cos3x}\right)$$

Differentiate:

$$\frac{dy}{dx}=\frac{1}{2}\left[\frac{d}{dx}\ln(1+\cos3x)-\frac{d}{dx}\ln(1-\cos3x)\right]$$

Since $\frac{d}{dx}\cos3x=-3\sin3x$, we have

$$\frac{dy}{dx}=\frac{1}{2}\left[\frac{-3\sin3x}{1+\cos3x}-\frac{3\sin3x}{1-\cos3x}\right] =-\frac{3\sin3x}{2}\left[\frac{1}{1+\cos3x}+\frac{1}{1-\cos3x}\right]$$

Combine the two fractions:

$$\frac{1}{1+\cos3x}+\frac{1}{1-\cos3x}=\frac{(1-\cos3x)+(1+\cos3x)}{1-\cos^2 3x} =\frac{2}{\sin^2 3x}\quad (\text{since }1-\cos^2 3x=\sin^2 3x)$$

Therefore,

$$\frac{dy}{dx}=-\frac{3\sin3x}{2}\cdot\frac{2}{\sin^2 3x}=-\frac{3}{\sin3x}$$

Thus,

$$\frac{dy}{dx}=-\frac{3}{\sin3x}$$

(3) Given:

$$y=4^{\log_{2}(\sin x)}+9^{\log_{3}(\cos x)}$$

Use the identity $a^{\log_b c}=c^{\log_b a}$. Note that

$$4^{\log_2(\sin x)}=(\sin x)^{\log_2 4}=(\sin x)^2,\quad \text{since } \log_2 4=2,$$

and

$$9^{\log_{3}(\cos x)}=(\cos x)^{\log_3 9}=(\cos x)^2,\quad \text{since } \log_3 9=2.$$

Thus,

$$y=\sin^2 x+\cos^2 x=1 \quad \Rightarrow \quad \frac{dy}{dx}=0.$$

So,

$$\frac{dy}{dx}=0$$

(4) Given:

$$y=\log\Bigl(x+\sqrt{x^2+a^2}\Bigr)$$

Differentiate:

$$\frac{dy}{dx}=\frac{1}{x+\sqrt{x^2+a^2}} \Biggl(1+\frac{x}{\sqrt{x^2+a^2}}\Biggr) =\frac{1}{\sqrt{x^2+a^2}}.$$

Thus,

$$\frac{dy}{dx}=\frac{1}{\sqrt{x^2+a^2}}$$

(5) Given:

$$y=\tan^{-1}\left(\frac{\cos5x+\sin5x}{\cos5x-\sin5x}\right)$$

Notice that writing in terms of tangent:

$$\frac{\cos5x+\sin5x}{\cos5x-\sin5x}=\frac{1+\tan5x}{1-\tan5x}=\tan\Bigl(5x+\frac{\pi}{4}\Bigr)$$

Thus,

$$y=\tan^{-1}\left(\tan\Bigl(5x+\frac{\pi}{4}\Bigr)\right)=5x+\frac{\pi}{4}\quad (\text{within the principal value})$$

Differentiate:

$$\frac{dy}{dx}=5$$

(6) Given:

$$y=\tan^{-1}\left(\frac{\cos x}{1+\sin x}\right)$$

Express in half-angle form. Write $t=\tan\frac{x}{2}$. Then using the half-angle formulas:

$$\sin x=\frac{2t}{1+t^2},\quad \cos x=\frac{1-t^2}{1+t^2}$$

it can be shown that

$$\frac{\cos x}{1+\sin x}=\frac{1-t}{1+t}=\tan\left(\frac{\pi}{4}-\frac{x}{2}\right)$$

Thus,

$$y=\tan^{-1}\left(\tan\left(\frac{\pi}{4}-\frac{x}{2}\right)\right) =\frac{\pi}{4}-\frac{x}{2}$$

Differentiate:

$$\frac{dy}{dx}=-\frac{1}{2}$$

(7) Given:

$$y=\csc^{-1}\Biggl(\frac{5}{3\cos x+4\sin x}\Biggr)$$

Recognize that $3\cos x+4\sin x$ can be written as $5\sin(x+\alpha)$ where

$$\sin\alpha=\frac{3}{5},\quad \cos\alpha=\frac{4}{5} \quad (\alpha=\sin^{-1}(3/5)).$$

Thus,

$$\frac{5}{3\cos x+4\sin x}=\frac{5}{5\sin(x+\alpha)}=\csc(x+\alpha)$$

and so

$$y=\csc^{-1}(\csc(x+\alpha))=x+\alpha,$$

up to the principal branch. Therefore,

$$\frac{dy}{dx}=1$$

(8) Given:

$$y=\sec^{-1}\Biggl(\frac{1+\frac{e^{2\sqrt{x}}}{\sqrt{x}}}{2}\Biggr)$$

Let

$$u(x)=\frac{1+\frac{e^{2\sqrt{x}}}{\sqrt{x}}}{2}.$$

The derivative formula for $y=\sec^{-1}(u)$ is

$$\frac{dy}{dx}=\frac{u'(x)}{\left|u(x)\right|\sqrt{u(x)^2-1}}.$$

To find $u'(x)$, first write:

$$u(x)=\frac{1}{2}+\frac{1}{2}\,\frac{e^{2\sqrt{x}}}{\sqrt{x}}.$$

Differentiate the second term using the product (or quotient) rule. Writing it as

$$v(x)=e^{2\sqrt{x}}\,x^{-1/2},$$

its derivative is:

$$v'(x)=x^{-1/2}\cdot\frac{d}{dx}\Bigl(e^{2\sqrt{x}}\Bigr)+e^{2\sqrt{x}}\cdot\frac{d}{dx}\Bigl(x^{-1/2}\Bigr).$$

Since

$$\frac{d}{dx}\Bigl(e^{2\sqrt{x}}\Bigr)=e^{2\sqrt{x}}\cdot\frac{d}{dx}(2\sqrt{x}) =e^{2\sqrt{x}}\cdot\frac{1}{\sqrt{x}},$$

and

$$\frac{d}{dx}(x^{-1/2})=-\frac{1}{2}x^{-3/2},$$

we get

$$v'(x)=\frac{e^{2\sqrt{x}}}{x}+ \left(-\frac{e^{2\sqrt{x}}}{2x^{3/2}}\right) =\frac{e^{2\sqrt{x}}}{x}-\frac{e^{2\sqrt{x}}}{2x^{3/2}}.$$

Thus,

$$u'(x)=\frac{1}{2}\,v'(x) =\frac{e^{2\sqrt{x}}}{2x}-\frac{e^{2\sqrt{x}}}{4x^{3/2}}.$$

Finally, the derivative is:

$$\frac{dy}{dx}=\frac{\frac{e^{2\sqrt{x}}}{2x}-\frac{e^{2\sqrt{x}}}{4x^{3/2}}}{\left|\frac{1+\frac{e^{2\sqrt{x}}}{\sqrt{x}}}{2}\right|\sqrt{\left(\frac{1+\frac{e^{2\sqrt{x}}}{\sqrt{x}}}{2}\right)^2-1}}.$$

(One may leave the answer in this implicit form.)

(9) Given:

$$y=\tan^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)$$

A known identity shows that

$$\tan^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)=\frac{1}{2}\sin^{-1}x.$$

Differentiate:

$$\frac{dy}{dx}=\frac{1}{2}\cdot\frac{1}{\sqrt{1-x^2}} =\frac{1}{2\sqrt{1-x^2}}.$$

(10) Given:

$$y=\cot^{-1}\left(\frac{5-4\tan x}{4+5\tan x}\right)$$

Express in terms of $\tan^{-1}$ by using the identity $\cot^{-1}(z)=\tan^{-1}(1/z)$:

$$y=\tan^{-1}\left(\frac{4+5\tan x}{5-4\tan x}\right).$$

Notice that

$$\tan(x+\theta)=\frac{\tan x+\tan\theta}{1-\tan x\tan\theta}.$$

Choosing $\tan\theta=\frac{4}{5}$ we have:

$$\tan\left(x+\tan^{-1}\frac{4}{5}\right) =\frac{\tan x+\frac{4}{5}}{1-\tan x\frac{4}{5}} =\frac{5\tan x+4}{5-4\tan x},$$

which matches the expression. Therefore,

$$y=x+\tan^{-1}\frac{4}{5}\quad \Rightarrow \quad \frac{dy}{dx}=1.$$

Thus,

$$\frac{dy}{dx}=1.$$