⇒Derivative

Correct Option is (C)

$\begin{array}{rl}& \text{ Let }y=\mathrm{f}(\sec x)\text{ and }\mathrm{z}=\mathrm{g}(\tan x)\\ & \frac{\mathrm{d}y}{\mathrm{d}x}={\mathrm{f}}^{\prime }(\sec x)\cdot \sec x\tan x\\ & \frac{dz}{dx}=g^{\prime }(\tan x)\cdot {\sec }^{2}x\\ & \text{ Now, }\frac{\mathrm{d}y}{\mathrm{dz}}=\frac{{\mathrm{f}}^{\prime }(\sec x)\sec x\tan x}{{\mathrm{g}}^{\prime }(\tan x){\sec }^{2}x}\\ & \frac{dy}{dz}=\frac{f^{\prime }(\sec x)\tan x}{g^{\prime }(\tan x)\cdot \sec x}\\ & {\frac{\mathrm{d}y}{\mathrm{dz}}|}_{x=\frac{\pi }{4}}=\frac{{\mathrm{f}}^{\prime }(\sec \frac{\pi }{4})\tan \frac{\pi }{4}}{{\mathrm{g}}^{\prime }(\tan \frac{\pi }{4})\sec \frac{\pi }{4}}\\ & =\frac{f^{\prime }(\sqrt{2})\cdot (1)}{g^{\prime }(1)\cdot \sqrt{2}}\Rightarrow \frac{4\times 1}{2\sqrt{2}}\\ & =\sqrt{2}\end{array}$

Correct Option is (D)

Given the equation :

$y=\log \tan \left(\frac{x}{2}\right)+{\sin }^{-1}(\cos x)$

The goal is to find $\frac{dy}{dx}$.

Step 1 : Convert ${\sin }^{-1}(\cos x)$ using the complementary angle identity

From trigonometric identities, we know :

${\sin }^{-1}(\cos x)=\frac{\pi }{2}-x$

This is because $\sin (\frac{\pi }{2}-x)=\cos (x)$.

So the equation becomes :

$y=\log \tan \left(\frac{x}{2}\right)+\frac{\pi }{2}-x$

Step 2 : Differentiate term by term

For the term $\log \tan \left(\frac{x}{2}\right)$ :

Using the chain rule and properties of logarithmic differentiation :

$\frac{d}{dx}\log \tan \left(\frac{x}{2}\right)=\frac{1}{\tan \left(\frac{x}{2}\right)}\times {\sec }^2\left(\frac{x}{2}\right)\times \frac{1}{2}$

Where :

• $\frac{1}{\tan \left(\frac{x}{2}\right)}$ is the derivative of the logarithmic function.
• ${\sec }^2\left(\frac{x}{2}\right)$ is the derivative of $\tan \left(\frac{x}{2}\right)$.
• $\frac{1}{2}$ is the derivative of $\frac{x}{2}$.

For the term $\frac{\pi }{2}-x$ :

The derivative is simply -1, since the derivative of a constant is 0 and the derivative of x is 1.

Combining the two terms, we get :

$\frac{dy}{dx}=\frac{1}{\tan \left(\frac{x}{2}\right)}\times {\sec }^2\left(\frac{x}{2}\right)\times \frac{1}{2}-1$

Step 3 : Simplify

$\frac{1}{\tan \left(\frac{x}{2}\right)}$ can be written as $\frac{\cos \frac{x}{2}}{\sin \frac{x}{2}}$.

And ${\sec }^2\left(\frac{x}{2}\right)$ is $\frac{1}{{\cos }^2\frac{x}{2}}$.

Multiplying these terms :

$\frac{dy}{dx}=\frac{\cos \frac{x}{2}}{\sin \frac{x}{2}{\cos }^2\frac{x}{2}}\times \frac{1}{2}-1$

Step 4 : Use trigonometric identities to further simplify

Using the identity $\sin 2\theta =2\sin \theta \cos \theta$ :

$\sin x=2\sin \frac{x}{2}\cos \frac{x}{2}$

The expression can then be simplified as :

$\frac{dy}{dx}=\frac{1}{2\sin \frac{x}{2}\cos \frac{x}{2}}-1$

Substitute the value from the double angle identity :

$\frac{dy}{dx}=\frac{1}{\sin x}-1$

Which is :

$\frac{dy}{dx}=\mathrm{cosec}x-1$

Thus, the solution for $\frac{dy}{dx}$ is $\mathrm{cosec}x-1$.

Correct Option is (B)

$h(x)=\sqrt{4f(x)+3g(x)}$ ..... (1)

$\therefore h(1)=\sqrt{4(4)+3(3)}=5$ .... (2)

Squaring (1), we get

$[\mathrm{h}(\mathrm{x}){]}^2=4\mathrm{f}(\mathrm{x})+3\mathrm{g}(\mathrm{x})$

Differentiating w.r.t. $\mathrm{x}$, we get $2\mathrm{h}(\mathrm{x}){\mathrm{h}}^{\prime }(\mathrm{x})=4{\mathrm{f}}^{\prime }(\mathrm{x})+3{\mathrm{g}}^{\prime }(\mathrm{x})$

At $x=1$, we get

$2(5){\mathrm{h}}^{\prime }(1)=4(3)+3(4)=24$ ... [From (2) and data given]

$\therefore {\mathrm{h}}^{\prime }(1)=\frac{24}{10}=\frac{12}{5}$