Question

Let $t=\frac{2\sqrt{2}-(1+\sqrt{3})}{\sqrt{3}-1}$ and $f\left(x\right)=\frac{2x}{1-x^2},g\left(x\right)=\frac{3x-x^3}{1-3x^2}$, then $\frac{d}{dt}\left\{f\right(g\left(t\right)\left)\right\}=$

• A. $1$
• B. $2$
• C. $\left(\frac{12+3\sqrt{6}+3\sqrt{2}}{2}\right)$
• D. $-1$

Answer 1

The correct option is C $\left(\frac{12+3\sqrt{6}+3\sqrt{2}}{2}\right)$

We know that,

$\tan 2x=\frac{2\tan x}{1-{\tan }^{2}x}$ & $\tan 3x=\frac{3\tan x-{\tan }^{3}x}{1-3{\tan }^{2}x}$

So, we clearly have

$f\left(\tan x\right)=\tan 2x\&g\left(\tan x\right)=\tan 3x$

let us simplify the value of $t$

$t=\frac{2\sqrt{2}-(1+\sqrt{3})}{\sqrt{3}-1}$

rationalizing the denominator,

$t=\frac{(2\sqrt{2}-(1+\sqrt{3}))(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}$

after simplification,

$t=\sqrt{2}-\sqrt{3}-\sqrt{4}+\sqrt{6}$

$⟹t=\tan 7.5$ (here $7.5$ is in degrees)

Now, to find $\frac{d}{dt}\left\{f\left(g\left(t\right)\right)\right\}$ let us assume

$t=\tan x$ (where $x=7.5$)

differentiating on both sides, we have

$⟹dt={\sec }^{2}xdx$

$⟹\frac{dx}{dt}={\cos }^{2}x$

So,

$⟹\frac{d}{dt}\left\{f\left(g\left(t\right)\right)\right\}=\frac{d}{dt}\left\{f\left(g\left(\tan x\right)\right)\right\}$$=\frac{d}{dt}\left\{f\left(\tan 3x\right)\right\}$$=\frac{d}{dt}\left\{\tan 6x\right\}$

$=\left({\sec }^{2}6x\right)\left(6\right)\left(\frac{dx}{dt}\right)$$=\left({\sec }^{2}6x\right)\left(6\right)\left({\cos }^{2}x\right)$

put $x=7.5$,

we get,

$=\left(12\right){\cos }^{2}7.5$

The value becomes

$\left(\frac{12+3\sqrt{6}+3\sqrt{2}}{2}\right)$.