Question

If $x^{3x}-2x^x\cot y-1=0,$ then the value of $y\text{'}\left(1\right)$ is

• A. $-\ln 2$
• B. $-\frac{3}{2}$
• C. $1$
• D. $\ln 2$

Answer 1

The correct option is B $-\frac{3}{2}$

Given : $x^{3x}-2x^x\cot y-1=0\cdots \left(1\right)$
Put $x=1$ in the equation $\left(1\right)$
$1-2\cot y-1=0$
$\Rightarrow \cot y=0\Rightarrow y=\frac{\pi }{2}$
We know that
$\frac{d}{dx}\left(x^x\right)=x^x(1+\ln x)\cdots \left(2\right)$
Similarly,
$\frac{d}{dx}\left(x^{3x}\right)=\frac{d}{dx}(x^x{)}^3$
$=3\left(x^{2x}\right)\frac{d}{dx}\left(x^x\right)$
$=3\left(x^{2x}\right)x^x(1+\ln x)$ [From $\left(2\right)$]
$\therefore \frac{d}{dx}\left(x^{3x}\right)=3x^{3x}(1+\ln x)\cdots \left(3\right)$

Now, differentiating the equation $\left(1\right)$ w.r.t. $x$
$3x^{3x}(1+\ln x)-2\left[x^x\right(1+\ln x)\cdot \cot y-x^x\cdot {\text{cosec}}^{2}y\cdot y^{\prime }]=0$
Put $x=1$ and $y=\frac{\pi }{2}$ in the above equation
$3\cdot 1\cdot (1+\ln 1)-2[1\cdot (1+\ln 1)\cdot \cot \frac{\pi }{2}-1\cdot {\text{cosec}}^2\frac{\pi }{2}\cdot y^{\prime }]=0$
$\Rightarrow 3-2[0-y^{\prime }]=0$
$\Rightarrow y^{\prime }\left(1\right)=-\frac{3}{2}$