Question

If y is an implicit function of x defined by $x^{2x}-{2x}^{x}coty-1=0.$ Then, $y^{\prime }\left(1\right)$ is equal to

• A. $-1$
• B. $1$
• C. $\log 2$
• D. $-\log 2$

Answer 1

The correct option is A $-1$

$x^{2x}-{2x}^{x}coty-1=0.$ ...............[1]

At $x=1$; we have

$1-\cot y-1=0$

$⟹y=\frac{\pi }{2}$

Differentiating w.r.t $x$, we get:

$2x^{2x}(1+\ln x)-2\left[x^x\right(-cose{c}^{2}y\frac{dy}{dx}+\cot y{x}^x(1+\ln x)\left)\right]=0$

At $P(1,\frac{\pi }{2})$, we have

$2(1+\ln 1)-2\left[1\right(-1)\frac{dy}{dx}{|}_P+0]=0$

$⟹2+2\frac{dy}{dx}{|}_P=0$

$⟹\frac{dy}{dx}{|}_P=-1$

Hence, $y^{\prime }\left(1\right)=-1$