Question

If $y$ be an implicit function of $x$ defined by $x^{2x}-2x^x\cot \left(y\right)-1=0$, then $y\text{'}\left(1\right)$ is equal to

• A. $-1$
• B. $1$
• C. $\log \left(2\right)$
• D. $-\log \left(2\right)$

Answer 1

The correct option is A

$-1$

Explanation for the correct option.

Step 1. Find the value of $y$ at $x=1$.

In the given equation $x^{2x}-2x^x\cot \left(y\right)-1=0$, substitute $1$ for $x$ and solve for $y$.

$$\begin{gathered} 1^{2·1}-2·1^1\cot \left(y\right)-1=0 \\ \Rightarrow 1-2\cot \left(y\right)-1=0 \\ \Rightarrow -2\cot \left(y\right)=0 \\ \Rightarrow \cot \left(y\right)=0 \\ \Rightarrow y=\frac{\mathrm{\pi }}{2}\left[\cot \left(\frac{\mathrm{\pi }}{2}\right)=0\right] \end{gathered}$$

Step 2. Divide the given implicit function into three function.

The given implicit function $x^{2x}-2x^x\cot \left(y\right)-1=0$ can be divided into three parts as $u=x^{2x}$, $v=-2x^x\cot \left(y\right)$ and $w=-1$ such that $u+v+w=0$ and so $\frac{du}{dx}+\frac{dv}{dx}+\frac{dw}{dx}=0$.

Step 3. Find the value of $\frac{du}{dx}$.

For the function $u=x^{2x}$, take logarithm both sides.

$$\begin{gathered} \log \left(u\right)=\log \left(x^{2x}\right) \\ \Rightarrow \log \left(u\right)=2x\log \left(x\right) \end{gathered}$$

Now, differentiate it with respect to $x$.

$$\begin{gathered} \frac{d}{dx}\left(\log \left(u\right)\right)=\frac{d}{dx}\left(2x\log \left(x\right)\right) \\ \Rightarrow \frac{1}{u}\frac{du}{dx}=2\log x+2x\times \frac{1}{x} \\ \Rightarrow \frac{du}{dx}=u\times 2\left[\log x+1\right] \\ \Rightarrow \frac{du}{dx}=2x^{2x}\left[\log x+1\right]\left[u=x^{2x}\right] \end{gathered}$$

Step 4. Find the value of $\frac{dv}{dx}$.

For the function $v=-2x^x\cot \left(y\right)$, differentiate both sides with respect to $x$.

$\begin{array}{rcl}\frac{dv}{dx}& =& \frac{d}{dx}\left(-2x^x\cot \left(y\right)\right)\\ & =& -2\left[x^x\left(-cose{c}^{2}y\right)\frac{dy}{dx}+\cot \left(y\right)\frac{d{x}^x}{dx}\right]\\ & =& -2\left[x^x\left(-cose{c}^{2}y\right)\frac{dy}{dx}+\cot \left(y\right)x^x\left(1+\log x\right)\right]\\ & =& 2x^{x}cose{c}^{2}y\frac{dy}{dx}-2x^x\left(1+\log x\right)\cot \left(y\right)\end{array}$

Step 5. Find the value of $\frac{dw}{dx}$.

For the function $w=-1$, differentiate both sides with respect to $x$.

$\begin{array}{rcl}\frac{dw}{dx}& =& \frac{d}{dx}(-1)\\ & =& 0\end{array}$

Step 6. Find the value of $\frac{dy}{dx}$ at $x=1$.

In the equation $\frac{du}{dx}+\frac{dv}{dx}+\frac{dw}{dx}=0$, substitute the found values of differentiation.

$$\begin{gathered} \frac{du}{dx}+\frac{dv}{dx}+\frac{dw}{dx}=0 \\ \Rightarrow 2x^{2x}\left[\log x+1\right]+\begin{array}{l}2\end{array}{x}^{x}cose{c}^{2}y\begin{array}{l}\frac{dy}{dx}\end{array}\begin{array}{l}-\end{array}2x^x\begin{array}{l}\left(1+\log x\right)\end{array}\begin{array}{l}\cot \end{array}\begin{array}{l}\left(y\right)\end{array}\begin{array}{l}+\end{array}\begin{array}{l}0\end{array}\begin{array}{rcl}& =& \end{array}\begin{array}{l}0\end{array} \end{gathered}$$

Now at $x=1$ the value of $y$ is $\frac{\mathrm{\pi }}{2}$. So substitute these values and find the value of $\frac{dy}{dx}$ at $x=1$.

$$\begin{gathered} 2·1^{2·1}\left[\log 1+1\right]+\begin{array}{rcl}& & 2\end{array}·1^1{\csc }^2\frac{\mathrm{\pi }}{2}\begin{array}{rcl}& & \frac{dy}{dx}\end{array}\begin{array}{rcl}& & -\end{array}\begin{array}{rcl}& & 2\end{array}·1^1\begin{array}{rcl}& & \left(1+\log 1\right)\end{array}\begin{array}{rcl}& & \cot \end{array}\begin{array}{rcl}& & \left(\frac{\mathrm{\pi }}{2}\right)\end{array}\begin{array}{rcl}& & +\end{array}\begin{array}{rcl}& & 0\end{array}\begin{array}{rcl}& =& \end{array}\begin{array}{rcl}& & 0\end{array} \\ \Rightarrow 2\left[0+1\right]+2\times 1^2\begin{array}{rcl}& & \frac{dy}{dx}\end{array}-2\left(1+0\right)\times 0=0 \\ \Rightarrow 2+2\begin{array}{rcl}& & \frac{dy}{dx}\end{array}=0 \\ \Rightarrow 2\begin{array}{rcl}& & \frac{dy}{dx}\end{array}=-2 \\ \Rightarrow \begin{array}{rcl}& & \frac{dy}{dx}\end{array}=-1 \end{gathered}$$

So the value of $y\text{'}\left(1\right)$ is $-1$.

Hence, the correct option is A.